JEE Atomic Structure Questions

To evaluate these statements based on Bohr's atomic model, we must first establish the fundamental proportionalities for an electron's velocity ($v$) and its orbital radius ($r$):

$$v \propto \frac{Z}{n}$$

$$r \propto \frac{n^2}{Z}$$

Every other physical quantity regarding the electron can be derived entirely by combining these two basic mathematical relationships.

Looking at the first statement regarding kinetic energy ($KE$), we know that $KE = \frac{1}{2}mv^2$. Since the mass ($m$) is constant, the kinetic energy is directly proportional to the square of the velocity. By substituting our established velocity proportionality, we get:

$$KE \propto v^2 \propto \left(\frac{Z}{n}\right)^2 \propto \frac{Z^2}{n^2}$$

This perfectly aligns with the given statement, making statement (A) correct.

The second statement asks us to evaluate the product of the electron's velocity ($v$) and its principal quantum number ($n$). Based on our foundational rule, we multiply the velocity expression by $n$:

$$v \cdot n \propto \left(\frac{Z}{n}\right) \cdot n \propto Z$$

The $n$ in the numerator completely cancels out the one in the denominator, leaving the product strictly proportional to $Z$. The statement suggests this product is proportional to $Z^2$, rendering statement (B) incorrect.

For the third statement involving the frequency of revolution ($f$), we calculate it by dividing the orbital velocity by the circumference of the orbit ($2\pi r$). Because circumference is directly tied to the radius, frequency is proportional to velocity divided by radius:

$$f \propto \frac{v}{r} \propto \frac{\frac{Z}{n}}{\frac{n^2}{Z}} \propto \frac{Z^2}{n^3}$$

The statement provided incorrectly lists the numerator as $Z^3$, making statement (C) a false claim.

Finally, the fourth statement examines the Coulombic force ($F$) of attraction between the positively charged nucleus and the negatively charged electron. According to Coulomb's law, this electrostatic force is directly proportional to the atomic number ($Z$) and inversely proportional to the square of the distance, or radius ($r^2$):

$$F \propto \frac{Z}{r^2} \propto \frac{Z}{\left(\frac{n^2}{Z}\right)^2} \propto \frac{Z}{\frac{n^4}{Z^2}} \propto \frac{Z^3}{n^4}$$

This confirms that the electrostatic force is indeed proportional to $Z^3$ divided by $n^4$, making statement (D) completely accurate. Therefore, only statements (A) and (D) represent correct applications of Bohr's theory.

First we recall the basic relations of Bohr’s model for a hydrogen-like (one-electron) species of nuclear charge $$+Ze$$.

Bohr postulated that in the stationary orbit of principal quantum number $$n$$, the necessary centripetal force for circular motion of the electron is provided by the Coulombic attraction between the nucleus and the electron. Hence we write

$$\frac{m\,v_n^{\,2}}{r_n}= \frac{1}{4\pi\varepsilon_0}\;\frac{Z\,e^{2}}{r_n^{\,2}},$$

where $$m$$ is the electron mass, $$v_n$$ its speed in the $$n^{\text{th}}$$ orbit and $$r_n$$ the radius of that orbit.

Rearranging the above expression, we get

$$m\,v_n^{\,2}= \frac{1}{4\pi\varepsilon_0}\;\frac{Z\,e^{2}}{r_n}.$$

Bohr also gave the radius of the $$n^{\text{th}}$$ orbit as

$$r_n = \frac{\varepsilon_0\,h^{2}}{\pi\,m\,e^{2}}\; \frac{n^{2}}{Z},$$

where $$h$$ is Planck’s constant.

Substituting this $$r_n$$ into the previous force balance, we obtain

$$m\,v_n^{\,2}= \frac{1}{4\pi\varepsilon_0}\; \frac{Z\,e^{2}}{ \dfrac{\varepsilon_0\,h^{2}}{\pi\,m\,e^{2}}\; \dfrac{n^{2}}{Z}}.$$ Cancelling and simplifying step by step, we have

$$m\,v_n^{\,2}= \frac{1}{4\pi\varepsilon_0}\; \frac{Z^{2}e^{4}\pi\,m}{\varepsilon_0\,h^{2}\,n^{2}}.$$

Dividing both sides by $$m$$,

$$v_n^{\,2}= \frac{1}{4\pi\varepsilon_0}\; \frac{Z^{2}e^{4}\pi}{\varepsilon_0\,h^{2}\,n^{2}}.$$

Taking square root,

$$v_n = \frac{Z\,e^{2}}{2\varepsilon_0\,h}\;\frac{1}{n}.$$

Thus we have derived the well-known proportionality

$$v_n \propto \frac{Z}{n}.$$ Now we analyse the two statements in the light of the above relation.

• For Statement I we examine the dependence on $$Z$$. The formula shows that when $$Z$$ decreases (we move to a nucleus with fewer positive charges), the numerator $$Z$$ becomes smaller, so $$v_n$$ becomes smaller. Therefore the magnitude of velocity actually decreases with a decrease in positive nuclear charge. Hence Statement I is false.

• For Statement II we examine the dependence on $$n$$. Because velocity is inversely proportional to $$n$$, a decrease in the principal quantum number (i.e., moving to an inner orbit) makes $$n$$ smaller and hence $$v_n$$ larger. Therefore the magnitude of velocity indeed increases with decrease in $$n$$. Hence Statement II is true.

We have found that Statement I is false while Statement II is true. Among the given choices, this corresponds to Option A.

Hence, the correct answer is Option A.

Statement I says that Bohr's theory accounts for the stability and line spectrum of $$\text{Li}^+$$. The ion $$\text{Li}^+$$ has atomic number 3 and charge +1, meaning it retains 2 electrons. Since Bohr's theory applies only to single-electron (hydrogen-like) species such as $$\text{H}$$, $$\text{He}^+$$, and $$\text{Li}^{2+}$$, it cannot explain the spectrum of the two-electron $$\text{Li}^+$$ ion. Statement I is false.

Statement II says that Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field, known as the Zeeman effect. This is correct — Bohr's model lacks the quantum numbers needed to account for this splitting. Statement II is true.

Since Statement I is false and Statement II is true, the correct answer is option (2).

We begin by recalling what Rutherford’s nuclear model proposed. In this model electrons were pictured as orbiting the tiny, massive nucleus just as planets orbit the Sun. There was no assumption of quantisation of either radius or energy. According to classical electrodynamics, a charged particle moving in a curved path must continuously radiate energy, so an electron in Rutherford’s orbit would lose energy steadily and spiral into the nucleus. The radiation produced during this continuous energy loss would itself be continuous in wavelength. Hydrogen, however, actually shows a set of sharply defined spectral lines, for example the Balmer series. Rutherford’s model therefore fails to account for the line spectrum.

So, for Statement I we observe that Rutherford’s gold-foil experiment - and the atomic model derived from it - indeed cannot explain the observed discrete spectral lines of hydrogen. Hence Statement I is true.

Next, let us recall Bohr’s postulates. Bohr assumed that the electron in a hydrogen atom moves in a circular orbit of fixed radius $$r_n$$ with a definite linear momentum $$p_n = m v_n$$ such that the angular momentum is quantised:

$$m v_n r_n = n \hbar, \qquad n = 1,2,3,\dots$$

Knowing $$r_n$$ and $$v_n$$ simultaneously means we can in principle know the electron’s position and momentum with arbitrary precision. But quantum mechanics sets a fundamental limit given by Heisenberg’s uncertainty principle:

$$\Delta x \,\Delta p \ge \frac{h}{4\pi}.$$

In Bohr’s picture, $$\Delta x \to 0$$ (because the orbit radius is fixed) and $$\Delta p \to 0$$ (because the speed is fixed), making the product $$\Delta x\,\Delta p \to 0$$, which violates the inequality. Therefore the idealised circular-orbit picture of Bohr is incompatible with the uncertainty principle discovered later. Hence Statement II is also true.

Since both statements are correct, the option that asserts the truth of both is selected.

Hence, the correct answer is Option D.

In the Thomson “plum-pudding’’ model the atom is imagined as a sphere of positive charge whose magnitude is spread out uniformly over the whole atomic volume, while the negatively charged electrons are embedded like small “plums’’ inside this continuous positive “pudding”.

Because the positive charge is not concentrated at a point but is smeared out uniformly, an incoming $$\alpha$$-particle (which itself carries a positive charge $$+2e$$) would experience two effects:

1. All along its path inside the atom it would feel the repulsive electrostatic force due to this diffuse positive charge. Since the force acts continuously, the $$\alpha$$-particle would lose a little kinetic energy throughout the traversal, therefore its speed would become slightly less on emerging from the foil.

2. The direction of the electrostatic force would keep changing gradually as the $$\alpha$$-particle moves, because at each instant the net electric field inside the uniform positive sphere points toward the centre (this follows from Gauss’s law for a uniformly charged sphere). Consequently, instead of one sharp “kick’’ the particle would suffer a series of tiny deflections that, when summed, amount to only a small overall angular deviation from its original straight-line path.

Mathematically, inside a uniformly charged sphere of radius $$R$$ and total positive charge $$+Ze$$, Gauss’s law gives the magnitude of the electric field at a distance $$r<R$$ from the centre as

$$E(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{Z e\, r}{R^3},$$

which is directly proportional to $$r$$. Hence the force on an $$\alpha$$-particle of charge $$+2e$$ is

$$F(r)=2e\,E(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{2 Z e^{2}\, r}{R^{3}}.$$

This force always points radially outward, so while the $$\alpha$$-particle traverses the atom it is slowed down (its kinetic energy decreases) and its trajectory bends slightly outward, but because the field grows only linearly with $$r$$ the maximum force it ever feels is modest. Therefore the cumulative deflection angle is small, and there is no possibility of a large-angle ( >$$90^\circ$$ ) scattering or a full $$180^\circ$$ rebound.

Putting the two conclusions together, the Thomson model predicts that:

• every $$\alpha$$-particle will pass through the gold foil;
• each particle will be deflected only by a small angle;
• each particle will emerge with a speed a little less than the incident speed because of the continual but weak electrostatic repulsion inside the diffuse positive charge.

This description is precisely what is stated in Option D: “$$\alpha$$-particles pass through the gold foil deflected by small angles and with reduced speed.” The other options conflict with one or more of the above inferences. In particular, Option A ignores the expected slight energy loss, Option B speaks of large-angle scattering, and Option C predicts total back-scattering—none of which align with the consequences of a diffuse positive charge distribution.

Hence, the correct answer is Option D.

A radial node is strictly defined as a point at a finite distance from the nucleus where the radial probability density drops to exactly zero. Mathematically, it is a place where the electron's actual wavefunction changes sign (crossing from positive to negative, or vice versa). There are $$n-l-1$$ radial nodes. The total nodes for $$3s=$$ $$3 - 0 -1=2$$. Therefore the right option is C.

We start by recalling that every isotope is written in the form $$\;^{A}_{Z}\text{X}\;$$ where $$A$$ is the mass number (total protons + neutrons) and $$Z$$ is the atomic number (only protons, which equals the number of electrons in a neutral atom).

The radioactive isotope of hydrogen is tritium, written as $$\;^{3}_{1}\text{H}\;$$ because hydrogen has atomic number $$Z=1$$ and tritium has mass number $$A=3$$.

We now apply the formula for neutrons:

Number of neutrons $$=A-Z$$

Substituting $$A=3$$ and $$Z=1$$, we get

$$\text{Neutrons}=3-1=2$$

Since the atom is neutral, the number of electrons equals the atomic number:

$$\text{Electrons}=Z=1$$

So, tritium contains $$2$$ neutrons and $$1$$ electron.

Among the given options, this corresponds to Option C.

Hence, the correct answer is Option C.

We need to identify which form of hydrogen emits low energy $$\beta^-$$ particles.

Among the isotopes of hydrogen: Protium ($$_1^1H$$) has 1 proton and 0 neutrons, Deuterium ($$_1^2H$$) has 1 proton and 1 neutron, and Tritium ($$_1^3H$$) has 1 proton and 2 neutrons.

Protium and Deuterium are stable isotopes and are not radioactive. The proton $$H^+$$ is simply an ion and is also not radioactive.

Tritium ($$_1^3H$$) is the only radioactive isotope of hydrogen. It undergoes beta decay: $$_1^3H \to \, _2^3He + \beta^- + \bar{\nu}_e$$. The maximum energy of the emitted $$\beta^-$$ particles is about 18.6 keV, which is very low compared to beta particles from other radioactive nuclei.

The correct answer is Option (3): Tritium $$_1^3H$$.

Hydrogen has three isotopes: protium ($$^1\text{H}$$), deuterium ($$^2\text{H}$$), and tritium ($$^3\text{H}$$). Among these, protium and deuterium are stable and non-radioactive, so they do not emit any radioactive particles.

Tritium ($$^3\text{H}$$) is radioactive. It undergoes $$\beta^-$$ decay: $$^3_1\text{H} \to \, ^3_2\text{He} + \beta^- + \bar{\nu}_e$$. The $$\beta^-$$ particles emitted by tritium have a very low maximum energy of about 18.6 keV, making them low-energy $$\beta^-$$ emitters. The half-life of tritium is approximately 12.3 years, which is greater than 12 years.

Therefore, the isotope of hydrogen that emits low-energy $$\beta^-$$ particles with a half-life greater than 12 years is tritium.

The answer is option (2).

The ionization energy of metal A equals the energy of the photon with wavelength 663 nm that is just sufficient to ionise the atom. The energy of a single photon is given by $$E = \frac{hc}{\lambda}$$.

Substituting the given values: $$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{663 \times 10^{-9}}$$.

Computing the numerator: $$6.63 \times 10^{-34} \times 3.00 \times 10^8 = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$$ J.

Computing the energy per atom: $$E = \frac{1.989 \times 10^{-25}}{663 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{6.63 \times 10^{-7}} = 3.00 \times 10^{-19}$$ J.

To convert to kJ mol$$^{-1}$$, we multiply by Avogadro's number and divide by 1000: $$IE = \frac{3.00 \times 10^{-19} \times 6.02 \times 10^{23}}{1000}$$.

$$IE = \frac{1.806 \times 10^{5}}{1000} = 180.6$$ kJ mol$$^{-1}$$.

Rounding to the nearest integer, the ionization energy is $$\boxed{181}$$ kJ mol$$^{-1}$$.

The magnetic quantum number $$m_l$$ tells us the orientation of an orbital.
For any subshell with azimuthal quantum number $$l$$, the possible $$m_l$$ values are $$-l,\;-(l-1),\dots ,0,\dots ,+(l-1),\; +l$$.
Hence every subshell that has $$l \ge 0$$ always contains exactly one orbital with $$m_l = 0$$.

The question asks for the number of such $$m_l = 0$$ orbitals that are completely filled (contain two electrons) in the ground-state configuration of germanium $$(Z = 32)$$.

Step 1: Write the ground-state electronic configuration of Ge (using the Aufbau, Hund and Pauli rules).
$$\text{Ge}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^2$$

Step 2: Examine each occupied subshell.
For every subshell, decide whether its $$m_l = 0$$ orbital is fully occupied (2 electrons) or not.

s-subshell ⇒ $$l = 0$$ ⇒ only one orbital ($$m_l = 0$$). It has 2 e⁻ ⇒ count 1.

Same reasoning ⇒ count 1 more (total 2).

p-subshell ⇒ three orbitals ($$m_l = -1,0,+1$$). Six electrons fill all three, so the $$m_l = 0$$ orbital is filled ⇒ count 1 (total 3).

s-subshell ⇒ count 1 (total 4).

p-subshell completely filled ⇒ count 1 (total 5).

s-subshell ⇒ count 1 (total 6).

d-subshell ⇒ five orbitals ($$m_l = -2,-1,0,+1,+2$$). Ten electrons fill all five, so the $$m_l = 0$$ orbital is filled ⇒ count 1 (total 7).

This subshell contains only two electrons. According to Hund’s rule they occupy two different orbitals with parallel spins, leaving the third orbital half-filled. Thus the $$m_l = 0$$ orbital holds at most one electron and is not completely filled ⇒ do not count.

Step 3: Add all contributions.
Total completely filled $$m_l = 0$$ orbitals $$= 7$$.

Hence, $$x = 7$$.

Gallium (Ga) has atomic number 31. Its ground state electronic configuration is [Ar] 3d¹⁰ 4s² 4p¹.

Ga⁺ ion is formed by removal of one electron from the neutral Ga atom. The electron is removed from the outermost subshell, which is 4p¹. So the configuration of Ga⁺ is [Ar] 3d¹⁰ 4s².

The valence electrons of Ga⁺ are those in the outermost (valence) shell, which is the 4th shell. The electrons present in the 4th shell are the 4s² electrons. The 3d electrons, though filled after 3p, are considered inner (core) electrons once the 4s is filled, but for Ga⁺ the outermost subshell with electrons is 4s.

For the 4s electrons, the azimuthal (angular momentum) quantum number $$\ell = 0$$.

Therefore, the azimuthal quantum number for the valence electrons of Ga⁺ is $$\ell = 0$$.

We begin by recalling the atomic number of zinc. Elemental zinc has $$Z = 30$$, therefore a neutral zinc atom possesses $$30$$ electrons.

The ground-state electronic configuration of a neutral zinc atom, written in the usual energy order $$1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d$$, is

$$\text{Zn} : [\text{Ar}]\,3d^{10}\,4s^{2}$$

where $$[\text{Ar}]$$ represents the filled inner core $$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$$.

Now we form the ion $$\text{Zn}^{+}$$. Making a cation means removing electrons, and in transition-metal atoms the electrons are always lost first from the subshell with the highest principal quantum number $$n$$. Here the $$4s$$ electrons (with $$n = 4$$) are removed before any $$3d$$ electrons (with $$n = 3$$).

Consequently, removing one electron from the neutral zinc atom takes away one of the two $$4s$$ electrons:

$$\text{Zn}^{+} : [\text{Ar}]\,3d^{10}\,4s^{1}$$

The single remaining outermost electron therefore resides in the $$4s$$ orbital.

For an $$s$$ subshell we know the azimuthal (orbital) quantum number is

$$l = 0$$

The magnetic quantum number $$m_l$$ can take all integer values from $$-l$$ to $$+l$$, inclusive. Stating the rule explicitly:

$$ m_l = -l,\, -(l-1),\,\dots,\,0,\,\dots,\,(l-1),\, +l $$

When $$l = 0$$ we have only one possible value, because the range collapses to a single integer:

$$m_l = 0$$

This sole value characterises every $$s$$ orbital, including the $$4s$$ orbital that contains the outermost electron of $$\text{Zn}^{+}$$.

Hence, the correct answer is Option 0.

The de Broglie wavelength of an electron accelerated through a potential difference $$V$$ is given by:

$$\lambda = \frac{h}{\sqrt{2meV}}$$

Substituting the given values: $$h = 6.63 \times 10^{-34}$$ Js, $$m = 9.1 \times 10^{-31}$$ kg, $$e = 1.6 \times 10^{-19}$$ C, and $$V = 40{,}000$$ V:

$$2meV = 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40000$$

$$= 2 \times 9.1 \times 1.6 \times 4 \times 10^{-31-19+4}$$

$$= 2 \times 58.24 \times 10^{-46} = 116.48 \times 10^{-46} = 1.1648 \times 10^{-44} \text{ kg}^2\text{m}^2\text{s}^{-2}$$

$$\sqrt{2meV} = \sqrt{1.1648 \times 10^{-44}} = 1.0793 \times 10^{-22} \text{ kg m s}^{-1}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{1.0793 \times 10^{-22}} = 6.14 \times 10^{-12} \text{ m}$$

So $$\lambda = X \times 10^{-12}$$ m, where $$X \approx 6$$.

The value of X is $$\mathbf{6}$$.

We have a bulb whose power is given as $$P = 50 \text{ W}$$. Power tells us the energy supplied per unit time, that is $$1 \text{ W} = 1 \text{ J s}^{-1}$$, so the bulb emits $$50 \text{ J}$$ of energy every second.

The light is monochromatic with wavelength $$\lambda = 795 \text{ nm}$$. First convert the wavelength into metres:

$$\lambda = 795 \text{ nm} = 795 \times 10^{-9} \text{ m} = 7.95 \times 10^{-7} \text{ m}.$$

The energy of a single photon is obtained from the Planck-Einstein relation, which we state explicitly:

Energy of one photon: $$E = \dfrac{h c}{\lambda},$$ where $$h = 6.63 \times 10^{-34} \text{ J s}$$ is Planck’s constant and $$c = 3.0 \times 10^{8} \text{ m s}^{-1}$$ is the speed of light.

Substituting the given values, we get

$$E = \dfrac{(6.63 \times 10^{-34}) (3.0 \times 10^{8})}{7.95 \times 10^{-7}} = \dfrac{1.989 \times 10^{-25}}{7.95 \times 10^{-7}}.$$

Now divide the mantissas and handle the powers of ten separately:

$$\dfrac{1.989}{7.95} = 0.25,$$ $$10^{-25} \div 10^{-7} = 10^{-25 + 7} = 10^{-18}.$$

So

$$E = 0.25 \times 10^{-18} \text{ J} = 2.5 \times 10^{-19} \text{ J}.$$

Let $$N$$ be the number of photons emitted per second. The total energy emitted in one second is the power $$P$$, so

$$P = N E \quad \Longrightarrow \quad N = \dfrac{P}{E}.$$

Substituting $$P = 50 \text{ J s}^{-1}$$ and $$E = 2.5 \times 10^{-19} \text{ J},$$ we get

$$N = \dfrac{50}{2.5 \times 10^{-19}} = \left(\dfrac{50}{2.5}\right) \times 10^{19} = 20 \times 10^{19} = 2.0 \times 10^{20}.$$

The question writes the result as $$x \times 10^{20}$$, so we compare:

$$N = 2.0 \times 10^{20} \; \Longrightarrow \; x = 2.$$

So, the answer is $$2$$.

We use the Heisenberg uncertainty principle: $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$, which gives $$\Delta x = \frac{h}{4\pi \cdot m \cdot \Delta v}$$.

The mass of the ball is $$m = 10$$ g $$= 0.01$$ kg. The velocity is 90 m/s and the uncertainty in velocity is 5%, so $$\Delta v = 0.05 \times 90 = 4.5$$ m/s.

Substituting the values: $$\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 0.01 \times 4.5}$$.

Computing the denominator: $$4 \times 3.14 \times 0.01 \times 4.5 = 4 \times 3.14 \times 0.045 = 4 \times 0.1413 = 0.5652$$.

Therefore $$\Delta x = \frac{6.63 \times 10^{-34}}{0.5652} = 11.73 \times 10^{-34} = 1.173 \times 10^{-33}$$ m.

Rounded off to the nearest integer, the uncertainty in position is $$1 \times 10^{-33}$$ m, so the answer is $$1$$.

We are given that a proton and a $$Li^{3+}$$ nucleus are accelerated by the same potential. We need to find $$\frac{\lambda_{Li}}{\lambda_p}$$.

The de Broglie wavelength of a charged particle accelerated through potential $$V$$ is given by:

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

where $$h$$ is Planck's constant, $$m$$ is the mass, $$q$$ is the charge, and $$V$$ is the accelerating potential.

For the proton: $$\lambda_p = \frac{h}{\sqrt{2 m_p \cdot e \cdot V}}$$

For $$Li^{3+}$$: $$\lambda_{Li} = \frac{h}{\sqrt{2 m_{Li} \cdot 3e \cdot V}}$$

Here $$m_{Li} = 8.3 \, m_p$$ and the charge on $$Li^{3+}$$ is $$3e$$.

Taking the ratio:

$$\frac{\lambda_{Li}}{\lambda_p} = \frac{\sqrt{2 m_p \cdot e \cdot V}}{\sqrt{2 m_{Li} \cdot 3e \cdot V}} = \sqrt{\frac{m_p \cdot e}{m_{Li} \cdot 3e}}$$

$$= \sqrt{\frac{m_p}{8.3 \, m_p \times 3}} = \sqrt{\frac{1}{24.9}}$$

$$= \frac{1}{\sqrt{24.9}} = \frac{1}{4.99} \approx 0.2$$

We are told the ratio equals $$x \times 10^{-1}$$. So:

$$0.2 = x \times 10^{-1}$$

$$x = 2$$

So, the answer is $$2$$.

We begin by noting the data given in the problem. The source supplies a total energy of 1000 J in 10 s, therefore the energy delivered each second (the power) is

$$P \;=\; \frac{1000\ \text{J}}{10\ \text{s}} \;=\; 100\ \text{J s}^{-1}.$$

This means that every second, 100 J of radiant energy strike the sodium surface.

Next we must find the energy carried by one photon of the monochromatic light. For a photon of wavelength $$\lambda$$ the energy is given by the Planck relation

$$E_{\text{photon}} \;=\; \frac{h\,c}{\lambda},$$

where $$h = 6.626 \times 10^{-34}\ \text{J s}$$ and $$c = 3.00 \times 10^{8}\ \text{m s}^{-1}.$$ The wavelength is

$$\lambda \;=\; 400\ \text{nm} \;=\; 400 \times 10^{-9}\ \text{m}.$$

Substituting these numbers,

$$E_{\text{photon}} = \frac{(6.626 \times 10^{-34}\,\text{J s}) (3.00 \times 10^{8}\,\text{m s}^{-1})} {400 \times 10^{-9}\,\text{m}}.$$

Simplifying step by step, first multiply the numerators:

$$6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 19.878 \times 10^{-26}\ \text{J m}.$$

Rewrite the denominator clearly:

$$400 \times 10^{-9}\ \text{m} = 4.00 \times 10^{2} \times 10^{-9}\ \text{m} = 4.00 \times 10^{-7}\ \text{m}.$$

Now divide:

$$E_{\text{photon}} = \frac{19.878 \times 10^{-26}}{4.00 \times 10^{-7}} = 4.9695 \times 10^{-19}\ \text{J}.$$

Thus, each photon carries approximately

$$E_{\text{photon}} \;\approx\; 4.97 \times 10^{-19}\ \text{J}.$$

The number of photons incident every second is then

$$N_{\text{photon\;per\;sec}} = \frac{\text{energy per second}}{\text{energy per photon}} = \frac{100\ \text{J s}^{-1}}{4.97 \times 10^{-19}\ \text{J}} = 2.01 \times 10^{20}\ \text{photons s}^{-1}.$$

The wavelength is sufficient for photo-emission, and we assume a one-to-one correspondence between photons absorbed and electrons ejected. Therefore

$$N_{\text{electron\;per\;sec}} = N_{\text{photon\;per\;sec}} \approx 2.01 \times 10^{20}\ \text{electrons s}^{-1}.$$

This can be written in the form $$x \times 10^{20}$$ electrons per second, giving

$$x \;\approx\; 2.01.$$

Rounding to the nearest integer, $$x = 2.$$

So, the answer is $$2$$.

We have an electron of mass $$m = 9.1 \times 10^{-31}\,{\rm kg}$$ moving with a speed $$v = 5 \times 10^{6}\,{\rm m\,s^{-1}}$$. The problem states that the speed is known only to within an uncertainty of 0.02 %. First we convert this percentage uncertainty into an absolute uncertainty in speed.

Since $$0.02\% = \dfrac{0.02}{100} = 0.0002,$$ the uncertainty in speed is

$$\Delta v = 0.0002 \times v = 0.0002 \times 5 \times 10^{6} = 1.0 \times 10^{3}\,{\rm m\,s^{-1}}.$$

Next, we connect this speed uncertainty to the momentum uncertainty. Momentum is defined by $$p = mv,$$ so its uncertainty is

$$\Delta p = m \,\Delta v = 9.1 \times 10^{-31}\,{\rm kg}\; \times 1.0 \times 10^{3}\,{\rm m\,s^{-1}} = 9.1 \times 10^{-28}\,{\rm kg\,m\,s^{-1}}.$$

To find the corresponding position uncertainty we invoke the Heisenberg uncertainty principle, which in its standard form reads

$$\Delta x\,\Delta p \;\ge\; \frac{h}{4\pi},$$

where $$h = 6.63 \times 10^{-34}\,{\rm J\,s}$$ and $$\pi = 3.14.$$ Assuming the minimum allowed product (equality case) gives

$$\Delta x = \frac{h}{4\pi\,\Delta p}.$$

Substituting the known numbers,

$$\Delta x = \frac{6.63 \times 10^{-34}} {4 \times 3.14 \times 9.1 \times 10^{-28}} = \frac{6.63 \times 10^{-34}} {12.56 \times 9.1 \times 10^{-28}}.$$

First multiply the constants in the denominator:

$$12.56 \times 9.1 = 114.296.$$

Hence,

$$\Delta x = \frac{6.63 \times 10^{-34}} {1.14296 \times 10^{-26}} = \left(\frac{6.63}{1.14296}\right) \times 10^{(-34) - (-26)} = 5.8 \times 10^{-8}\,{\rm m}.$$

To express this in the requested form $$x \times 10^{-9}\,{\rm m},$$ rewrite

$$5.8 \times 10^{-8}\,{\rm m} = 58 \times 10^{-9}\,{\rm m}.$$

So the nearest integer value of $$x$$ is $$58$$.

Hence, the correct answer is Option 58.

Vanadium has atomic number $$Z = 23$$. Its electronic configuration is $$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^3\, 4s^2$$.

The electrons in p-orbitals come from the $$2p$$ and $$3p$$ subshells. The $$2p$$ subshell holds 6 electrons and the $$3p$$ subshell holds 6 electrons, giving a total of $$6 + 6 = 12$$ electrons in p-orbitals.

The answer is $$12$$.

We need to find the number of orbitals with principal quantum number $$n = 5$$ and magnetic quantum number $$m_l = +2$$.

For a given value of $$n$$, the azimuthal quantum number $$l$$ can take values from $$0$$ to $$n - 1$$. For $$n = 5$$, we have $$l = 0, 1, 2, 3, 4$$.

The magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$. For $$m_l = +2$$ to be valid, we need $$l \geq 2$$. So the allowed values of $$l$$ are $$l = 2$$ (d-orbital), $$l = 3$$ (f-orbital), and $$l = 4$$ (g-orbital).

Each combination of $$(n, l, m_l)$$ corresponds to one unique orbital. Therefore, there are 3 orbitals with $$n = 5$$ and $$m_l = +2$$: the $$5d$$, $$5f$$, and $$5g$$ orbitals (each with $$m_l = +2$$).

The answer is $$3$$.

We have been given a monochromatic infrared source whose power is $$P = 1\ \text{mW}$$. First, we convert this power into SI units. Since $$1\ \text{mW} = 10^{-3}\ \text{W}$$, we can write $$P = 1 \times 10^{-3}\ \text{W}$$.

The device is operated for a time interval of $$t = 0.1\ \text{s}$$. In physics, the total energy $$E_{\text{total}}$$ radiated in a given time is obtained from the very definition of power, which states

$$P = \frac{\text{Energy}}{\text{Time}} \; \Longrightarrow \; \text{Energy} = P \times t.$$

Substituting the numerical values, we get

$$E_{\text{total}} = \bigl(1 \times 10^{-3}\ \text{W}\bigr)\,\bigl(0.1\ \text{s}\bigr) = 1 \times 10^{-4}\ \text{J}.$$

Next, we calculate the energy carried by a single photon. The well-known Einstein-Planck relation gives the energy $$E_{\text{photon}}$$ of one photon as

$$E_{\text{photon}} = h \, \nu = \frac{h\,c}{\lambda},$$

where $$h = 6.63 \times 10^{-34}\ \text{J\,s}$$ is Planck’s constant, $$c = 3.00 \times 10^{8}\ \text{m\,s}^{-1}$$ is the speed of light in vacuum, and $$\lambda$$ is the wavelength. The given wavelength is $$\lambda = 1000\ \text{nm}$$. First we express this wavelength in metres:

$$1000\ \text{nm} = 1000 \times 10^{-9}\ \text{m} = 1 \times 10^{-6}\ \text{m}.$$

Now we substitute all values into the photon-energy formula:

$$E_{\text{photon}} = \frac{\bigl(6.63 \times 10^{-34}\ \text{J\,s}\bigr)\,\bigl(3.00 \times 10^{8}\ \text{m\,s}^{-1}\bigr)}{1 \times 10^{-6}\ \text{m}}.$$

Multiplying the numerators and then handling the powers of ten carefully, we obtain

$$E_{\text{photon}} = \frac{6.63 \times 3.00 \times 10^{-34 + 8}}{1 \times 10^{-6}}\ \text{J} = \frac{19.89 \times 10^{-26}}{10^{-6}}\ \text{J}.$$

Dividing by $$10^{-6}$$ is equivalent to multiplying by $$10^{6}$$, so

$$E_{\text{photon}} = 19.89 \times 10^{-26 + 6}\ \text{J} = 19.89 \times 10^{-20}\ \text{J}.$$

For convenience we rewrite this in standard scientific notation by keeping only one non-zero digit before the decimal:

$$E_{\text{photon}} = 1.989 \times 10^{-19}\ \text{J}.$$

Now, the number of photons $$N$$ emitted is the ratio of the total energy radiated to the energy of one photon:

$$N = \frac{E_{\text{total}}}{E_{\text{photon}}} = \frac{1 \times 10^{-4}\ \text{J}}{1.989 \times 10^{-19}\ \text{J}}.$$

Carrying out the division step by step, we first divide the numerical coefficients:

$$\frac{1}{1.989} \approx 0.503.$$

Next we subtract the exponents of ten: $$10^{-4} / 10^{-19} = 10^{(-4) - (-19)} = 10^{15}.$$ Combining these two parts, we get

$$N \approx 0.503 \times 10^{15} = 5.03 \times 10^{14}.$$

The question asks us to express this answer in the form $$x \times 10^{13}$$. Notice that

$$5.03 \times 10^{14} = 50.3 \times 10^{13}.$$

Rounding $$50.3$$ to the nearest integer gives $$50$$. Therefore,

$$x = 50.$$

So, the answer is $$50$$.

We are given $$n = 4$$ and $$m_l = -3$$. Since the magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$, having $$m_l = -3$$ requires $$l \geq 3$$. For $$n = 4$$, the allowed values of $$l$$ are 0, 1, 2, and 3, so the minimum (and only possible) value is $$l = 3$$. This corresponds to the 4f orbital.

The number of radial nodes is given by the formula $$n - l - 1$$. Substituting the values, we get $$4 - 3 - 1 = 0$$.

Therefore, the 4f orbital has 0 radial nodes.

We have to find how many electrons of neptunium (atomic number $$Z=93$$) occupy any subshell of type $$f$$ in its ground-state electronic configuration. To do this we first write the complete configuration and then simply count the electrons that sit in all the $$f$$ subshells that appear.

The rule governing the order in which subshells are filled is called the Aufbau principle. It can be expressed through the $$(n+\ell)$$ rule:

For two subshells, the one with smaller }(n+\ell)\text{ is filled first. If }(n+\ell)\text{ is the same, the subshell with the smaller }n\text{ is filled first.

Using this filling order we proceed step by step, allotting a total of $$93$$ electrons.

We start with the lowest available subshells:

$$1s^2$$, $$\;2s^2$$, $$\;2p^6$$, $$\;3s^2$$, $$\;3p^6$$, $$\;4s^2$$, $$\;3d^{10}$$, $$\;4p^6$$, $$\;5s^2$$, $$\;4d^{10}$$, $$\;5p^6$$, $$\;6s^2$$, $$\;4f^{14}$$, $$\;5d^{10}$$, $$\;6p^6$$, $$\;7s^2$$

Up to this point the number of electrons already assigned is

$$2+2+6+2+6+2+10+6+2+10+6+2+14+10+6+2=86.$$

The first $$86$$ electrons complete the configuration of radon, so we write

$$\text{Neptunium}: [\mathrm{Rn}]\,\cdots$$

Now there are $$93-86 = 7$$ electrons still to be placed. According to the same $$(n+\ell)$$ rule the next subshells in line are $$5f$$, then $$6d$$.

Following experimental evidence for the actinide series, the most stable (ground-state) distribution for neptunium is

$$5f^4\,6d^1\,7s^2.$$

Combining this with the radon core gives the full ground-state configuration:

$$\boxed{[\mathrm{Rn}]\,5f^{4}\,6d^{1}\,7s^{2}}.$$

Now we count all electrons that reside in any $$f$$ subshell:

• The $$4f$$ subshell is completely filled inside the radon core, so it contains $$14$$ electrons.
• The $$5f$$ subshell just written contains $$4$$ electrons.

Adding these, the total number of $$f$$ electrons is

$$14 + 4 = 18.$$

Hence, the correct answer is Option 18.

We are given that light of wavelength $$\lambda = 248$$ nm falls on a metal with threshold energy (work function) $$\phi = 3.0$$ eV. We need to find the de-Broglie wavelength of the emitted electrons.

First, we calculate the energy of the incident photon. Using $$E = \frac{hc}{\lambda}$$, we get $$E = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9}}$$. The numerator is $$6.63 \times 3.0 \times 10^{-34+8} = 19.89 \times 10^{-26}$$ J m. Dividing by $$248 \times 10^{-9}$$ m gives $$E = \frac{19.89 \times 10^{-26}}{248 \times 10^{-9}} = 0.08020 \times 10^{-17} = 8.02 \times 10^{-19}$$ J.

Converting this to electron volts: $$E = \frac{8.02 \times 10^{-19}}{1.6 \times 10^{-19}} = 5.01 \approx 5.0$$ eV.

By Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is $$KE = E - \phi = 5.0 - 3.0 = 2.0$$ eV. Converting to joules: $$KE = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$$ J.

Now, the de-Broglie wavelength is given by $$\lambda_{dB} = \frac{h}{p}$$, where $$p = m_e v$$ is the momentum of the electron. Since $$KE = \frac{p^2}{2m_e}$$, we have $$p = \sqrt{2 \times m_e \times KE}$$.

Substituting the values: $$p = \sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}$$. Computing the product inside the square root: $$2 \times 9.1 \times 3.2 = 58.24$$, so $$p = \sqrt{58.24 \times 10^{-50}}$$. We can write this as $$p = \sqrt{58.24} \times 10^{-25}$$.

To evaluate $$\sqrt{58.24}$$, we note that $$58.24 = 4 \times 14.56 = 4 \times 3 \times 4.853$$. Alternatively, $$7.6^2 = 57.76$$ and $$7.63^2 \approx 58.22$$. So $$\sqrt{58.24} \approx 7.63$$. Therefore $$p \approx 7.63 \times 10^{-25}$$ kg m s$$^{-1}$$.

The de-Broglie wavelength is $$\lambda_{dB} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{7.63 \times 10^{-25}} = \frac{6.63}{7.63} \times 10^{-34+25} = 0.8689 \times 10^{-9}$$ m $$= 8.689 \times 10^{-10}$$ m $$= 8.689$$ angstrom.

Rounding off to the nearest integer, the de-Broglie wavelength of the emitted electrons is $$\mathbf{9}$$ angstrom.

For an electron moving in the $$n^{\text{th}}$$ Bohr orbit of a hydrogen atom, two standard relations are used repeatedly:

1. (Coulomb’s law equals centripetal force) $$\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}}$$

2. (Bohr’s quantisation of angular momentum) $$m v_{n} r_{n}=n\frac{h}{2\pi}$$

Eliminating $$v_{n}$$ from these two equations gives the well-known expression for the radius of the $$n^{\text{th}}$$ orbit,

$$r_{n}=n^{2}a_{0},$$

where the Bohr radius $$a_{0}$$ is

$$a_{0}= \frac{\varepsilon_{0}h^{2}}{\pi m e^{2}}.$$

We need the kinetic energy in the second orbit ($$n=2$$). From mechanics,

$$\text{K.E.}=\frac{1}{2}m v_{n}^{2}.$$

First we find $$v_{n}^{2}$$ from the force balance. Using the first relation,

$$\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}} \;\;\Longrightarrow\;\; v_{n}^{2}= \frac{e^{2}}{4\pi\varepsilon_{0}m r_{n}}.$$

Now substitute $$r_{n}=n^{2}a_{0}$$:

$$v_{n}^{2}= \frac{e^{2}}{4\pi\varepsilon_{0}m n^{2}a_{0}}.$$

Hence the kinetic energy is

$$\text{K.E.}= \frac12 m v_{n}^{2} =\frac12 m\left(\frac{e^{2}}{4\pi\varepsilon_{0}m n^{2}a_{0}}\right) =\frac{e^{2}}{8\pi\varepsilon_{0} n^{2} a_{0}}.$$

To convert the factor $$\dfrac{e^{2}}{4\pi\varepsilon_{0}}$$ into a form containing $$h^{2}$$ and $$a_{0}$$, we use the definition of $$a_{0}$$ written above. Starting with

$$a_{0}= \frac{\varepsilon_{0}h^{2}}{\pi m e^{2}},$$

rearrange to obtain

$$e^{2}= \frac{\varepsilon_{0}h^{2}}{\pi m a_{0}} \;\;\Longrightarrow\;\; \frac{e^{2}}{4\pi\varepsilon_{0}} =\frac{h^{2}}{4\pi^{2} m a_{0}}.$$

Insert this expression into the kinetic‐energy formula:

$$\text{K.E.}= \frac{1}{2 n^{2} a_{0}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) = \frac{1}{2 n^{2} a_{0}} \left(\frac{h^{2}}{4\pi^{2} m a_{0}}\right) =\frac{h^{2}}{8\pi^{2} n^{2} m a_{0}^{2}}.$$

For the second orbit, $$n=2\;$$ so $$n^{2}=4$$:

$$\text{K.E.}= \frac{h^{2}}{8\pi^{2}(4) m a_{0}^{2}} =\frac{h^{2}}{32\pi^{2} m a_{0}^{2}}.$$

This matches the required form $$\text{K.E.}= \frac{h^{2}}{x\,m a_{0}^{2}},$$ so

$$x = 32\pi^{2}.$$

Using $$\pi = 3.14$$ (as given),

$$\pi^{2}=3.14^{2}=9.8596,$$

$$x = 32 \times 9.8596 = 315.5072.$$

Now the question asks for $$10x$$ and its nearest integer value:

$$10x = 10 \times 315.5072 = 3155.072 \approx 3155.$$\

Hence, the correct answer is Option 3155.

We have been told that monochromatic radiation of wavelength $$\lambda = 500\ \text{nm}$$ (nanometres) is incident on a metal surface. The corresponding photon frequency is obtained from the fundamental relation between wavelength, frequency and the speed of light, namely $$c = \nu \lambda$$, where $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light.

Re-arranging, the frequency of the incident photons is

$$ \nu \;=\; \frac{c}{\lambda}. $$

Substituting $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ and $$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.0 \times 10^{-7}\ \text{m},$$ we get

$$ \nu \;=\; \frac{3.0 \times 10^{8}}{5.0 \times 10^{-7}} \;=\; 6.0 \times 10^{14}\ \text{Hz}. $$

The metal has a threshold (cut-off) frequency $$\nu_0 = 4.3 \times 10^{14}\ \text{Hz}$$. According to Einstein’s photoelectric equation, the maximum kinetic energy $$K_{\text{max}}$$ of an ejected electron is given by

$$ K_{\text{max}} = h(\nu - \nu_0), $$

where $$h = 6.63 \times 10^{-34}\ \text{J s}$$ is Planck’s constant.

First find the excess frequency above the threshold:

$$ \nu - \nu_0 = 6.0 \times 10^{14} - 4.3 \times 10^{14} = 1.7 \times 10^{14}\ \text{Hz}. $$

Now calculate the kinetic energy:

$$ K_{\text{max}} = h(\nu - \nu_0) = (6.63 \times 10^{-34}\ \text{J s})(1.7 \times 10^{14}\ \text{Hz}) = 1.1271 \times 10^{-19}\ \text{J}. $$

For an electron, this kinetic energy is related to its speed $$v$$ through the classical formula $$K = \tfrac12 m_e v^{2}$$, where the electron mass is $$m_e = 9.0 \times 10^{-31}\ \text{kg}$$. Stating the formula first:

$$ \frac12 m_e v^{2} = K_{\text{max}}. $$

Solving for $$v$$ gives

$$ v = \sqrt{\frac{2K_{\text{max}}}{m_e}}. $$

Substituting the numerical values:

$$ v = \sqrt{\frac{2(1.1271 \times 10^{-19})}{9.0 \times 10^{-31}}} = \sqrt{\frac{2.2542 \times 10^{-19}}{9.0 \times 10^{-31}}}. $$

Divide the mantissas and handle the powers of ten separately:

$$ \frac{2.2542}{9.0} = 0.250466\quad\text{and}\quad 10^{-19} \div 10^{-31} = 10^{12}. $$

Thus

$$ v = \sqrt{0.250466 \times 10^{12}} = \sqrt{2.50466 \times 10^{11}}. $$

Taking the square root:

$$ v \approx 5.0 \times 10^{5}\ \text{m s}^{-1}. $$

Because the question asks for the velocity expressed as “_________ $$\times 10^{5}$$ m s$$^{-1}$$ (nearest integer)”, we read off the factor as $$5$$.

So, the answer is $$5$$.

We recall the general Rydberg-Balmer relation for any line in the emission spectrum of hydrogen:

$$\bar v \;=\; R_H\left\{ \frac1{n_1^{\,2}} - \frac1{n_2^{\,2}}\right\},\qquad n_2 \gt n_1,\; n_1 = 1,2,3,\ldots$$

In the particular case of the Balmer series we always have the lower, or stationary, level fixed at $$n_1 = 2$$ while the upper level can have the successive integral values $$n_2 = 3,4,5,\ldots$$ Substituting this fixed value of $$n_1$$ in the formula we obtain

$$\bar v_{\text{Balmer}} = R_H \left\{ \frac1{2^{\,2}} - \frac1{n_2^{\,2}}\right\} \;=\; R_H\left\{ \frac14 - \frac1{n_2^{\,2}}\right\}.$$

Now we test the four given statements one by one.

Statement (I): “As wavelength decreases, the lines in the series converge.”

Because $$n_2$$ appears in the denominator, increasing $$n_2$$ makes $$\dfrac1{n_2^{\,2}}$$ steadily smaller, so the bracket $$\left\{\dfrac14 - \dfrac1{n_2^{\,2}}\right\}$$ becomes larger and hence $$\bar v$$ (the wave number) increases. Since wavelength $$\lambda$$ is the reciprocal of wave number $$\bar v$$, viz. $$\lambda = 1/\bar v,$$ an increase in $$\bar v$$ implies a decrease in $$\lambda$$. Meanwhile the difference in $$\bar v$$ between successive $$n_2$$ values keeps shrinking, so the spectral lines crowd together. Thus the lines do indeed converge as the wavelength becomes shorter. Statement (I) is correct.

Statement (II): “The integer $$n_1$$ is equal to 2.”

We have already substituted $$n_1 = 2$$ for the Balmer series, so Statement (II) is correct.

Statement (III): “The lines of the longest wavelength correspond to $$n_2 = 3.$$”

The longest wavelength means the smallest wave number. Inspection of $$\bar v = R_H\!\left\{\dfrac14 - \dfrac1{n_2^{\,2}}\right\}$$ shows that the smallest possible value of $$\bar v$$ occurs for the smallest allowable $$n_2$$, which is $$n_2 = 3$$. Therefore the first Balmer line (from $$n_2=3$$ to $$n_1=2$$) has the greatest wavelength. Statement (III) is correct.

Statement (IV): “The ionization energy of hydrogen can be calculated from the wave number of these lines.”

The ionization energy is the energy required to remove the electron completely from the ground state $$n=1$$ to $$n=\infty$$. Its value is

$$E_{\text{ion}} = h c R_H.$$ Indeed, one can find $$R_H$$ if the ground-state Lyman lines (with $$n_1 = 1$$) or the series limit of any complete series are available. However, the Balmer series deals with transitions that start from $$n_1 = 2$$, so the Balmer data alone do not yield the energy gap between $$n=1$$ and $$n=\infty$$ unless additional information is supplied. Hence Statement (IV) is not correct when only Balmer lines are considered.

Summarising, the true statements are (I), (II) and (III), while (IV) is false.

Hence, the correct answer is Option B.

We know that naturally occurring chlorine is a mixture of two stable isotopes, $$^{35}\text{Cl}$$ and $$^{37}\text{Cl}$$. Their atomic (isotopic) masses are, to a very good approximation, $$35\,\text{u}$$ and $$37\,\text{u}$$ respectively.

The weighted (average) molar mass that appears on the periodic table is obtained by taking a mole-weighted mean of the individual isotopic masses. The general formula for the average molar mass $$\overline{M}$$ of a two-isotope mixture is

$$\overline{M}=\frac{n_{1}M_{1}+n_{2}M_{2}}{n_{1}+n_{2}},$$

where

$$n_{1},\,n_{2}\;=\;$$ number of moles (or any proportional measure) of the two isotopes $$,$$

$$M_{1},\,M_{2}\;=\;$$ their respective molar masses $$.$$

In our case

$$M_{1}=35\;\text{g mol}^{-1},$$ $$M_{2}=37\;\text{g mol}^{-1},$$ $$\overline{M}=35.5\;\text{g mol}^{-1}.$$

Let the amount of $$^{35}\text{Cl}$$ be $$n_{1}$$ and that of $$^{37}\text{Cl}$$ be $$n_{2}$$. Substituting the known values into the formula gives

$$35.5=\frac{35\,n_{1}+37\,n_{2}}{n_{1}+n_{2}}.$$

Because only the ratio is required, we may set $$n_{2}=1$$ (any convenient reference) and denote $$n_{1}=r$$, so that the sought ratio $$^{35}\text{Cl}:^{37}\text{Cl}=r:1$$. Rewriting the equation with these symbols, we have

$$35.5=\frac{35\,r+37\cdot1}{r+1}.$$

Now we clear the denominator by multiplying both sides by $$r+1$$:

$$35.5\,(r+1)=35\,r+37.$$

Expanding the left side,

$$35.5\,r+35.5=35\,r+37.$$

Next we collect like terms, bringing all expressions involving $$r$$ to the left and constants to the right:

$$35.5\,r-35\,r=37-35.5.$$

The left side simplifies to $$0.5\,r$$ and the right side to $$1.5$$, so

$$0.5\,r=1.5.$$

Dividing both sides by $$0.5$$ yields

$$r=\frac{1.5}{0.5}=3.$$

Therefore the ratio of $$^{35}\text{Cl}$$ to $$^{37}\text{Cl}$$ is

$$^{35}\text{Cl}:^{37}\text{Cl}=3:1.$$

Among the options provided, the ratio $$3:1$$ corresponds to Option B.

Hence, the correct answer is Option B.

We begin with the definition of probability density in quantum mechanics. For any hydrogen-like orbital, the probability density at a point in space is given by the square of the magnitude of the wave-function:

$$\text{Probability density}=|\Psi_{n,l,m}(r,\theta,\phi)|^{2}.$$

Because we are squaring an absolute value, we immediately obtain the inequality

$$|\Psi_{n,l,m}|^{2}\;\ge\;0,$$

so the probability density can never be negative at any finite point in space.

Next we recall the nodal rules for hydrogen-like orbitals. The total number of nodes possessed by an orbital with quantum numbers $$n$$ and $$l$$ is $$n-1.$$ These nodes are of two kinds:

• Radial (spherical) nodes, counted by the well-known formula

$$\text{Number of radial nodes}=n-l-1,$$

• Angular (planar/conical) nodes, whose number equals $$l.$$ At every node the wave-function $$\Psi$$ changes sign, so its value is exactly zero; consequently the probability density, being $$|\Psi|^{2},$$ is also zero at that position.

We now examine each option one by one, using the above facts.

Option A: 1s orbital has $$n=1,\;l=0.$$ The number of radial nodes is $$1-0-1 = 0,$$ and the number of angular nodes is $$0.$$ Thus, except at $$r\to\infty,$$ the wave-function is never zero, so the probability density is never zero within finite space. Hence Option A is incorrect.

Option B: We have already proved that $$|\Psi|^{2}$$ can never be negative, because it is a squared modulus. Therefore probability density cannot be negative for a 2p orbital or for any other orbital. Option B is incorrect.

Option C: 3p orbital has $$n=3,\;l=1.$$ First we compute the radial nodes: $$n-l-1 = 3-1-1 = 1.$$ In addition, because $$l=1,$$ there is $$1$$ angular node (a nodal plane). Both the radial node and the angular node are genuine points (or surfaces) where $$\Psi=0,$$ so the probability density $$|\Psi|^{2}$$ is zero there. Thus, for a 3p orbital, the probability density can indeed be zero at finite distances from the nucleus. Option C is correct.

Option D: 2s orbital has $$n=2,\;l=0.$$ The number of radial nodes is $$2-0-1 = 1,$$ so there exists a spherical surface inside which $$\Psi=0,$$ making the probability density zero on that surface. Therefore the statement that it can “never be zero” is false, and Option D is incorrect.

Out of the four alternatives, only Option C survives careful scrutiny.

Hence, the correct answer is Option C.

For an electron moving in a Bohr orbit, the de Broglie hypothesis tells us that its orbital circumference must accommodate an integral number of wavelengths. Mathematically, this statement is written as $$2\pi r_n = n\lambda,$$ where $$r_n$$ is the radius of the $$n^{\text{th}}$$ orbit, $$\lambda$$ is the de Broglie wavelength, and $$n$$ is the principal quantum number.

We want an explicit expression for $$\lambda,$$ so we solve the above equation for $$\lambda$$ by dividing both sides by $$n$$:

$$\lambda = \frac{2\pi r_n}{n}.$$

Next, we must substitute the Bohr‐model formula for the radius of the $$n^{\text{th}}$$ orbit of a hydrogen-like atom. That formula is $$r_n = n^2 a_0,$$ where $$a_0$$ is the Bohr radius (a universal constant for hydrogen).

Substituting $$r_n = n^2 a_0$$ into the wavelength expression, we obtain

$$\lambda = \frac{2\pi(n^2 a_0)}{n}.$$

We simplify the fraction by cancelling one factor of $$n$$ in the numerator and the denominator:

$$\lambda = 2\pi n a_0.$$

Now we place the electron specifically in the $$4^{\text{th}}$$ Bohr orbit, which means we set $$n = 4.$$ Putting this value into the simplified formula gives

$$\lambda = 2\pi(4)a_0 = 8\pi a_0.$$

Thus the de Broglie wavelength of an electron in the fourth Bohr orbit is $$8\pi a_0.$$

Hence, the correct answer is Option D.

For any hydrogen-like species, Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n=\frac{n^{2}a_0}{Z}$$

where $$a_0$$ is the Bohr radius and $$Z$$ is the atomic number of the nucleus.

We first apply this formula to the ion $$\text{Li}^{2+}$$. For lithium, $$Z=3$$. Therefore

$$r_n(\text{Li}^{2+})=\frac{n^{2}a_0}{3}.$$

Putting $$n=4$$, we get

$$r_4(\text{Li}^{2+})=\frac{4^{2}a_0}{3}=\frac{16a_0}{3}.$$

Putting $$n=3$$, we get

$$r_3(\text{Li}^{2+})=\frac{3^{2}a_0}{3}=\frac{9a_0}{3}=3a_0.$$

Now the required difference for lithium is

$$\Delta R_1=r_4-r_3=\frac{16a_0}{3}-3a_0=\frac{16a_0}{3}-\frac{9a_0}{3}=\frac{7a_0}{3}.$$

Next we turn to the ion $$\text{He}^+$$. For helium, $$Z=2$$. The same formula gives

$$r_n(\text{He}^+)=\frac{n^{2}a_0}{2}.$$

Putting $$n=4$$, we get

$$r_4(\text{He}^+)=\frac{4^{2}a_0}{2}=\frac{16a_0}{2}=8a_0.$$

Putting $$n=3$$, we get

$$r_3(\text{He}^+)=\frac{3^{2}a_0}{2}=\frac{9a_0}{2}=4.5a_0.$$

The required difference for helium is therefore

$$\Delta R_2=r_4-r_3=8a_0-4.5a_0=3.5a_0=\frac{7a_0}{2}.$$

To find the desired ratio, we write

$$\frac{\Delta R_1}{\Delta R_2}=\frac{\dfrac{7a_0}{3}}{\dfrac{7a_0}{2}}=\frac{7a_0}{3}\times\frac{2}{7a_0}=\frac{2}{3}.$$

Hence

$$\Delta R_1:\Delta R_2=2:3.$$

Hence, the correct answer is Option C.

We need to determine which of the provided figures cannot be explained as a direct manifestation of the quantum nature of atoms.

Phenomena Guided by Quantum Nature:

The quantum nature of matter and radiation is fundamentally characterized by discrete energy levels, localized packet-like behaviors (photons), and deviations from classical continuous physical principles:

• Black-body Radiation Curve: The intensity vs. wavelength distribution cannot be justified using classical wave theory. Its accurate explanation relies strictly on Planck's Quantum Hypothesis, where energy is emitted or absorbed in discrete packets called quanta.
• Photoelectric Effect: The linear graph of the kinetic energy of photoelectrons versus the frequency of incident radiation exhibits a clear threshold behavior. This is directly validated by Einstein's photon model of light, demonstrating its quantized particle nature.
• Absorption/Emission Line Spectra: The presence of precise, distinct dark lines against a continuous wavelength background occurs because electrons transition exclusively between fixed, quantized electronic energy states within an atom.

Classical Manifestation (The Exception):

Let's evaluate the behavior of the internal energy of Argon ($$\text{Ar}$$) as a function of temperature:

• Argon is a monoatomic noble gas. According to the Classical Kinetic Theory of Gases and the law of equipartition of energy, its internal energy ($$\text{U}$$) is composed solely of continuous translational motion components:
$$U = \frac{3}{2}RT$$
• As the temperature increases, the internal energy scales up perfectly linearly and continuously, without involving quantized transitions or discrete quantum atomic configurations. Consequently, it represents a standard classical macroscopic thermodynamic relationship rather than a quantum phenomenon.

Conclusion:

Unlike the other options that strictly call for a quantum mechanical description, the linear progression of the internal energy of argon with temperature is fully explained by classical kinetic theory.

Answer: The option showing the graph of Internal energy of Ar vs Temperature (K).

We begin by recalling the ranges allowed by the quantum-number rules of the hydrogen-like atom.

First, the azimuthal quantum number $$l$$ can take all integral values from $$0$$ up to $$n-1$$. In symbols, we write the rule

$$0 \le l \le n-1.$$

Second, the magnetic quantum number $$m_l$$ (here simply written as $$m$$) can take every integral value from $$-l$$ to $$+l$$. That rule is written as

$$-l \le m \le +l.$$

Now we substitute the principal quantum number given in the question, namely $$n = 4$$. Using the first rule we obtain every permissible $$l$$ value for this $$n$$:

$$l = 0, \; 1, \; 2, \; 3.$$ (We have four possibilities because $$l$$ must be less than $$n = 4$$.)

The question also specifies the magnetic quantum number $$m = -2$$. From the second rule we know that a particular $$l$$ value is acceptable only if it satisfies

$$|m| \le l.$$

Here $$|m| = |-2| = 2$$, so we must have $$l \ge 2$$. Looking back at the list $$l = 0, 1, 2, 3$$, the values that satisfy $$l \ge 2$$ are

$$l = 2 \quad\text{and}\quad l = 3.$$

Each distinct pair $$(n, l)$$ corresponds to one subshell (for example, $$l = 2$$ with $$n = 4$$ is the $$4d$$ subshell, and $$l = 3$$ with $$n = 4$$ is the $$4f$$ subshell). Because we have exactly two qualifying $$l$$ values, we have two distinct subshells that can possess the magnetic quantum number $$m = -2$$ while still belonging to principal shell $$n = 4$$.

Hence the total number of subshells associated with $$n = 4$$ and $$m = -2$$ is

$$2.$$

Hence, the correct answer is Option B.

We begin by recalling the general formula for the wavelength of any spectral line in the hydrogen atom. For a transition that ends in an energy level of principal quantum number $$n_l$$ and starts from a higher level $$n_h$$ (where $$n_h > n_l$$), the well-known Balmer-Rydberg formula is stated as

$$\frac{1}{\lambda}=R\left(\frac{1}{n_l^{\,2}}-\frac{1}{n_h^{\,2}}\right),$$

where $$\lambda$$ is the wavelength of the emitted photon and $$R$$ is the Rydberg constant $$R\approx1.097\times10^{7}\,\text{m}^{-1}.$$

Balmer himself chose $$n_l = 2$$, so every line of the Balmer series corresponds to electrons falling to the second orbit. Therefore, for the Balmer series we have

$$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n_h^{\,2}}\right)=R\left(\frac{1}{4}-\frac{1}{n_h^{\,2}}\right),\qquad n_h = 3,4,5,\ldots$$

Now let us see what orders of magnitude for $$\lambda$$ this formula yields. We may take a few values of $$n_h$$ just to convince ourselves numerically:

For $$n_h = 3$$ (the first Balmer line, called Hα):

$$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=R\left(\frac{5}{36}\right).$$

Substituting $$R=1.097\times10^{7}\,\text{m}^{-1}$$, we get

$$\frac{1}{\lambda}=1.097\times10^{7}\times\frac{5}{36}=1.523\times10^{6}\,\text{m}^{-1},$$

so

$$\lambda=\frac{1}{1.523\times10^{6}}=6.57\times10^{-7}\,\text{m}=657\,\text{nm}.$$

This is clearly inside the range $$400 \text{ nm} \le \lambda \le 700 \text{ nm}$$, which we recognize as the visible region.

For $$n_h = 4$$ (Hβ):

$$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\left(\frac{4-1}{16}\right)=R\left(\frac{3}{16}\right).$$

So

$$\frac{1}{\lambda}=1.097\times10^{7}\times\frac{3}{16}=2.055\times10^{6}\,\text{m}^{-1},$$

yielding

$$\lambda=\frac{1}{2.055\times10^{6}}=4.87\times10^{-7}\,\text{m}=486\,\text{nm},$$

again comfortably in the visible band.

Proceeding to still higher $$n_h$$ values only pushes the wavelength a little further toward the violet end but always within or very near the visible limits before merging into the ultraviolet convergence limit at $$\lambda\approx364\,\text{nm}$$. The main intense Balmer lines lie wholly in the visible window.

Because the Balmer series wavelengths fall between roughly $$364\,\text{nm}$$ and $$656\,\text{nm}$$, the region of the electromagnetic spectrum in which these lines are observed is indeed the visible region.

Hence, the correct answer is Option A.

For any hydrogen-like species, the line-spectrum obeys the Rydberg formula

$$\frac{1}{\lambda}=R\,Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right),\qquad n_{2}>n_{1},$$

where $$R$$ is the Rydberg constant, $$Z$$ is the atomic number, $$n_{1}$$ is the lower level and $$n_{2}$$ is the upper level.

We first connect this formula with the given quantity $$\lambda_{1}$$, the shortest wavelength of the Lyman series of the hydrogen atom (H, so $$Z=1$$). In the Lyman series $$n_{1}=1$$, and the shortest wavelength arises when $$n_{2}\rightarrow\infty$$, because the term $$\frac{1}{n_{2}^{2}}$$ then vanishes:

$$\frac{1}{\lambda_{1}}=R\,(1)^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=R\,(1-0)=R.$$

Thus we have the handy relation

$$R=\frac{1}{\lambda_{1}}.$$

Next we deal with the Balmer series of the singly ionised helium atom He$$^{+}$$, whose nuclear charge is $$Z=2$$. For the Balmer series $$n_{1}=2$$. The longest wavelength in any series corresponds to the smallest energy (and hence wavenumber) difference, which occurs for the transition having the smallest possible $$n_{2}$$, i.e. $$n_{2}=3$$.

Substituting $$Z=2,\;n_{1}=2,\;n_{2}=3$$ in the Rydberg formula we obtain

$$\frac{1}{\lambda_{\,\text{Balmer}}}=R\,(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right).$$

Carrying out every arithmetic step,

$$\frac{1}{\lambda_{\,\text{Balmer}}}=R\cdot4\left(\frac{1}{4}-\frac{1}{9}\right).$$

Inside the brackets, bring to a common denominator 36:

$$\frac{1}{4}-\frac{1}{9}=\frac{9}{36}-\frac{4}{36}=\frac{5}{36}.$$

So

$$\frac{1}{\lambda_{\,\text{Balmer}}}=4R\left(\frac{5}{36}\right)=\frac{20R}{36}=\frac{5R}{9}.$$

Now replace $$R$$ by $$\dfrac{1}{\lambda_{1}}$$ using the earlier relation:

$$\frac{1}{\lambda_{\,\text{Balmer}}}=\frac{5}{9}\,\frac{1}{\lambda_{1}}.$$

Taking reciprocals on both sides gives the required wavelength explicitly in terms of $$\lambda_{1}$$:

$$\lambda_{\,\text{Balmer}}=\frac{9\,\lambda_{1}}{5}.$$

This matches Option C.

Hence, the correct answer is Option C.

We have been told that the principal quantum number is $$n = 5$$. The principal quantum number determines the main shell of the electron and, very importantly, fixes the possible values of the azimuthal quantum number $$\ell$$.

First, recall the rule for $$\ell$$: for any fixed $$n$$, $$\ell$$ can take all integral values from $$0$$ up to $$n-1$$. Therefore, when $$n = 5$$ we write

$$\ell = 0, 1, 2, 3, 4.$$

Now, for each specific value of $$\ell$$, the magnetic quantum number $$m_\ell$$ is allowed to vary from $$-\ell$$ to $$+\ell$$ in steps of one. Thus the total number of distinct $$m_\ell$$ values for a given $$\ell$$ is

$$2\ell + 1.$$

If we want the total number of orbitals for a fixed $$n$$, we must add up all the $$m_\ell$$ values for every possible $$\ell$$. Symbolically, that total is

$$\sum_{\ell = 0}^{n-1} (2\ell + 1) = n^{2}.$$

We now substitute $$n = 5$$ to get

$$\text{Number of orbitals for } n = 5 = 5^{2} = 25.$$

Each of these orbitals may be occupied by electrons having either of the two possible spin quantum numbers $$m_s = +\tfrac{1}{2}$$ or $$m_s = -\tfrac{1}{2}$$. The question explicitly fixes $$m_s = +\tfrac{1}{2}$$. When we restrict ourselves to a single spin orientation, we do not change the count of orbitals; we merely say that, out of the two spin possibilities for every orbital, we are picking just one.

Therefore, the number of orbitals that possess the principal quantum number $$n = 5$$ and the spin quantum number $$m_s = +\tfrac{1}{2}$$ remains

$$25.$$

Hence, the correct answer is Option B.

For any one-electron species, Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n=\frac{n^2h^2\varepsilon_0}{\pi m_e e^2Z}.$$

In hydrogen, where $$Z=1$$ and $$n=1$$, this radius is called the Bohr radius $$a_0$$, so we write

$$a_0=\frac{h^2\varepsilon_0}{\pi m_e e^2}.$$

Dividing the two expressions term by term, the constant factors cancel, and we obtain the general relation

$$r_n=\frac{n^2a_0}{Z}.$$

For the doubly-ionised lithium ion, $$\text{Li}^{2+}$$, the nuclear charge is $$Z=3$$ because the nucleus has three protons. The question asks for the radius of the second Bohr orbit, so we put $$n=2$$. Substituting these values into the formula just derived gives

$$r_2=\frac{(2)^2a_0}{3}=\frac{4a_0}{3}.$$

This matches Option C.

Hence, the correct answer is Option C.

For any nucleus, the relation between mass number and the numbers of protons and neutrons is stated first:

$$A \;=\; Z \;+\; N,$$

where $$A$$ is the mass number (total nucleons), $$Z$$ is the atomic number (number of protons) and $$N$$ is the number of neutrons.

In every isotope of hydrogen, the atomic number is the same, $$Z = 1,$$ because each hydrogen atom possesses one proton. The three isotopes differ only in their mass numbers, so we now examine them one by one.

Isotope (A) is protium and has mass number $$A = 1.$$ Substituting in the formula, we have

$$N = A - Z = 1 - 1 = 0.$$ So, $$x = 0.$$

Isotope (B) is deuterium and its mass number is $$A = 2.$$ Again, using the same formula,

$$N = A - Z = 2 - 1 = 1.$$ Hence, $$y = 1.$$

Isotope (C) is tritium with mass number $$A = 3.$$ Applying the formula once more,

$$N = A - Z = 3 - 1 = 2.$$ Therefore, $$z = 2.$$

We now add the three neutron numbers:

$$x + y + z = 0 + 1 + 2 = 3.$$

Thus, the required sum is $$3.$$

Hence, the correct answer is Option A.

For photoelectric emission we use Einstein’s photoelectric equation, which states

$$E_{\text{photon}} = \phi \;+\; K.E.$$

where $$E_{\text{photon}}$$ is the energy of the incident photon, $$\phi$$ is the work-function of the metal, and $$K.E.$$ is the kinetic energy of the emitted electron.

First we calculate the photon energy from its wavelength. The energy-wavelength relation is

$$E_{\text{photon}} \;=\; \dfrac{h\,c}{\lambda}.$$

We have the numerical values

$$h = 6.63 \times 10^{-34}\,\text{J s}, \qquad c = 3.00 \times 10^{8}\,\text{m s}^{-1}, \qquad \lambda = 300\,\text{nm} = 300 \times 10^{-9}\,\text{m}.$$

Substituting these in the formula gives

$$E_{\text{photon}} = \dfrac{(6.63 \times 10^{-34})\,(3.00 \times 10^{8})}{300 \times 10^{-9}}\;\text{J}.$$

Multiplying the numerators,

$$6.63 \times 3.00 = 19.89, \quad\text{and}\quad 10^{-34}\times 10^{8}=10^{-26},$$

so the numerator becomes

$$19.89 \times 10^{-26}\;\text{J m}.$$

For the denominator, write

$$300 \times 10^{-9} = (3.00 \times 10^{2}) \times 10^{-9} = 3.00 \times 10^{-7}\;\text{m}.$$

Now divide:

$$E_{\text{photon}} = \dfrac{19.89 \times 10^{-26}}{3.00 \times 10^{-7}}\;\text{J} = \left(\dfrac{19.89}{3.00}\right) \times 10^{-26+7}\;\text{J}.$$

Calculating the numeric and exponent parts separately,

$$\dfrac{19.89}{3.00} = 6.63, \qquad -26 + 7 = -19,$$

hence

$$E_{\text{photon}} = 6.63 \times 10^{-19}\;\text{J}.$$

The work-function of sodium is given as

$$\phi = 4.41 \times 10^{-19}\;\text{J}.$$

Using Einstein’s equation, the kinetic energy of the photoelectrons is

$$K.E. = E_{\text{photon}} - \phi = (6.63 \times 10^{-19}) - (4.41 \times 10^{-19})\;\text{J}.$$

Subtracting the coefficients while keeping the common power of ten,

$$6.63 - 4.41 = 2.22,$$

so

$$K.E. = 2.22 \times 10^{-19}\;\text{J}.$$

Finally, the problem asks for the answer in units of $$\times 10^{-21}\;\text{J}.$$ To rewrite the kinetic energy accordingly, note that

$$2.22 \times 10^{-19}\;\text{J} = 2.22 \times 10^{2} \times 10^{-21}\;\text{J} = 222 \times 10^{-21}\;\text{J}.$$

So, the answer is $$222$$.

We begin with the well-known Rydberg formula for the spectrum of atomic hydrogen. In terms of wave number (the reciprocal of wavelength) the formula is stated as

$$\bar{\nu}=R_H\left(\dfrac{1}{n_f^{\,2}}-\dfrac{1}{n_i^{\,2}}\right)$$

Here $$n_i$$ is the principal quantum number of the initial (higher) level and $$n_f$$ is that of the final (lower) level. For the emission line mentioned in the problem we have the fixed initial level $$n_i = 8$$ and the variable final level $$n_f = n$$, where obviously $$n<8$$ because energy is being released.

Substituting $$n_i = 8$$ into the formula we get

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}-\dfrac{1}{8^{2}}\right)$$

Since $$8^{2}=64$$, the expression simplifies step by step as follows:

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}-\dfrac{1}{64}\right)$$

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}\right)-R_H\left(\dfrac{1}{64}\right)$$

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}\right)-\dfrac{R_H}{64}$$

Let us now introduce the variable

$$x=\dfrac{1}{n^{2}}$$

Re-writing the result in terms of this new variable gives

$$\bar{\nu}=R_H\,x-\dfrac{R_H}{64}$$

We can see clearly that the right-hand side is of the form $$y=mx+c$$, where the dependent variable is $$y=\bar{\nu}$$, the independent variable is $$x=\dfrac{1}{n^{2}}$$, the slope (or gradient) is $$m=R_H$$, and the y-intercept is $$c=-\dfrac{R_H}{64}$$.

Thus, the graph of $$\bar{\nu}$$ versus $$\dfrac{1}{n^{2}}$$ is a straight line (i.e. linear), and the slope of this straight line is positive and equal to $$R_H$$.

Hence, the correct answer is Option C.

We have to decide where the very last (71st) electron of the element with atomic number $$Z = 71$$ will go. For this we apply the Aufbau (building-up) principle, which states:

Rule stated first: “Electrons occupy orbitals in the increasing order of $$n+\ell$$. When two orbitals have the same $$n+\ell$$ value, the one with the lower principal quantum number $$n$$ is filled first.”

Listing the subshells in the sequence dictated by this rule gives the familiar order

$$ 1s \;\rightarrow\; 2s \;\rightarrow\; 2p \;\rightarrow\; 3s \;\rightarrow\; 3p \;\rightarrow\; 4s \;\rightarrow\; 3d \;\rightarrow\; 4p \;\rightarrow\; 5s \;\rightarrow\; 4d \;\rightarrow\; 5p \;\rightarrow\; 6s \;\rightarrow\; 4f \;\rightarrow\; 5d \;\rightarrow\; 6p \;\rightarrow\; 7s \;\ldots $$

Now we successively place electrons until we reach the 71st one, keeping track of the capacity of each subshell ($$s:2,\; p:6,\; d:10,\; f:14$$):

$$ \begin{aligned} 1s^2 &\quad (\text{total }2)\\ 2s^2 &\quad (\text{total }4)\\ 2p^6 &\quad (\text{total }10)\\ 3s^2 &\quad (\text{total }12)\\ 3p^6 &\quad (\text{total }18)\\ 4s^2 &\quad (\text{total }20)\\ 3d^{10} &\quad (\text{total }30)\\ 4p^6 &\quad (\text{total }36)\\ 5s^2 &\quad (\text{total }38)\\ 4d^{10} &\quad (\text{total }48)\\ 5p^6 &\quad (\text{total }54)\; \Bigl(\text{configuration of }^{54}\text{Xe}\Bigr)\\[4pt] 6s^2 &\quad (\text{total }56)\\ 4f^{14} &\quad (\text{total }70) \end{aligned} $$

At this stage we have accommodated 70 electrons. We still need to place one more—the 71st electron.

The next subshell in the Aufbau list after $$4f$$ is $$5d$$ because $$5d$$ has $$n+\ell = 5+2 = 7$$, which is the same as $$6p$$ ( $$6+1 = 7$$ ), but its principal quantum number $$n = 5$$ is smaller than that of $$6p$$ ($$n = 6$$). Therefore $$5d$$ is lower in energy than $$6p$$ and is filled first.

So the 71st electron must enter the $$5d$$ subshell, giving the outer portion of the configuration

$$\boxed{6s^2\,4f^{14}\,5d^1}$$

Hence, the correct answer is Option A.

First we recall the empirical (n + l) rule, also called the Madelung rule, which states: “For multi-electron atoms the sub-shell having smaller $$n + l$$ value possesses lower energy. If two sub-shells have the same $$n + l$$ value, the one with smaller principal quantum number $$n$$ is lower in energy.”

Now we evaluate the quantity $$n + l$$ for every given electron:

For electron I we have $$n = 4$$ and $$l = 2$$. So $$ n + l = 4 + 2 = 6. $$

For electron II we have $$n = 3$$ and $$l = 2$$. So $$ n + l = 3 + 2 = 5. $$

For electron III we have $$n = 4$$ and $$l = 1$$. So $$ n + l = 4 + 1 = 5. $$

For electron IV we have $$n = 3$$ and $$l = 1$$. So $$ n + l = 3 + 1 = 4. $$

Next we arrange the sub-shells in ascending order of their $$n + l$$ values:

Electron IV has $$n + l = 4$$, the smallest of all, so it is the lowest in energy.

Electrons II and III both have $$n + l = 5$$. For such a tie we compare their principal quantum numbers $$n$$. Electron II has $$n = 3$$ while electron III has $$n = 4$$. Because the smaller $$n$$ corresponds to lower energy, electron II precedes electron III.

Electron I has the largest value, $$n + l = 6$$, so it is the highest in energy.

Collecting all these comparisons we obtain the increasing energy order:

$$\text{IV} \;<\; \text{II} \;<\; \text{III} \;<\; \text{I}.$$

This sequence matches exactly with Option B.

Hence, the correct answer is Option B.

First, recall Einstein’s photo-electric equation, which links the energy of the incident photon to the kinetic energy of the emitted electron and the work function $$\phi$$ of the metal:

$$h\nu = \dfrac{1}{2} m v^{2} + \phi$$

Here $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the light, $$m$$ is the mass of the electron, and $$v$$ is the speed with which the photo-electron emerges.

Because the wavelength $$\lambda$$ of the light is given, we first change it to frequency using the relation $$\nu = \dfrac{c}{\lambda}$$. Stating the values supplied in the question, we have

$$h = 6.626 \times 10^{-34}\ \text{J s}, \quad c = 3.0 \times 10^{8}\ \text{m s}^{-1}, \quad \lambda = 4000\ \text{\AA} = 4000 \times 10^{-10}\ \text{m} = 4.0 \times 10^{-7}\ \text{m}. $$

So the frequency is

$$ \nu = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^{8}}{4.0 \times 10^{-7}} = 0.75 \times 10^{15}\ \text{s}^{-1} = 7.5 \times 10^{14}\ \text{s}^{-1}. $$

The photon energy $$h\nu$$ is therefore

$$ h\nu = (6.626 \times 10^{-34})(7.5 \times 10^{14}) = 4.9695 \times 10^{-19}\ \text{J}. $$

Next we calculate the kinetic energy of the photo-electron. The electron’s mass is given as $$m = 9.0 \times 10^{-31}\ \text{kg}$$ and its speed as $$v = 6.0 \times 10^{5}\ \text{m s}^{-1}$$. Writing the classical kinetic-energy formula,

$$ \text{K.E.} = \dfrac{1}{2} m v^{2} = \dfrac{1}{2}(9.0 \times 10^{-31}) (6.0 \times 10^{5})^{2}. $$

We square the velocity first:

$$ (6.0 \times 10^{5})^{2} = 36 \times 10^{10} = 3.6 \times 10^{11}. $$

Substituting this value,

$$ \text{K.E.} = \dfrac{1}{2}(9.0 \times 10^{-31})(3.6 \times 10^{11}) = 4.5 \times 10^{-31} \times 3.6 \times 10^{11} = 16.2 \times 10^{-20}\ \text{J} = 1.62 \times 10^{-19}\ \text{J}. $$

Now we insert $$h\nu$$ and the kinetic energy into Einstein’s equation to isolate the work function $$\phi$$:

$$ \phi = h\nu - \text{K.E.} = 4.9695 \times 10^{-19}\ \text{J} - 1.62 \times 10^{-19}\ \text{J} = 3.3495 \times 10^{-19}\ \text{J}. $$

To express $$\phi$$ in electron-volts (eV), we use the definition $$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$$. Dividing by this factor gives

$$ \phi = \dfrac{3.3495 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} \approx 2.09\ \text{eV}. $$

Rounding to the usual one-decimal accuracy used in the options, we write

$$ \phi \approx 2.1\ \text{eV}. $$

Hence, the correct answer is Option D.

To decide in which atom the $$2s$$ orbital possesses the lowest (most negative) energy, we compare how strongly the nucleus attracts an electron in that same $$2s$$ orbital for each element.

First recall the general principle that governs one-electron energies in multi-electron atoms: the energy of a given subshell becomes more negative as the effective nuclear charge $$Z_{\text{eff}}$$ felt by the electron increases. The mathematical statement is that, after accounting for shielding, the approximate one-electron energy varies as

$$E_{n,\ell}\;\propto\;-\dfrac{\left(Z_{\text{eff}}\right)^{2}}{n^{2}},$$

where $$n$$ is the principal quantum number and $$\ell$$ the azimuthal quantum number. For all the atoms listed the quantum numbers of the orbital of interest are the same: $$n = 2$$ and $$\ell = 0$$ (because it is an $$s$$ orbital). Hence, the only factor that can change the energy is $$Z_{\text{eff}}$$.

Now $$Z_{\text{eff}}$$ increases with the actual nuclear charge $$Z$$ but is slightly reduced by shielding from the inner electrons. Even after shielding, a higher $$Z$$ still means a larger $$Z_{\text{eff}}$$ and therefore a more negative energy.

Let us list the atomic numbers:

$$\begin{aligned} \text{H}: &\; Z = 1,\\ \text{Li}: &\; Z = 3,\\ \text{Na}: &\; Z = 11,\\ \text{K}: &\; Z = 19. \end{aligned}$$

As we move from hydrogen to potassium, $$Z$$ clearly increases. Shielding does grow because more inner shells appear (for example, Na has the complete $$1s^{2}2s^{2}2p^{6}$$ core; K has even more inner electrons), yet the growth of nuclear charge is still greater than the added shielding. Consequently, the net $$Z_{\text{eff}}$$ experienced by a $$2s$$ electron rises in the order

$$\text{H} \lt \text{Li} \lt \text{Na} \lt \text{K}.$$

Because the energy is proportional to $$-Z_{\text{eff}}^{2}$$, a larger $$Z_{\text{eff}}$$ makes the energy more negative, i.e. lower. Thus the $$2s$$ orbital is most stabilised in potassium.

Hence, the correct answer is Option A.

We recall the general Rydberg formula for hydrogen:

$$\bar{v}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{n_2^{\,2}}\right),\qquad n_2>n_1,$$

where $$\bar{v}$$ is the wave-number, $$R$$ is the Rydberg constant and $$n_1$$ is fixed for a given spectral series while $$n_2$$ varies.

For any series, the maximum wave-number $$\bar{v}_{\text{max}}$$ occurs when the electron is removed to infinity, i.e. $$n_2\to\infty$$. In that limit $$1/n_2^{\,2}\to0$$, so

$$\bar{v}_{\text{max}}=R\left(\frac{1}{n_1^{\,2}}-0\right)=\frac{R}{n_1^{\,2}}.$$

The minimum wave-number $$\bar{v}_{\text{min}}$$ occurs for the first possible transition in that series, i.e. when $$n_2=n_1+1$$. Therefore

$$\bar{v}_{\text{min}}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{(n_1+1)^{2}}\right).$$

We define $$\Delta\bar{v}=\bar{v}_{\text{max}}-\bar{v}_{\text{min}}$$. Substituting the two expressions just written, we get

$$\Delta\bar{v}= \frac{R}{n_1^{\,2}}-R\left(\frac{1}{n_1^{\,2}}-\frac{1}{(n_1+1)^{2}}\right) = R\left[\frac{1}{n_1^{\,2}}-\frac{1}{n_1^{\,2}}+\frac{1}{(n_1+1)^{2}}\right] = \frac{R}{(n_1+1)^{2}}.$$

Thus in every series the difference $$\Delta\bar{v}$$ depends only on $$n_1+1$$:

$$\Delta\bar{v}=\frac{R}{(n_1+1)^{2}}.$$

Lyman series. Here $$n_1=1$$, so $$n_1+1=2$$. Therefore

$$\Delta\bar{v}_{\text{Lyman}}=\frac{R}{2^{2}}=\frac{R}{4}.$$

Balmer series. Here $$n_1=2$$, so $$n_1+1=3$$. Hence

$$\Delta\bar{v}_{\text{Balmer}}=\frac{R}{3^{2}}=\frac{R}{9}.$$

We now form the required ratio:

$$\frac{\Delta\bar{v}_{\text{Lyman}}}{\Delta\bar{v}_{\text{Balmer}}} =\frac{R/4}{R/9} =\frac{9}{4}.$$

Therefore the ratio is $$9:4$$.

Hence, the correct answer is Option D.

We are told that the therapeutic radiation used for heat treatment has a wavelength of about $$900\ \text{nm}$$. Our task is to look among the listed hydrogen‐atom spectral lines and find the one whose wavelength matches this value.

For the hydrogen atom, the wavenumber (that is, the reciprocal of the wavelength in centimetres) of any spectral line is given by the Rydberg formula, which we first state clearly:

$$\bar\nu = R_H\!\left(\dfrac1{n_1^{\,2}} - \dfrac1{n_2^{\,2}}\right)$$

Here

$$n_2 \gt n_1,\qquad R_H = 1 \times 10^{5}\ \text{cm}^{-1},$$

and

$$\bar\nu = \dfrac1{\lambda}\quad(\text{with }\lambda\text{ expressed in cm}).$$

The four options involve four different pairs of $$n_1$$ and $$n_2$$. We shall examine each one in turn, calculating its wavelength step by step, and then compare it with the required $$900\ \text{nm}$$.

Option A: Paschen, $$\infty \to 3$$

For a transition that starts from $$n_2 = \infty$$ and ends at $$n_1 = 3$$, the term $$1/n_2^{\,2}$$ becomes $$0$$. Substituting into the Rydberg formula we have

$$\bar\nu_A = R_H\!\left(\dfrac1{3^{2}} - 0\right) = 1 \times 10^{5}\ \text{cm}^{-1}\!\left(\dfrac1{9}\right) = \dfrac{1 \times 10^{5}}{9}\ \text{cm}^{-1}.$$

Carrying out the division,

$$\bar\nu_A = 1.1111 \times 10^{4}\ \text{cm}^{-1}.$$

Now the wavelength in centimetres is the reciprocal:

$$\lambda_A = \dfrac1{\bar\nu_A} = \dfrac1{1.1111 \times 10^{4}}\ \text{cm} = 9.000 \times 10^{-5}\ \text{cm}.$$

To convert this to nanometres we note that $$1\ \text{cm} = 10^{7}\ \text{nm}$$, so

$$\lambda_A = 9.000 \times 10^{-5}\ \text{cm}\ \times 10^{7}\ \dfrac{\text{nm}}{\text{cm}} = 9.000 \times 10^{2}\ \text{nm} = 900\ \text{nm}.$$

This is exactly the required $$900\ \text{nm}$$. Nevertheless, for completeness, we shall still evaluate the other three options.

Option B: Paschen, $$5 \to 3$$

Here $$n_2 = 5$$ and $$n_1 = 3$$. Using the Rydberg formula:

$$\bar\nu_B = R_H\!\left(\dfrac1{3^{2}} - \dfrac1{5^{2}}\right) = 1 \times 10^{5}\ \text{cm}^{-1} \left(\dfrac1{9} - \dfrac1{25}\right).$$

Calculating the bracket first,

$$\dfrac1{9} - \dfrac1{25} = \dfrac{25 - 9}{225} = \dfrac{16}{225}.$$

Hence

$$\bar\nu_B = 1 \times 10^{5}\ \text{cm}^{-1}\ \times \dfrac{16}{225} = \dfrac{1.6 \times 10^{6}}{225}\ \text{cm}^{-1} = 7.111 \times 10^{3}\ \text{cm}^{-1}.$$

The wavelength is therefore

$$\lambda_B = \dfrac1{7.111 \times 10^{3}}\ \text{cm} = 1.406 \times 10^{-4}\ \text{cm} = 1.406 \times 10^{-4}\ \text{cm}\ \times 10^{7}\ \dfrac{\text{nm}}{\text{cm}} = 1.406 \times 10^{3}\ \text{nm} = 1406\ \text{nm}.$$

This is far larger than $$900\ \text{nm}$$, hence Option B cannot be the required line.

Option C: Balmer, $$\infty \to 2$$

Here $$n_2 = \infty$$ and $$n_1 = 2$$, so

$$\bar\nu_C = R_H\!\left(\dfrac1{2^{2}} - 0\right) = 1 \times 10^{5}\ \text{cm}^{-1}\!\left(\dfrac1{4}\right) = 2.5 \times 10^{4}\ \text{cm}^{-1}.$$

The corresponding wavelength is

$$\lambda_C = \dfrac1{2.5 \times 10^{4}}\ \text{cm} = 4.0 \times 10^{-5}\ \text{cm} = 4.0 \times 10^{-5}\ \text{cm}\ \times 10^{7}\ \dfrac{\text{nm}}{\text{cm}} = 4.0 \times 10^{2}\ \text{nm} = 400\ \text{nm}.$$

This is in the visible region and again is not close to $$900\ \text{nm}$$.

Option D: Lyman, $$\infty \to 1$$

With $$n_2 = \infty$$ and $$n_1 = 1$$ we obtain

$$\bar\nu_D = R_H\!\left(\dfrac1{1^{2}} - 0\right) = 1 \times 10^{5}\ \text{cm}^{-1}.$$

Thus

$$\lambda_D = \dfrac1{1 \times 10^{5}}\ \text{cm} = 1.0 \times 10^{-5}\ \text{cm} = 1.0 \times 10^{-5}\ \text{cm}\ \times 10^{7}\ \dfrac{\text{nm}}{\text{cm}} = 1.0 \times 10^{2}\ \text{nm} = 100\ \text{nm}.$$

This is in the ultraviolet region and again does not match $$900\ \text{nm}$$.

Only Option A, the Paschen line with transition $$\infty \to 3$$, yields a wavelength of exactly $$900\ \text{nm}$$, which is the requirement for heat treatment of muscular pain.

Hence, the correct answer is Option A.

We begin by recalling the two standard Bohr‐model relations for a hydrogenic (one-electron) atom with nuclear charge $$Z$$.

First, the radius of the $$n^{\text{th}}$$ orbit is given by the well-known formula

$$r_n \;=\; \dfrac{n^{2}a_0}{Z},$$

where $$a_0$$ is the Bohr radius for the hydrogen atom.

Second, de Broglie’s standing-wave condition states that an integral number of wavelengths must fit exactly into the circumference of the circular orbit. Mathematically,

$$2\pi r_n \;=\; n\lambda,$$

where $$\lambda$$ is the de Broglie wavelength of the electron in that orbit.

We are told that this wavelength has the numerical value

$$\lambda \;=\; 1.5\pi a_0 \;=\; \dfrac{3}{2}\pi a_0.$$

Our task is to find the ratio $$\dfrac{n}{Z}$$.

From the standing-wave condition we first isolate $$\lambda$$:

$$\lambda \;=\; \dfrac{2\pi r_n}{n}.$$

Now we substitute the explicit expression for $$r_n$$ from the radius formula:

$$\lambda \;=\; \dfrac{2\pi}{n}\,\Bigl(\dfrac{n^{2}a_0}{Z}\Bigr) \;=\; 2\pi\,\dfrac{n a_0}{Z}.$$

This theoretical value of $$\lambda$$ must equal the given value. Hence we set

$$2\pi\,\dfrac{n a_0}{Z} \;=\; \dfrac{3}{2}\pi a_0.$$

We can now simplify step by step. First cancel the common factors $$\pi$$ and $$a_0$$ on both sides:

$$2\,\dfrac{n}{Z} \;=\; \dfrac{3}{2}.$$

Multiplying both sides by $$Z$$ and, simultaneously, by $$2$$ to clear the fraction, we have

$$4n \;=\; 3Z.$$

Finally, dividing both sides by $$4Z$$ gives the desired ratio:

$$\dfrac{n}{Z} \;=\; \dfrac{3}{4} \;=\; 0.75.$$

Hence, the correct answer is Option 4.

We start with Einstein’s photoelectric equation, which relates the energy of the incident photon to the work function and the kinetic energy of the emitted photo-electron:

$$h\nu \;=\; h\nu_0 \;+\; \dfrac12\,m\,v^2$$

Here $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident radiation, $$\nu_0$$ is the threshold frequency of the metal, $$m$$ is the mass of the electron and $$v$$ is the speed of the ejected photo-electron.

First we isolate the kinetic energy term:

$$\dfrac12\,m\,v^2 \;=\; h\nu \;-\; h\nu_0$$

Now we solve for $$v^2$$ by multiplying both sides by $$\dfrac{2}{m}$$:

$$v^2 \;=\; \dfrac{2}{m}\,\bigl(h\nu \;-\; h\nu_0\bigr)$$

Taking the square root on both sides gives the speed $$v$$:

$$v \;=\; \sqrt{\dfrac{2}{m}\,\bigl(h\nu \;-\; h\nu_0\bigr)}$$

Next, we recall the de Broglie wavelength formula, which connects a particle’s momentum to its wavelength:

$$\lambda \;=\; \dfrac{h}{p}, \quad\text{where } p = m v$$

Substituting $$p = m v$$ into the formula gives:

$$\lambda \;=\; \dfrac{h}{m v}$$

Now we substitute the expression we have just obtained for $$v$$:

$$\lambda \;=\; \dfrac{h}{m}\;\dfrac{1}{\sqrt{\dfrac{2}{m}\,\bigl(h\nu \;-\; h\nu_0\bigr)}}$$

We rewrite the denominator clearly, keeping the square root intact:

$$\lambda \;=\; \dfrac{h}{m}\;\dfrac{1}{\sqrt{\dfrac{2h}{m}\,(\nu - \nu_0)}}$$

Since $$\dfrac{h}{m}$$ is a constant for the electron, we can collect all the constants together. Inside the square root, both $$2h$$ and $$m$$ are constants as well. Hence, after grouping constants, the wavelength can be seen to have the form

$$\lambda \;=\; \dfrac{\text{constant}}{\sqrt{\nu - \nu_0}}$$

Therefore, the de Broglie wavelength is inversely proportional to the square root of $$\nu - \nu_0$$:

$$\boxed{\lambda \;\propto\; \dfrac{1}{(\nu - \nu_0)^{1/2}}}$$

Among the given choices, this dependence corresponds to Option D.

Hence, the correct answer is Option D.

Because the magnitude (absolute value) of the wavefunction is at its maximum at these two points. Since the probability of finding an electron depends on [Psi(x)]^2, the negative sign at point c disappears upon squaring, making the probability density at both peak positions equally high.

• For an s-orbital, the radial wave function has a maximum, non-zero value at the nucleus.
• For p, d, and f orbitals, the wave function value at the nucleus is always zero because of the angular nodes 
• Looking at the graph, the curve starts at a very high value on the y-axis when r = 0, meaning it must be an s-orbital.

2. Finding the Number of Radial Nodes

• A radial node is a region in space where the probability of finding an electron drops exactly to zero..
• In the given graph, the curve touches the x-axis exactly once before rising again and gradually tapering off to infinity.
• Therefore, the number of radial nodes = 1.

3. Determining the Principal Quantum Number 

The formula to calculate the number of radial nodes for any given orbital is:

Radial Nodes = n - l - 1

Where:

• n = Principal quantum number
• l = Azimuthal quantum number (for an s-orbital, $l = 0$)

Substituting our known values into the equation:

1 = n - 0 - 1 = 2

We are dealing with hydrogen-like (single-electron) species. For any such species, the energy of an electron in the $$n^{\text{th}}$$ stationary state is given by the well-known expression

$$E_n = -13.6\;{\rm eV}\;\frac{Z^2}{n^2},$$

where $$Z$$ is the atomic number of the nucleus and $$n$$ is the principal quantum number. The factor $$-13.6\;{\rm eV}$$ is the ground-state energy of the hydrogen atom ($$Z = 1,\; n = 1$$) that has been provided in the question.

For the helium ion He$$^+$$ we have only one electron but the nuclear charge is double that of hydrogen, so

$$Z = 2.$$

Next we interpret the phrase “second excited state.” The counting of states goes as follows:

• Ground state → $$n = 1$$
• First excited state → $$n = 2$$
• Second excited state → $$n = 3$$

Hence we must use $$n = 3$$ for our calculation.

Substituting $$Z = 2$$ and $$n = 3$$ into the energy formula gives

$$E_3 = -13.6\;{\rm eV}\;\frac{(2)^2}{(3)^2}.$$

We now carry out the algebra step by step:

$$E_3 = -13.6\;{\rm eV}\;\frac{4}{9}.$$

Multiplying numerator and denominator we get

$$E_3 = -13.6 \times \frac{4}{9}\;{\rm eV}.$$

First compute the product in the numerator:

$$13.6 \times 4 = 54.4.$$

Therefore,

$$E_3 = -\frac{54.4}{9}\;{\rm eV}.$$

Finally, dividing $$54.4$$ by $$9$$ yields

$$E_3 = -6.044\ldots\;{\rm eV} \approx -6.04\;{\rm eV}.$$

Looking at the given options, this value matches Option B.

Hence, the correct answer is Option B.

We have to compare the shortest (limit) wavelengths of two spectral series in hydrogen. For any transition in hydrogen, the Rydberg formula gives

$$\frac{1}{\lambda}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{n_2^{\,2}}\right),$$

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower (fixed) orbit for the series, and $$n_2$$ is the upper orbit with $$n_2\gt n_1$$.

The “shortest wavelength” (i.e. the series limit) is obtained by letting the electron fall from $$n_2\to\infty$$. In that limit the term $$1/n_2^{\,2}$$ vanishes, so

$$\frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_1^{\,2}}-0\right)=\frac{R}{n_1^{\,2}}.$$

Taking the reciprocal we obtain an explicit expression for the shortest wavelength in a series:

$$\lambda_{\min}= \frac{n_1^{\,2}}{R}.$$

Now we are told that the ratio of the shortest wavelengths of two different series is about $$9$$. Let the two lower quantum numbers be $$n_{1a}$$ and $$n_{1b}$$. Then

$$\frac{\lambda_{\min,a}}{\lambda_{\min,b}} =\frac{\dfrac{n_{1a}^{\,2}}{R}}{\dfrac{n_{1b}^{\,2}}{R}} =\frac{n_{1a}^{\,2}}{n_{1b}^{\,2}} =9.$$

Cancelling the common factor $$R$$ and taking the square root on both sides, we find

$$\frac{n_{1a}}{n_{1b}} = \sqrt{9}=3.$$

Hence the two series must have lower quantum numbers whose ratio is exactly $$3:1$$.

The known values of $$n_1$$ for the named hydrogen series are:

$$\text{Lyman: }n_1=1,\quad \text{Balmer: }n_1=2,\quad \text{Paschen: }n_1=3,\quad \text{Brackett: }n_1=4,\quad \text{Pfund: }n_1=5.$$

To satisfy $$n_{1a}/n_{1b}=3$$, the only possible pair from this list is $$n_1=3$$ (Paschen) and $$n_1=1$$ (Lyman). Indeed,

$$\frac{\lambda_{\min,\text{Paschen}}}{\lambda_{\min,\text{Lyman}}} =\frac{3^{2}/R}{1^{2}/R}=9,$$

exactly as required.

Therefore, the two spectral series are Lyman and Paschen.

Hence, the correct answer is Option C.

First, recall that in quantum mechanics four quantum numbers -$$n,\,l,\,m_l,\,m_s$$- describe an electron in an atom. Out of these, the principal quantum number $$n$$ mainly controls the general size (average distance from the nucleus), while the azimuthal or orbital angular-momentum quantum number $$l$$ decides the shape and the angular momentum of the orbital.

For the magnitude of orbital angular momentum, wave mechanics gives the exact formula

$$L=\sqrt{l(l+1)}\,\hbar,$$

where $$\hbar=\dfrac{h}{2\pi}$$ is the reduced Planck constant. We now examine each given statement in the light of this accepted theory.

Examining statement (A): “An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.”

When $$l$$ is larger (for the same $$n$$) the centrifugal term in the Schrödinger radial equation, $$\dfrac{l(l+1)\hbar^{2}}{2mr^{2}},$$ becomes larger. This term pushes the electron probability density outward; equivalently, the most probable radius $$r_{mp}$$ increases with $$l$$. Therefore electrons in p, d … orbitals (larger $$l$$) are on the average farther from the nucleus than electrons in an s orbital ($$l=0$$). So statement (A) is true.

Examining statement (B): “For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.”

As just reasoned, when $$l$$ increases, $$r_{mp}$$ also increases; there is no inverse proportionality such as $$r\propto\dfrac1l$$. Instead, the trend is the opposite (direct, not inverse). Hence statement (B) is false.

Examining statement (C): “According to wave mechanics, the ground-state angular momentum is equal to $$\dfrac{h}{2\pi}$$.”

In the ground state of hydrogen (and hydrogen-like atoms) we have $$n=1$$ and, by quantum rules, $$l=0$$. Substituting $$l=0$$ in the angular-momentum formula gives

$$L=\sqrt{0(0+1)}\,\hbar=0.$$

Thus wave mechanics predicts zero orbital angular momentum for the ground state, not $$\hbar=\dfrac{h}{2\pi}$$. The quoted value corresponds to the Bohr model, not to the Schrödinger description. Therefore statement (C) is false.

Examining statement (D): “The plot of $$\psi$$ vs $$r$$ for various azimuthal quantum numbers shows peak shifting towards higher $$r$$ value.”

For a fixed principal quantum number $$n$$, increasing $$l$$ moves the maximum of the radial wave function $$R_{nl}(r)$$ outward, as discussed for (A). Hence the peak of $$\psi$$ (or of the radial probability distribution) indeed shifts toward a larger $$r$$. Statement (D) is true.

Combining our results:

$$\text{(A) = True},$$ $$\text{(B) = False},$$ $$\text{(C) = False},$$ $$\text{(D) = True}.$$

The only option that lists statements (A) and (D) together is Option 4.

Hence, the correct answer is Option D.

We need to identify which statement about an electron in the 1s orbital of a hydrogen atom is incorrect.

Option A: "The total energy of the electron is maximum when it is at a distance $$a_0$$ from the nucleus."

For the hydrogen atom, the most probable distance of the electron from the nucleus in the 1s orbital is the Bohr radius $$a_0$$. At this distance, the electron is in its ground state (n = 1), which is the state of minimum total energy (most negative energy), not maximum. The total energy at this distance is $$E_1 = -13.6$$ eV, which is the lowest possible energy. Therefore, this statement is incorrect.

Option B: "The electron can be found at a distance $$2a_0$$ from the nucleus."

The probability density function $$|\psi_{1s}|^2$$ for the 1s orbital is non-zero for all distances $$r > 0$$. Although the probability decreases exponentially with distance, there is still a non-zero probability of finding the electron at $$r = 2a_0$$. This statement is correct.

Option C: "The probability density of finding the electron is maximum at the nucleus."

For the 1s orbital, $$\psi_{1s} = \frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0}$$. The probability density $$|\psi_{1s}|^2$$ is maximum at $$r = 0$$ (at the nucleus), since $$e^{-r/a_0}$$ is maximum when $$r = 0$$. Note: the radial probability distribution $$4\pi r^2|\psi|^2$$ peaks at $$r = a_0$$, but the probability density itself is maximum at the nucleus. This statement is correct.

Option D: "The magnitude of the potential energy is double that of its kinetic energy on an average."

By the virial theorem for a Coulombic potential, $$\langle V \rangle = 2\langle E \rangle$$ and $$\langle T \rangle = -\langle E \rangle$$. Therefore $$|\langle V \rangle| = 2|\langle T \rangle|$$. The magnitude of the average potential energy is indeed double the average kinetic energy. This statement is correct.

The incorrect statement is Option A.

For a photon of wavelength $$\lambda$$ incident on a metal surface, the energy carried by that photon is obtained from the Planck-Einstein relation

$$E_{\text{photon}}=\dfrac{hc}{\lambda}.$$

According to the photoelectric equation, this energy is used to do two things: overcome the work function $$\phi$$ of the metal and provide kinetic energy $$K$$ to the ejected electron:

$$\dfrac{hc}{\lambda}=K+\phi.$$

In the present question we are told that the kinetic energy of the electron is “very high in comparison to the work function.” Hence $$K\gg\phi$$, so we may safely neglect $$\phi$$ in comparison with $$K$$. The entire photon energy is therefore effectively converted into kinetic energy:

$$K\;\approx\;\dfrac{hc}{\lambda}.$$

The kinetic energy of a non-relativistic particle is related to its linear momentum $$p$$ by the classical formula

$$K=\dfrac{p^{2}}{2m},$$

where $$m$$ is the electron mass. Equating this expression for $$K$$ with the photon energy we have

$$\dfrac{p^{2}}{2m}=\dfrac{hc}{\lambda}.$$ Multiplying both sides by $$2m$$ gives

$$p^{2}=2m\,\dfrac{hc}{\lambda}.$$

Now, imagine we want the ejected electron to have a momentum equal to $$1.5p$$. Let the new wavelength required to achieve that larger momentum be $$\lambda'$$. Repeating the same energy-momentum relationship for this new situation, we write

$$\dfrac{(1.5p)^{2}}{2m}=\dfrac{hc}{\lambda'}.$$

Expanding the square on the left side,

$$(1.5p)^{2}=2.25\,p^{2},$$

so the equation becomes

$$\dfrac{2.25\,p^{2}}{2m}=\dfrac{hc}{\lambda'}.$$

But from our earlier work we already know that $$\dfrac{p^{2}}{2m}=\dfrac{hc}{\lambda}$$. Substituting this into the new equation yields

$$2.25\;\dfrac{hc}{\lambda}=\dfrac{hc}{\lambda'}.$$

We can cancel the common factor $$hc$$ on both sides:

$$2.25\;\dfrac{1}{\lambda}=\dfrac{1}{\lambda'}.$$

Taking reciprocals of both sides gives

$$\lambda'=\dfrac{\lambda}{2.25}.$$

Recognising that $$2.25=2\!\!.25=\dfrac{9}{4},$$ we rewrite the fraction clearly:

$$\lambda'=\dfrac{4}{9}\,\lambda.$$

Hence, the correct answer is Option A.

We recall that species are said to be isoelectronic when each of them possesses exactly the same total number of electrons.

For any neutral atom the number of electrons equals its atomic number $$Z$$. When the species is an ion, we adjust this count as follows:

For a cation having charge $$+n$$, the ion has lost $$n$$ electrons, so

$$\text{electrons in cation}=Z-n.$$

For an anion carrying charge $$-n$$, the ion has gained $$n$$ electrons, therefore

$$\text{electrons in anion}=Z+n.$$

Now we write the atomic numbers of all the elements that appear in the four options:

$$\begin{aligned} Z(\text{Li})&=3,\\ Z(\text{N}) &=7,\\ Z(\text{O}) &=8,\\ Z(\text{F}) &=9,\\ Z(\text{Na})&=11,\\ Z(\text{Mg})&=12. \end{aligned}$$

Using the above relationships we calculate the electron count for every ion mentioned in the options:

$$\begin{aligned} \text{Li}^+ &: 3-1=2,\\[2pt] \text{N}^{3-}&: 7+3=10,\\[2pt] \text{O}^{2-}&: 8+2=10,\\[2pt] \text{F}^- &: 9+1=10,\\[2pt] \text{Na}^+ &: 11-1=10,\\[2pt] \text{Mg}^{2+}&: 12-2=10. \end{aligned}$$

With these values in hand we examine each option one by one.

Option A: $$\text{F}^- (10),\; \text{Li}^+ (2),\; \text{Na}^+ (10),\; \text{Mg}^{2+} (10)$$ — the value $$2$$ for $$\text{Li}^+$$ breaks the equality, so the set is not isoelectronic.

Option B: $$\text{N}^{3-} (10),\; \text{Li}^+ (2),\; \text{Mg}^{2+} (10),\; \text{O}^{2-} (10)$$ — again the $$2$$ electrons of $$\text{Li}^+$$ cause a mismatch.

Option C: $$\text{N}^{3-} (10),\; \text{O}^{2-} (10),\; \text{F}^- (10),\; \text{Na}^+ (10)$$ — every ion here has exactly $$10$$ electrons, so all four are indeed isoelectronic.

Option D: $$\text{Li}^+ (2),\; \text{Na}^+ (10),\; \text{O}^{2-} (10),\; \text{F}^- (10)$$ — the presence of $$\text{Li}^+$$ with only $$2$$ electrons destroys the uniformity.

Clearly, only Option C satisfies the requirement that each species in the list contain the same number of electrons.

Hence, the correct answer is Option C.

We begin by recalling the basic definition of an isotope. Two or more atoms are called isotopes of the same element when they possess the same atomic number $$Z$$ (same number of protons, thus belonging to the same element) but different mass numbers $$A$$ because their nuclei contain different numbers of neutrons.

Hydrogen has atomic number $$Z = 1$$, meaning every hydrogen nucleus must contain exactly one proton. However, the number of neutrons present in the nucleus can vary, and this leads to three distinct nuclides (nuclear species):

1. We have the nuclide with one proton and zero neutrons. Its mass number is

$$A = Z + \text{(number of neutrons)} = 1 + 0 = 1.$$

This isotope is written as $$^1_1\text{H}$$ and is traditionally called protium.

2. Next, we consider the nuclide that again has one proton but now possesses one neutron. Its mass number is

$$A = 1 + 1 = 2.$$

This isotope is written as $$^2_1\text{H}$$ and is called deuterium.

3. Finally, there is the nuclide with one proton and two neutrons. Its mass number becomes

$$A = 1 + 2 = 3.$$

This isotope is written as $$^3_1\text{H}$$ and is called tritium.

Thus, hydrogen naturally exists in three isotopic forms: protium, deuterium and tritium. No other combinations of neutron numbers are stable or commonly listed for hydrogen in introductory chemistry.

Looking at the options, only Option A lists all three—protium, deuterium and tritium—together, matching exactly with the set we have just derived.

Hence, the correct answer is Option A.

First, we recall the definition of isotopes: isotopes are nuclides that have the same atomic number $$Z$$ (same number of protons) but different mass numbers $$A$$ (different total number of protons and neutrons). In nuclear notation an isotope is written as $$^{A}_{Z}\text{X}$$, where $$\text{X}$$ is the chemical symbol of the element.

Hydrogen has atomic number $$Z = 1$$, so all of its isotopes must contain exactly one proton. The known naturally occurring isotopes of hydrogen are:

1. $$^{1}_{1}\text{H}$$, called protium. 2. $$^{2}_{1}\text{H}$$, called deuterium and sometimes written as $$\text{D}$$. 3. $$^{3}_{1}\text{H}$$, called tritium and sometimes written as $$\text{T}$$.

Now we count the total number of distinct isotopes listed above. We have clearly enumerated three different mass numbers: $$A = 1, 2, 3$$. So the total number of hydrogen isotopes is

$$\text{Total isotopes} = 3.$$

Next, we determine which of these isotopes are radioactive. A radioactive isotope is one whose nucleus is unstable and undergoes spontaneous decay, emitting radiation. We know from experimental nuclear data that

• Protium $$\left(^{1}_{1}\text{H}\right)$$ is perfectly stable. • Deuterium $$\left(^{2}_{1}\text{H}\right)$$ is also stable. • Tritium $$\left(^{3}_{1}\text{H}\right)$$ is unstable (radioactive) with a half-life of about $$12.3$$ years, undergoing $$\beta^{-}$$ decay to form helium-3.

Therefore, the number of radioactive isotopes among hydrogen’s three isotopes is

$$\text{Radioactive isotopes} = 1.$$

So we arrive at the ordered pair $$\bigl($$ total number of isotopes $$,\;$$ number of radioactive isotopes $$\bigr) = (3,\,1).$$

Looking at the options given in the question, Option C states “3 and 1,” which exactly matches our derived pair.

Hence, the correct answer is Option C.

We begin by recalling an important scientific fact: the ozone present in the upper stratosphere is most effective in absorbing the Sun’s ultraviolet (UV) radiation. Specifically, ozone molecules absorb radiation in the UV-B and part of the UV-C regions. These regions correspond to wavelengths shorter than the visible spectrum and fall roughly between $$200 \ \text{nm}$$ and $$315 \ \text{nm}.$$

Now, let us compare this essential wavelength range with the intervals given in the four options. We have:

Option A offers $$200 - 315 \ \text{nm},$$ which matches exactly with the range of UV radiation that the ozone layer blocks.

Option B gives $$600 - 750 \ \text{nm},$$ which lies in the red to near-infrared part of the visible spectrum and is therefore not the harmful UV range absorbed by ozone.

Option C lists $$400 - 550 \ \text{nm},$$ which corresponds to the violet-green visible light band, again not primarily stopped by ozone.

Option D proposes $$0.8 - 1.5 \ \text{nm},$$ which is in the X-ray region. Ozone does not principally target such high-energy, extremely short wavelengths; moreover, X-rays are mostly absorbed by the denser parts of Earth’s atmosphere, not by the ozone layer alone.

From this direct comparison, only Option A aligns perfectly with the scientifically established ozone-absorption window of $$200 - 315 \ \text{nm}.$$

Hence, the correct answer is Option A.

For photoelectric emission, we always begin with Einstein’s photoelectric equation, which states:

$$h\nu \;=\; \phi \;+\; K_{\text{max}}$$

Here $$h\nu$$ is the energy of the incident photon, $$\phi$$ is the work-function of the metal, and $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electron.

First we calculate the photon energy. The wavelength of the radiation is given as $$\lambda = 250\ \text{nm} = 250 \times 10^{-9}\ \text{m}.$$ The frequency $$\nu$$ is related to the wavelength by $$\nu = \dfrac{c}{\lambda},$$ so the photon energy is

$$E_{\text{photon}} = h\nu = \dfrac{hc}{\lambda}.$$

Substituting the values $$h = 6.626 \times 10^{-34}\ \text{J\,s},\; c = 3.00 \times 10^{8}\ \text{m/s},\; \lambda = 250 \times 10^{-9}\ \text{m},$$ we obtain

$$E_{\text{photon}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{250 \times 10^{-9}}\ \text{J}.$$

Now we multiply the two numbers in the numerator:

$$6.626 \times 3.00 = 19.878,$$ so

$$E_{\text{photon}} = \dfrac{19.878 \times 10^{-26}}{250 \times 10^{-9}}\ \text{J}.$$

Express the denominator in scientific notation: $$250 \times 10^{-9} = 2.5 \times 10^{-7}.$$ Hence

$$E_{\text{photon}} = \dfrac{19.878 \times 10^{-26}}{2.5 \times 10^{-7}}\ \text{J} = \left(\dfrac{19.878}{2.5}\right) \times 10^{-19}\ \text{J} = 7.951 \times 10^{-19}\ \text{J}.$$

To convert this energy to electron-volts we use $$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.$$ Thus

$$E_{\text{photon}} = \dfrac{7.951 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} \approx 4.96\ \text{eV}.$$

Next we compute the maximum kinetic energy of the emitted photoelectrons. It is stopped by a potential of $$V_s = 0.5\ \text{V},$$ and the stopping potential is related to kinetic energy by the formula

$$K_{\text{max}} = eV_s,$$

where $$e = 1.602 \times 10^{-19}\ \text{C}$$ is the elementary charge. Therefore

$$K_{\text{max}} = (1.602 \times 10^{-19}\ \text{C})(0.5\ \text{V}) = 0.801 \times 10^{-19}\ \text{J}.$$

Converting this to electron-volts as before:

$$K_{\text{max}} = \dfrac{0.801 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} = 0.500\ \text{eV}.$$

Now we apply Einstein’s equation in electron-volts:

$$\phi = E_{\text{photon}} - K_{\text{max}} = 4.96\ \text{eV} - 0.50\ \text{eV} = 4.46\ \text{eV}.$$

On rounding to the usual significant figures, the work function is $$\phi \approx 4.5\ \text{eV}.$$

Hence, the correct answer is Option C.

We begin with the de-Broglie relation, which states that the wavelength associated with a moving particle is given by

$$\lambda=\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p=mv$$ is the linear momentum of the particle of mass $$m$$ moving with speed $$v$$.

For an electron revolving around the nucleus in a Bohr orbit we additionally have Bohr’s angular-momentum quantisation condition, stated as

$$mvr=\frac{nh}{2\pi},$$

where $$r$$ is the radius of the orbit and $$n$$ is the principal quantum number (an integer).

Rearranging this Bohr condition to express the momentum $$mv$$ gives

$$mv=\frac{nh}{2\pi r}.$$

We now substitute this value of $$mv$$ into the de-Broglie formula. Doing every algebraic step explicitly we get

$$\lambda=\frac{h}{mv} =\frac{h}{\displaystyle\frac{nh}{2\pi r}} =h \times \frac{2\pi r}{nh} =\frac{2\pi r}{n}.$$

We are interested in the first Bohr orbit, so we put $$n=1$$. This immediately yields

$$\lambda=\frac{2\pi r}{1}=2\pi r_1,$$

where $$r_1$$ is the radius of the first Bohr orbit.

The well-known expression for the Bohr radius is

$$r_1=a_0=0.529\ \text{Å},$$

with $$1\ \text{Å}=10^{-10}\ \text{m}$$. No unit conversion is necessary because the options are already expressed in ångströms.

Substituting $$r_1=0.529\ \text{Å}$$ into the wavelength expression we obtain

$$\lambda = 2\pi \times 0.529\ \text{Å}.$$

This result matches exactly with Option B.

Hence, the correct answer is Option 2.

We begin by recalling the precise scientific facts that correspond to each option and then comparing them with the sentences given in the question. The statement that does not agree with accepted theory will be the false one.

Option A speaks about the splitting of spectral lines when an external electric field is applied. The well-known name for this phenomenon is indeed the Stark effect. Hence the sentence in Option A is in complete accordance with established spectroscopy and is therefore true.

Option B concerns the radiation emitted by a black body as its temperature changes. According to Wien’s displacement law we have

$$\lambda_{\text{max}}\,T = b,$$

where $$\lambda_{\text{max}}$$ is the wavelength at which the emission is most intense, $$T$$ is the absolute temperature and $$b$$ is a universal constant. Rearranging gives

$$\lambda_{\text{max}} = \frac{b}{T}.$$

So, as $$T$$ increases, $$\lambda_{\text{max}}$$ decreases; put in words, the peak of the spectrum shifts from longer to shorter wavelengths, i.e. toward the violet end of the spectrum, or equivalently to higher frequencies. The sentence in Option B says that the “frequency of emitted radiation … goes from a higher wavelength to lower wavelength as the temperature increases.” Although the wording mixes “frequency” and “wavelength” in one clause, its physical content is that the radiation peak moves from longer to shorter wavelengths as temperature rises, which matches Wien’s law. Therefore Option B is also true.

Option C asserts that a photon possesses both momentum and wavelength. The de Broglie relation for a photon is

$$p = \frac{h}{\lambda},$$

where $$p$$ is the momentum, $$h$$ is Planck’s constant and $$\lambda$$ is the wavelength. This clearly endows the photon with a definite momentum whenever its wavelength is finite. Hence Option C is true as well.

Option D states that the Rydberg constant has the unit of energy. To check this, recall the Rydberg formula for the wavenumbers of hydrogen spectral lines:

$$\bar{\nu} = \frac{1}{\lambda} = R\!\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right),$$

where $$\bar{\nu}$$ is the wavenumber, $$\lambda$$ is the wavelength, $$n_1$$ and $$n_2$$ are positive integers with $$n_2 > n_1$$, and $$R$$ is the Rydberg constant. In this equation the term inside the parentheses is dimensionless because it involves only the reciprocals of squared integers. Consequently, the dimensions of $$R$$ must be exactly the same as those of $$\bar{\nu}$$, namely the reciprocal of length:

$$[R] = \frac{1}{\text{length}} = \text{m}^{-1}.$$

A quantity that carries the dimension $$\text{m}^{-1}$$ cannot simultaneously possess the dimension of energy $$\text{(kg·m}^2\text{/s}^2)$$. Therefore the assertion in Option D that “Rydberg constant has unit of energy” is wrong.

Since Options A, B and C are all true and Option D is false, the false statement demanded by the question is Option D.

Hence, the correct answer is Option D.

For hydrogen-like atoms we use the Rydberg formula, stated first:

$$\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$

Here $$R$$ is the Rydberg constant, $$Z$$ is the atomic number, $$n_{1}$$ is the lower (final) orbit and $$n_{2}$$ is the higher (initial) orbit.

We are told that the shortest wavelength in the Lyman series of the hydrogen atom has the value $$A$$. In the Lyman series the electron always falls to $$n_{1}=1$$. The shortest wavelength occurs when the initial level is infinitely high, that is $$n_{2}\rightarrow\infty$$. Substituting $$Z=1,\,n_{1}=1,\,n_{2}\rightarrow\infty$$ we have

$$\frac{1}{\lambda_{\text{min(Lyman, H)}}}=R\cdot1^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right) =R\left(1-0\right)=R.$$

Thus

$$\lambda_{\text{min(Lyman, H)}}=\frac{1}{R}=A.$$

So we immediately obtain

$$R=\frac{1}{A}.$$

Now we turn to the Paschen series of the singly ionised helium atom $$\text{He}^{+}$$. This ion is hydrogen-like with nuclear charge $$Z=2$$. In the Paschen series the electron finally falls to $$n_{1}=3$$. The longest wavelength in any series corresponds to the smallest energy (and hence smallest frequency) transition within that series. Therefore, for Paschen, the initial level must be the next higher orbit, i.e. $$n_{2}=4$$. Using the same Rydberg formula with $$Z=2,\,n_{1}=3,\,n_{2}=4$$ we have

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=R\,(2)^{2}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right).$$

Now we evaluate the brackets first:

$$\frac{1}{3^{2}}-\frac{1}{4^{2}}=\frac{1}{9}-\frac{1}{16} =\frac{16-9}{144}=\frac{7}{144}.$$

Multiplying by $$Z^{2}=4$$:

$$4\left(\frac{7}{144}\right)=\frac{28}{144}=\frac{7}{36}.$$

Hence

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=R\cdot\frac{7}{36}.$$

Substituting $$R=\dfrac{1}{A}$$ obtained earlier, we get

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=\frac{1}{A}\cdot\frac{7}{36} =\frac{7}{36A}.$$

Taking the reciprocal to find the wavelength:

$$\lambda_{\text{max(Paschen, He}^{+})}=\frac{36A}{7}.$$

Hence, the correct answer is Option B.

For any hydrogenic atom, Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n = n^2 a_0,$$

where $$a_0$$ is the Bohr radius. Numerically,

$$a_0 = 0.529\;\text{Å} = 0.529 \times 10^{-10}\;\text{m} = 52.9\;\text{pm}.$$

We are told that after the electron finishes its transition, it is found in an orbit whose radius is

$$r_n = 211.6\;\text{pm}.$$

Using the formula $$r_n = n^2 a_0$$, we can solve for $$n$$:

$$n^2 = \dfrac{r_n}{a_0}.$$

Substituting the numerical values, we have

$$n^2 = \dfrac{211.6\;\text{pm}}{52.9\;\text{pm}}.$$

Carrying out the division,

$$n^2 = \dfrac{211.6}{52.9} = 4.00.$$

So,

$$n = \sqrt{4.00} = 2.$$

This means the electron finally lands in the second orbit (the $$n = 2$$ level) after completing its transition.

In the hydrogen spectrum, all transitions that terminate at $$n = 2$$ constitute the Balmer series. Therefore, the radiation observed from this transition belongs to the Balmer region.

Hence, the correct answer is Option B.

We begin with Bohr’s postulate for the stationary orbits of the hydrogen atom. The radius of the $$n^{\text{th}}$$ orbit is given by the expression

$$ r_n \;=\; \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}, $$

where

$$\varepsilon_0 = 8.854185 \times 10^{-12}\;\text{kg}^{-1}\,\text{m}^{-3}\,\text{A}^2,$$
$$h = 6.6262 \times 10^{-34}\;\text{J\,s},$$
$$m = 9.1091 \times 10^{-31}\;\text{kg},$$
$$e = 1.60210 \times 10^{-19}\;\text{C},$$
and $$n$$ is the principal quantum number.

For the first orbit ($$n = 1$$) this radius is usually denoted by $$a_0$$ and called the Bohr radius. Setting $$n = 1$$ in the above formula, we have

$$ a_0 \;=\; \frac{\varepsilon_0 h^2}{\pi m e^2}. $$

Now we substitute all the given constants one by one:

$$ a_0 \;=\; \frac{(8.854185 \times 10^{-12}) \,(6.6262 \times 10^{-34})^2}{\pi \,(9.1091 \times 10^{-31})\,(1.60210 \times 10^{-19})^2}. $$

First, we square Planck’s constant:

$$ h^2 \;=\; (6.6262 \times 10^{-34})^2 \;=\; 43.893 \times 10^{-68} \;=\; 4.3893 \times 10^{-67}\;\text{J}^2\text{s}^2. $$

Next, we square the electronic charge:

$$ e^2 \;=\; (1.60210 \times 10^{-19})^2 \;=\; 2.5667 \times 10^{-38}\;\text{C}^2. $$

Now we multiply $$\varepsilon_0$$ and $$h^2$$ in the numerator:

$$ \varepsilon_0 h^2 \;=\; (8.854185 \times 10^{-12}) \times (4.3893 \times 10^{-67}) \;=\; 3.885 \times 10^{-78}. $$

In the denominator we have $$\pi m e^2$$, so we first multiply $$m$$ and $$e^2$$:

$$ m e^2 \;=\; (9.1091 \times 10^{-31}) \times (2.5667 \times 10^{-38}) \;=\; 2.337 \times 10^{-68}. $$

Then we multiply by $$\pi$$:

$$ \pi m e^2 \;=\; 3.1416 \times 2.337 \times 10^{-68} \;=\; 7.343 \times 10^{-68}. $$

Forming the ratio gives

$$ a_0 \;=\; \frac{3.885 \times 10^{-78}}{7.343 \times 10^{-68}} \;=\; 5.292 \times 10^{-11}\;\text{m}. $$

We recognize that $$1\;\text{Å} = 10^{-10}\;\text{m}$$, so

$$ a_0 \;=\; 5.292 \times 10^{-11}\;\text{m} \;=\; 0.529\;\text{Å}. $$

Thus the first Bohr radius is $$0.529$$ Å. For the second Bohr orbit we set $$n = 2$$ in the general formula. Because $$r_n = a_0 n^2$$, we have

$$ r_2 \;=\; a_0 (2)^2 \;=\; a_0 \times 4. $$

Substituting the value of $$a_0$$ we just calculated:

$$ r_2 \;=\; 4 \times 0.529\;\text{Å} \;=\; 2.116\;\text{Å}. $$

Rounding appropriately, we obtain

$$ r_2 \approx 2.12\;\text{Å}. $$

The option that matches this value is 2.12 Å.

Hence, the correct answer is Option C.

First, let us write clearly the electronic configuration given in the question:

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6$$

This configuration contains a total of $$2+2+6+2+6 = 18$$ electrons. Thus any ion that possesses exactly eighteen electrons will have this noble-gas (argon) configuration.

To decide which salt does not give any ion with 18 electrons, we must examine every ion produced by each salt on dissolution in water and count its electrons one by one. We recall the simple fact:

Number of electrons in an ion $$=$$ atomic number of the element $$\pm$$ charge on the ion.

If the ion is positive (cation), we subtract electrons; if it is negative (anion), we add electrons.

Salt A  $$\bigl({\rm NaF}\bigr)$$

On dissolving $$NaF$$ we obtain $$Na^+$$ and $$F^-$$.

For $$Na^+$$: atomic number of Na is $$11$$, charge $$+1$$ ⇒ electrons $$= 11-1 = 10$$.

For $$F^-$$: atomic number of F is $$9$$, charge $$-1$$ ⇒ electrons $$= 9+1 = 10$$.

Neither ion possesses $$18$$ electrons.

Salt B  $$\bigl({\rm KBr}\bigr)$$

Dissolution gives $$K^+$$ and $$Br^-$$.

For $$K^+$$: atomic number $$19$$, charge $$+1$$ ⇒ electrons $$= 19-1 = 18$$.

We already obtain an 18-electron ion here, so $$KBr$$ does furnish the required configuration.

Salt C  $$\bigl({\rm NaCl}\bigr)$$

Dissolution gives $$Na^+$$ and $$Cl^-$$.

For $$Cl^-$$: atomic number $$17$$, charge $$-1$$ ⇒ electrons $$= 17+1 = 18$$.

Thus $$NaCl$$ also supplies an ion with the argon configuration.

Salt D  $$\bigl({\rm CaI_2}\bigr)$$

Dissolution gives $$Ca^{2+}$$ and two $$I^-$$ ions.

For $$Ca^{2+}$$: atomic number $$20$$, charge $$+2$$ ⇒ electrons $$= 20-2 = 18$$.

So $$CaI_2$$ likewise produces an 18-electron ion.

Let us summarise:

$$\begin{array}{lcl} NaF &:& \;10\;{\rm e^-,\;}10\;{\rm e^-} \\ KBr &:& \;18\;{\rm e^-} \\ NaCl &:& \;18\;{\rm e^-} \\ CaI_2 &:& \;18\;{\rm e^-} \end{array}$$

Only the aqueous solution of $$NaF$$ fails to contain any ion with $$18$$ electrons.

Hence, the correct answer is Option A.

First, recall the de-Broglie relation, which connects the momentum $$p$$ of any particle with its wavelength $$\lambda$$ by the formula

$$\lambda \;=\;\frac{h}{p}\;.$$

From this same relation, it follows directly that

$$\frac{h}{\lambda}\;=\;p\;.$$

Therefore, in order to find $$\dfrac{h}{\lambda}$$ for the electron beam, we only need to calculate the magnitude of the momentum $$p$$ that each electron acquires after being accelerated through the given potential difference $$V$$.

When an electron of charge $$e$$ is accelerated from rest through an electric potential difference $$V$$ (expressed here in electrostatic units, i.e. esu), the work done by the electric field becomes its kinetic energy. Hence we can write

$$eV \;=\; \text{kinetic energy of the electron}.$$

For a particle of mass $$m$$ moving with speed $$v$$, the classical kinetic-energy formula is

$$\text{K.E.}\;=\;\frac{1}{2}\,m v^{2}.$$

Equating the electrical work to the kinetic energy, we have

$$eV \;=\;\frac{1}{2}\,m v^{2}\;.$$

Now we solve step-by-step for the velocity $$v$$. Multiplying both sides by $$2$$ gives

$$2\,eV \;=\;m v^{2}\;.$$

Dividing both sides by $$m$$, we obtain

$$v^{2} \;=\;\frac{2\,eV}{m}\;.$$

Taking the positive square root (since speed is positive), we get

$$v \;=\;\sqrt{\frac{2\,eV}{m}}\;.$$

The linear momentum $$p$$ of the electron is given by the basic definition

$$p \;=\;m v\;.$$

Substituting the expression for $$v$$ that we have just derived, we find

$$p \;=\;m \,\sqrt{\frac{2\,eV}{m}} \;=\;\sqrt{m^{2}}\,\sqrt{\frac{2\,eV}{m}} \;=\;\sqrt{m}\,\sqrt{\frac{2\,eV\,m}{m}} \;=\;\sqrt{2\,m\,eV}\;.$$

(In the third step above, the mass $$m$$ inside and outside the square roots was combined so that all factors lie neatly under a single radical.) Thus the magnitude of the momentum acquired by each electron is

$$p \;=\;\sqrt{2\,m\,eV}\;.$$

But earlier we noted that $$\dfrac{h}{\lambda}=p$$. Therefore, substituting the value of $$p$$ obtained here, we finally have

$$\frac{h}{\lambda}\;=\;\sqrt{2\,m\,eV}\;.$$

Among the four listed options, the expression $$\sqrt{2\,meV}$$ matches exactly with the result we have derived.

Hence, the correct answer is Option B.

The question asks for the total number of orbitals that correspond to the principal quantum number $$n = 5$$. To find this, we will recall how quantum numbers determine orbitals and then add up every possibility one by one.

First, remember that the principal quantum number $$n$$ fixes the value of the azimuthal (or orbital‐angular‐momentum) quantum number $$l$$. The rule is:

$$l = 0,1,2,\dots,(n-1).$$

So, when $$n = 5$$ we have the following possible $$l$$ values:

$$l = 0,\;1,\;2,\;3,\;4.$$

Now, for each value of $$l$$ there is a magnetic quantum number $$m_l$$ that can take on integer values from $$-l$$ to $$+l$$ in steps of one. The total number of possible $$m_l$$ values (and hence the number of orbitals) for a given $$l$$ equals $$2l + 1$$. We will list them and add them all up.

For $$l = 0$$ (this is the 5s subshell):

Number of orbitals $$= 2(0) + 1 = 1.$$

For $$l = 1$$ (this is the 5p subshell):

Number of orbitals $$= 2(1) + 1 = 3.$$

For $$l = 2$$ (this is the 5d subshell):

Number of orbitals $$= 2(2) + 1 = 5.$$

For $$l = 3$$ (this is the 5f subshell):

Number of orbitals $$= 2(3) + 1 = 7.$$

For $$l = 4$$ (this is the 5g subshell):

Number of orbitals $$= 2(4) + 1 = 9.$$

Now we add them all together to get the total number of orbitals for the shell where $$n = 5$$:

$$1 + 3 + 5 + 7 + 9 = 25.$$

There is another quicker formula that gives the same result directly: the total number of orbitals in any shell with principal quantum number $$n$$ is $$n^2$$. Substituting $$n = 5$$ gives

$$n^2 = 5^2 = 25,$$

which matches the detailed counting above.

Hence, the correct answer is Option B.

The question asks for the correct sequence of filling electrons when the principal quantum number $$ n = 6 $$. According to the Aufbau principle, electrons fill subshells in order of increasing energy, determined by the $$ (n + l) $$ rule, where $$ n $$ is the principal quantum number and $$ l $$ is the azimuthal quantum number. If two subshells have the same $$ (n + l) $$ value, the subshell with the lower $$ n $$ is filled first.

For $$ n = 6 $$, the relevant subshells and their notations are:

• $$ ns $$ corresponds to $$ 6s $$
• $$ (n-1)d $$ corresponds to $$ (6-1)d = 5d $$
• $$ (n-2)f $$ corresponds to $$ (6-2)f = 4f $$
• $$ np $$ corresponds to $$ 6p $$

Now, calculate the $$ (n + l) $$ values for each subshell:

• For $$ 6s $$: $$ n = 6 $$, $$ l = 0 $$ (since s orbital has $$ l = 0 $$), so $$ n + l = 6 + 0 = 6 $$.
• For $$ 4f $$: $$ n = 4 $$, $$ l = 3 $$ (since f orbital has $$ l = 3 $$), so $$ n + l = 4 + 3 = 7 $$.
• For $$ 5d $$: $$ n = 5 $$, $$ l = 2 $$ (since d orbital has $$ l = 2 $$), so $$ n + l = 5 + 2 = 7 $$.
• For $$ 6p $$: $$ n = 6 $$, $$ l = 1 $$ (since p orbital has $$ l = 1 $$), so $$ n + l = 6 + 1 = 7 $$.

Comparing the $$ (n + l) $$ values:

• The $$ 6s $$ subshell has the lowest $$ (n + l) = 6 $$, so it is filled first.
• The subshells $$ 4f $$, $$ 5d $$, and $$ 6p $$ all have $$ (n + l) = 7 $$. Since they have the same $$ (n + l) $$ value, we compare their $$ n $$ values: $$ 4f $$ has $$ n = 4 $$, $$ 5d $$ has $$ n = 5 $$, and $$ 6p $$ has $$ n = 6 $$. The subshell with the smallest $$ n $$ is filled first, so $$ 4f $$ is filled before $$ 5d $$, and $$ 5d $$ is filled before $$ 6p $$.

Therefore, the sequence of filling is: $$ 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p $$.

In terms of the given notations:

• $$ 6s $$ is $$ ns $$
• $$ 4f $$ is $$ (n-2)f $$
• $$ 5d $$ is $$ (n-1)d $$
• $$ 6p $$ is $$ np $$

So the sequence becomes: $$ ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np $$.

Now, comparing this with the options:

• Option A: $$ ns \rightarrow (n-1)d \rightarrow (n-2)f \rightarrow np $$ corresponds to $$ 6s \rightarrow 5d \rightarrow 4f \rightarrow 6p $$, which is incorrect because $$ 4f $$ should come before $$ 5d $$.
• Option B: $$ ns \rightarrow np \rightarrow (n-1)d \rightarrow (n-2)f $$ corresponds to $$ 6s \rightarrow 6p \rightarrow 5d \rightarrow 4f $$, which is incorrect because $$ 6p $$ should be last.
• Option C: $$ ns \rightarrow (n-2)f \rightarrow np \rightarrow (n-1)d $$ corresponds to $$ 6s \rightarrow 4f \rightarrow 6p \rightarrow 5d $$, which is incorrect because $$ 5d $$ should come before $$ 6p $$.
• Option D: $$ ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np $$ corresponds to $$ 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p $$, which matches the correct sequence.

Hence, the correct answer is Option D.

We recall that in the Bohr model of the hydrogen atom the allowed (quantised) total energies of the electron are given by the well-known expression

$$E_n = -\dfrac{13.6\ \text{eV}}{n^2},$$

where $$n$$ is the principal quantum number and can take only positive integer values $$n = 1,2,3,\dots$$. The negative sign shows that the electron is bound to the nucleus; its energy is less than that of a free electron, which is taken to be zero.

Now we list the first few energies by substituting successive integer values of $$n$$ into the formula.

For the ground state we put $$n = 1$$. We have

$$E_1 = -\dfrac{13.6\ \text{eV}}{1^2} = -13.6\ \text{eV}.$$

For the first excited state we put $$n = 2$$. So

$$E_2 = -\dfrac{13.6\ \text{eV}}{2^2} = -\dfrac{13.6\ \text{eV}}{4} = -3.4\ \text{eV}.$$

For the second excited state we put $$n = 3$$. Thus

$$E_3 = -\dfrac{13.6\ \text{eV}}{3^2} = -\dfrac{13.6\ \text{eV}}{9} \approx -1.51\ \text{eV}.$$

We observe an important point: every legitimate energy level obtained from the formula is negative. Positive values like $$+6.8\ \text{eV}$$ or $$+13.6\ \text{eV}$$ cannot correspond to any bound state of hydrogen.

Next, let us inspect the proposed options one by one and compare them with the allowed values we have just derived.

A. $$+6.8\ \text{eV}$$ – This is positive, whereas all allowed energies are negative. Hence this cannot be an energy level.

B. $$+13.6\ \text{eV}$$ – Again positive, so not permissible.

C. $$-6.8\ \text{eV}$$ – This is negative, but we test whether it fits the formula. Setting

$$-6.8 = -\dfrac{13.6}{n^2}$$

gives $$n^2 = \dfrac{13.6}{6.8} = 2$$ and therefore $$n = \sqrt{2}$$, which is not an integer. Because $$n$$ must be an integer, $$-6.8\ \text{eV}$$ is not an allowed level.

D. $$-3.4\ \text{eV}$$ – We already found that for $$n = 2$$ the energy is exactly $$-3.4\ \text{eV}$$. Thus this value satisfies the quantisation condition and represents the first excited state.

Among the given choices, only $$-3.4\ \text{eV}$$ corresponds to an allowed excited state of hydrogen.

Hence, the correct answer is Option 4.

We start from the statement that, for any non-relativistic particle in thermal equilibrium at temperature $$T$$, the average kinetic energy is

$$K=\frac32\,kT.$$

For a particle of rest mass $$m$$ the kinetic energy is also related to the linear momentum $$p$$ by the classical formula

$$K=\frac{p^{2}}{2m}.$$

Solving this for the momentum gives

$$p=\sqrt{2mK}.$$

The de Broglie wavelength is defined as

$$\lambda=\frac{h}{p},$$

so after substituting $$p=\sqrt{2mK}$$ we obtain

$$\lambda=\frac{h}{\sqrt{2mK}}.$$

Now we insert $$K=\dfrac32 kT$$ to express the wavelength of a thermal particle explicitly in terms of its mass:

$$\lambda=\frac{h}{\sqrt{2m\left(\dfrac32 kT\right)}} =\frac{h}{\sqrt{3mkT}}.$$

From this final form we notice the crucial dependence

$$\lambda\propto\frac1{\sqrt{m}}.$$

In simple words, for particles that possess the same thermal kinetic energy, the lighter the mass, the longer the de Broglie wavelength.

Let us now deal with each object mentioned in the problem.

1. Thermal electron

The electron mass is

$$m_e\approx9.11\times10^{-31}\,\text{kg}.$$

Its thermal de Broglie wavelength at temperature $$T$$ is therefore

$$\lambda_e=\frac{h}{\sqrt{3m_e kT}}.$$

2. Thermal neutron

The neutron mass is

$$m_n\approx1.675\times10^{-27}\,\text{kg},$$

which is about $$1836$$ times heavier than the electron. Because $$\lambda\propto1/\sqrt{m}$$ we immediately see that

$$\lambda_n=\frac{h}{\sqrt{3m_n kT}}$$

is much smaller than $$\lambda_e$$. Quantitatively, the ratio is

$$\frac{\lambda_e}{\lambda_n}=\sqrt{\frac{m_n}{m_e}}\approx\sqrt{1836}\approx43,$$

so the electron wavelength exceeds the neutron wavelength by roughly two orders of magnitude.

3. Visible photon

A photon is massless; its wavelength is fixed by its energy through the Planck relation

$$E=\frac{hc}{\lambda_{\text{ph}}}.$$

For visible light, $$\lambda_{\text{ph}}$$ lies roughly between $$400\;\text{nm}$$ and $$700\;\text{nm}$$; a central value $$\lambda_{\text{ph}}\approx500\;\text{nm}=5\times10^{-7}\text{ m}$$ is adequate for comparison.

To see the numerical contrast, let us evaluate the actual thermal wavelengths at room temperature, say $$T=300\;\text{K}$$.

Thermal electron

$$K=\frac32 kT=\frac32(1.38\times10^{-23})(300) \approx6.21\times10^{-21}\,\text{J},$$

$$p_e=\sqrt{2m_eK} =\sqrt{2(9.11\times10^{-31})(6.21\times10^{-21})} \approx1.06\times10^{-25}\,\text{kg m s}^{-1},$$

$$\lambda_e=\frac{h}{p_e} =\frac{6.626\times10^{-34}}{1.06\times10^{-25}} \approx6.2\times10^{-9}\,\text{m}=6.2\;\text{nm}.$$

Thermal neutron

$$p_n=\sqrt{2m_nK} =\sqrt{2(1.675\times10^{-27})(6.21\times10^{-21})} \approx4.56\times10^{-24}\,\text{kg m s}^{-1},$$

$$\lambda_n=\frac{h}{p_n} =\frac{6.626\times10^{-34}}{4.56\times10^{-24}} \approx1.45\times10^{-10}\,\text{m}=0.145\;\text{nm}.$$

Visible photon

We already have $$\lambda_{\text{ph}}\approx5\times10^{-7}\,\text{m}=500\;\text{nm}.$$

Collecting the three wavelengths we find

$$\lambda_{\text{photon}}\;(\sim500\text{ nm}) \; > \; \lambda_{\text{electron}}\;(\sim6\text{ nm}) \; > \; \lambda_{\text{neutron}}\;(\sim0.15\text{ nm}).$$

Thus the required descending order is

Visible photon > thermal electron > thermal neutron.

Comparing with the options supplied in the question, this sequence corresponds exactly to Option A.

Hence, the correct answer is Option A.

The ionization energy given is 495.5 kJ mol⁻¹ for gaseous Na atoms. This energy is required to remove an electron from one mole of sodium atoms. However, to find the lowest frequency of light that ionizes a single sodium atom, we need the ionization energy per atom.

First, convert the ionization energy from kJ mol⁻¹ to J mol⁻¹. Since 1 kJ = 1000 J, multiply by 1000:

$$495.5 \text{kJ mol}^{-1} = 495.5 \times 1000 \text{J mol}^{-1} = 495500 \text{J mol}^{-1}.$$

Now, to find the energy per atom, divide by Avogadro's number, $$N_A = 6.022 \times 10^{23}$$ mol⁻¹, which gives the number of atoms per mole:

$$\text{Energy per atom} = \frac{495500 \text{J mol}^{-1}}{6.022 \times 10^{23} \text{atoms mol}^{-1}}.$$

Simplify the expression:

$$\text{Energy per atom} = \frac{495500}{6.022 \times 10^{23}} \text{J atom}^{-1}.$$

Calculate the numerical value. First, express 495500 in scientific notation: $$495500 = 4.955 \times 10^5$$. Then:

$$\text{Energy per atom} = \frac{4.955 \times 10^5}{6.022 \times 10^{23}} = \frac{4.955}{6.022} \times 10^{5 - 23} = \frac{4.955}{6.022} \times 10^{-18}.$$

Divide 4.955 by 6.022:

$$\frac{4.955}{6.022} \approx 0.8225.$$

So,

$$\text{Energy per atom} \approx 0.8225 \times 10^{-18} \text{J atom}^{-1} = 8.225 \times 10^{-19} \text{J atom}^{-1}.$$

The lowest frequency of light that can ionize the atom corresponds to a photon with energy equal to the ionization energy per atom. The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency. Therefore,

$$\nu = \frac{E}{h}.$$

Given $$h = 6.626 \times 10^{-34}$$ J s, substitute the values:

$$\nu = \frac{8.225 \times 10^{-19}}{6.626 \times 10^{-34}} \text{s}^{-1}.$$

Simplify:

$$\nu = \frac{8.225}{6.626} \times 10^{-19 - (-34)} = \frac{8.225}{6.626} \times 10^{15}.$$

Divide 8.225 by 6.626:

$$\frac{8.225}{6.626} \approx 1.2413.$$

So,

$$\nu \approx 1.2413 \times 10^{15} \text{s}^{-1}.$$

Rounding to three significant figures, as in the options, gives $$1.24 \times 10^{15}$$ s⁻¹.

Comparing with the options:

A. $$1.24 \times 10^{15}$$ s⁻¹

B. $$7.50 \times 10^4$$ s⁻¹

C. $$4.76 \times 10^{14}$$ s⁻¹

D. $$3.15 \times 10^{15}$$ s⁻¹

The value $$1.24 \times 10^{15}$$ s⁻¹ matches option A.

Hence, the correct answer is Option A.

We start with the photoelectric effect equation. The maximum kinetic energy of the ejected photoelectron is given by the difference between the energy of the incident photon and the work function of the metal. The work function is the minimum energy required to eject an electron, which corresponds to the threshold wavelength λ₀.

The energy of the incident photon with wavelength λ is $$ E = \frac{hc}{\lambda} $$, where h is Planck's constant and c is the speed of light. The work function φ is $$ \phi = \frac{hc}{\lambda_0} $$. Therefore, the kinetic energy (K.E.) is:

$$ K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} $$

Simplifying this expression:

$$ K.E. = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) $$

Note that $$ \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{\lambda_0 - \lambda}{\lambda \lambda_0} $$, so we can write:

$$ K.E. = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

This kinetic energy is also equal to $$ \frac{1}{2} m v^2 $$, where m is the mass of the electron and v is its velocity. Setting them equal:

$$ \frac{1}{2} m v^2 = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

Now, solve for v²:

$$ v^2 = \frac{2 hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

Taking the square root of both sides to find v:

$$ v = \sqrt{ \frac{2 hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) } $$

Comparing this with the given options, we see that it matches option C:

$$ \sqrt{\frac{2hc}{m}\left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)} $$

Option A lacks the factor c and has incorrect units. Option B has an extra (λ₀ - λ) without the denominator, leading to incorrect units. Option D has the reciprocal terms reversed and lacks c, making it incorrect both dimensionally and physically.

Hence, the correct answer is Option C.

We begin by recalling that any electron in an atom is completely described by four quantum numbers: the principal quantum number $$n$$, the azimuthal (orbital‐angular‐momentum) quantum number $$l$$, the magnetic quantum number $$m_l$$ and the spin quantum number $$m_s$$.

The principal quantum number $$n$$ tells us the main energy level (or shell) to which the electron belongs. The azimuthal quantum number $$l$$ takes integral values from $$0$$ to $$(n-1)$$ and designates the subshell: $$l = 0$$ corresponds to an $$s$$ subshell, $$l = 1$$ to a $$p$$ subshell, $$l = 2$$ to a $$d$$ subshell, and so on. For any fixed $$l$$, the magnetic quantum number $$m_l$$ can assume all integral values from $$-l$$ through $$0$$ to $$+l$$. Finally, the spin quantum number $$m_s$$ can be either $$+\dfrac{1}{2}$$ or $$-\dfrac{1}{2}$$, representing the two possible spin orientations.

Now, rubidium has atomic number $$Z = 37$$. To locate its valence electron, we must write the complete ground-state electronic configuration, building it up step by step according to the Aufbau (energy‐ordering) principle, the Pauli exclusion principle and Hund’s rule:

$$1s^2\;2s^2\;2p^6\;3s^2\;3p^6\;4s^2\;3d^{10}\;4p^6\;5s^1$$

Up to krypton $$[Kr]$$, which accounts for $$36$$ electrons, we have filled all lower subshells. The $$37^{\text{th}}$$ electron of rubidium therefore enters the next available subshell, which is the $$5s$$ subshell, giving the outer-electronic configuration $$[Kr]\,5s^1$$. Hence the valence (outermost) electron of rubidium occupies a $$5s$$ orbital.

For this $$5s$$ electron:

• The principal quantum number is simply the shell number: $$n = 5$$.

• Because it is in an $$s$$ subshell, we use the correspondence $$s \rightarrow l = 0$$. So we have $$l = 0$$.

• For any $$s$$ orbital (where $$l = 0$$), the only possible magnetic quantum number is $$m_l = 0$$, because $$m_l$$ ranges from $$-l$$ to $$+l$$ and that gives only one value when $$l = 0$$.

• The spin quantum number can be either $$+\dfrac{1}{2}$$ or $$-\dfrac{1}{2}$$. Conventionally, in listing a single electron we choose the $$+\dfrac{1}{2}$$ value, although the other value would also represent an allowed state. Here, every option in the question specifies $$+\dfrac{1}{2}$$, so we select $$m_s = +\dfrac{1}{2}$$.

Collecting these results, the complete set of quantum numbers for the valence electron of rubidium is

$$n = 5,\; l = 0,\; m_l = 0,\; m_s = +\dfrac{1}{2}.$$

We now inspect the given options:

A. $$5,\;0,\;0,\;+\dfrac{1}{2}$$

B. $$5,\;1,\;0,\;+\dfrac{1}{2}$$

C. $$5,\;1,\;1,\;+\dfrac{1}{2}$$

D. $$5,\;0,\;1,\;+\dfrac{1}{2}$$

Only Option A matches the required set $$\left(5,\;0,\;0,\;+\dfrac{1}{2}\right)$$ exactly. The other options are incorrect because they assign either an inappropriate azimuthal quantum number $$l$$ or an impossible magnetic quantum number $$m_l$$ for the $$s$$ subshell.

Hence, the correct answer is Option A.

The given equation is $$\Delta E = -2.0 \times 10^{-18} \, \text{J} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$$. This represents the change in energy when an electron transitions from an initial energy level $$n_1$$ to a final level $$n_2$$. For absorption, energy is absorbed to move the electron from a lower level to a higher level. Here, the electron is excited from $$n_1 = 1$$ to $$n_2 = 2$$.

Substitute $$n_1 = 1$$ and $$n_2 = 2$$ into the equation:

$$\Delta E = -2.0 \times 10^{-18} \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = -2.0 \times 10^{-18} \left( \frac{1}{4} - 1 \right)$$

Compute the expression inside the parentheses:

$$\frac{1}{4} - 1 = \frac{1}{4} - \frac{4}{4} = -\frac{3}{4}$$

Now substitute back:

$$\Delta E = -2.0 \times 10^{-18} \times \left( -\frac{3}{4} \right) = 2.0 \times 10^{-18} \times \frac{3}{4}$$

Calculate the multiplication:

$$2.0 \times 10^{-18} \times \frac{3}{4} = \frac{2.0 \times 3}{4} \times 10^{-18} = \frac{6.0}{4} \times 10^{-18} = 1.5 \times 10^{-18} \, \text{J}$$

The energy absorbed, $$\Delta E = 1.5 \times 10^{-18} \, \text{J}$$, corresponds to the energy of the photon absorbed. The energy of a photon is given by $$E = \frac{hc}{\lambda}$$, where $$h$$ is Planck's constant, $$c$$ is the speed of light, and $$\lambda$$ is the wavelength. Therefore,

$$\Delta E = \frac{hc}{\lambda}$$

Solving for $$\lambda$$:

$$\lambda = \frac{hc}{\Delta E}$$

Given $$h = 6.625 \times 10^{-34} \, \text{Js}$$ and $$c = 3 \times 10^8 \, \text{ms}^{-1}$$, substitute the values:

$$\lambda = \frac{(6.625 \times 10^{-34}) \times (3 \times 10^8)}{1.5 \times 10^{-18}}$$

First, multiply the numerator:

$$6.625 \times 10^{-34} \times 3 \times 10^8 = 6.625 \times 3 \times 10^{-34 + 8} = 19.875 \times 10^{-26}$$

Now divide by the denominator:

$$\lambda = \frac{19.875 \times 10^{-26}}{1.5 \times 10^{-18}} = \frac{19.875}{1.5} \times 10^{-26 - (-18)} = \frac{19.875}{1.5} \times 10^{-8}$$

Compute $$\frac{19.875}{1.5}$$:

$$19.875 \div 1.5 = 13.25$$

So,

$$\lambda = 13.25 \times 10^{-8} \, \text{m} = 1.325 \times 10^{-7} \, \text{m}$$

Comparing with the options, $$1.325 \times 10^{-7} \, \text{m}$$ corresponds to option A.

Hence, the correct answer is Option A.

The de-Broglie wavelength $$\lambda$$ of a particle is given by the formula $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle. Momentum $$p$$ is calculated as $$p = m \times v$$, where $$m$$ is the mass and $$v$$ is the velocity. So, the formula becomes $$\lambda = \frac{h}{m v}$$.

We are given the mass $$m = 6.63$$ grams and velocity $$v = 100$$ meters per second. However, the SI unit for mass is kilograms, so we must convert grams to kilograms. Since 1 kilogram = 1000 grams, we have:

$$m = \frac{6.63}{1000} = 0.00663 \text{ kg}$$

This can also be written as $$m = 6.63 \times 10^{-3}$$ kg. The velocity $$v = 100$$ m/s remains as is. Planck's constant $$h$$ is $$6.63 \times 10^{-34}$$ Joule seconds (J s), and since Joule is kg m² s⁻², the units are consistent.

Now substitute the values into the formula:

$$\lambda = \frac{6.63 \times 10^{-34}}{m \times v} = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-3}) \times 100}$$

First, compute the denominator: $$m \times v = (6.63 \times 10^{-3}) \times 100$$. Since $$100 = 10^2$$, we have:

$$m \times v = 6.63 \times 10^{-3} \times 10^2 = 6.63 \times 10^{-3 + 2} = 6.63 \times 10^{-1}$$

Now the expression is:

$$\lambda = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-1}}$$

Simplify by dividing the numerical coefficients and the powers of 10 separately. The numerical part is $$\frac{6.63}{6.63} = 1$$. The exponent part is $$\frac{10^{-34}}{10^{-1}} = 10^{-34 - (-1)} = 10^{-34 + 1} = 10^{-33}$$. Therefore:

$$\lambda = 1 \times 10^{-33} \text{ m}$$

So, the de-Broglie wavelength is $$10^{-33}$$ meters. Comparing with the options, this matches option A.

Hence, the correct answer is Option A.

The energy of an electron in a hydrogen-like atom is given by the formula:

$$ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} $$

where $$ E_n $$ is the energy in electron volts (eV), $$ Z $$ is the atomic number, and $$ n $$ is the principal quantum number.

For the hydrogen atom (H), the given energy in the first Bohr's orbit (where $$ n = 1 $$) is -13.6 eV. Since hydrogen has $$ Z = 1 $$, this matches the formula: $$ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} $$.

Now, we need to find the energy for the Li²⁺ ion. Lithium (Li) has an atomic number $$ Z = 3 $$. The Li²⁺ ion has lost two electrons, leaving only one electron, so it is a hydrogen-like system. Therefore, we can use the same energy formula.

The question asks for the energy in the first excited state. The first excited state corresponds to the second orbit, so $$ n = 2 $$.

Substituting $$ Z = 3 $$ and $$ n = 2 $$ into the formula:

$$ E_2 = -\frac{13.6 \times (3)^2}{(2)^2} $$

First, calculate $$ (3)^2 $$:

$$ (3)^2 = 3 \times 3 = 9 $$

Next, calculate $$ (2)^2 $$:

$$ (2)^2 = 2 \times 2 = 4 $$

Now substitute these values:

$$ E_2 = -\frac{13.6 \times 9}{4} $$

Multiply 13.6 by 9:

$$ 13.6 \times 9 = 122.4 $$

Now divide by 4:

$$ \frac{122.4}{4} = 30.6 $$

So the expression becomes:

$$ E_2 = -30.6 \text{ eV} $$

Therefore, the energy value of the electron in the first excited state of Li²⁺ is -30.6 eV.

Comparing with the options:

A. 27.2 eV

B. -30.6 eV

C. 30.6 eV

D. -27.2 eV

The value -30.6 eV corresponds to option B.

Hence, the correct answer is Option B.

The energy of a single photon is calculated using the formula $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency of the light.

Given:

• Frequency, $$\nu = 2.47 \times 10^{15}$$ Hz
• Planck's constant, $$h = 6.63 \times 10^{-34}$$ J s

Substitute the values into the formula: $$$ E = (6.63 \times 10^{-34}) \times (2.47 \times 10^{15}) $$$

First, multiply the coefficients: $$$ 6.63 \times 2.47 $$$

Break it down step by step: $$$ 6.63 \times 2.47 = 6.63 \times (2 + 0.4 + 0.07) $$$ $$$ = (6.63 \times 2) + (6.63 \times 0.4) + (6.63 \times 0.07) $$$ $$$ = 13.26 + 2.652 + 0.4641 $$$ $$$ = 16.3761 $$$

Now, multiply the powers of 10: $$$ 10^{-34} \times 10^{15} = 10^{-34 + 15} = 10^{-19} $$$

Combine the results: $$$ E = 16.3761 \times 10^{-19} \text{ J} $$$

Convert to standard scientific notation by adjusting the decimal: $$$ 16.3761 \times 10^{-19} = 1.63761 \times 10^1 \times 10^{-19} = 1.63761 \times 10^{-18} \text{ J} $$$

Consider significant figures. Both given values ($$6.63$$ and $$2.47$$) have three significant figures. Round $$1.63761 \times 10^{-18}$$ to three significant figures:

• The third digit is $$3$$ (in $$1.63$$), and the next digit $$7 \geq 5$$, so round up: $$1.64 \times 10^{-18}$$ J.

Compare with the options:

• A: $$8.041 \times 10^{-40}$$ J
• B: $$2.680 \times 10^{-19}$$ J
• C: $$1.640 \times 10^{-18}$$ J
• D: $$6.111 \times 10^{-17}$$ J

Hence, the correct answer is Option C.

The energy of an electron in a hydrogen-like atom is given in the statement as

$$E = -2.178 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right)\,\text{J}.$$

For the hydrogen atom we have $$Z = 1$$, so the formula simplifies to

$$E = -2.178 \times 10^{-18}\left(\frac{1}{n^{2}}\right)\,\text{J}.$$

The electron is to be promoted from the ground level $$n = 1$$ to the first excited level $$n = 2$$. Let us calculate the energy of each level separately.

For $$n = 1$$:

$$E_{1} = -2.178 \times 10^{-18}\left(\frac{1}{1^{2}}\right) = -2.178 \times 10^{-18}\ \text{J}.$$

For $$n = 2$$:

$$E_{2} = -2.178 \times 10^{-18}\left(\frac{1}{2^{2}}\right) = -2.178 \times 10^{-18}\left(\frac{1}{4}\right) = -\frac{2.178}{4} \times 10^{-18}\ \text{J} = -5.445 \times 10^{-19}\ \text{J}.$$

The energy that must be absorbed is the difference between the two levels:

$$\Delta E = E_{2} - E_{1}.$$

Substituting the values we have just obtained,

$$\Delta E = \left(-5.445 \times 10^{-19}\right) - \left(-2.178 \times 10^{-18}\right)$$ $$\Delta E = -5.445 \times 10^{-19} + 2.178 \times 10^{-18}$$ $$\Delta E = \left(2.178 - 0.5445\right) \times 10^{-18}$$ $$\Delta E = 1.6335 \times 10^{-18}\ \text{J}.$$

(A more rounded value, $$1.634 \times 10^{-18}\ \text{J}$$, will give the same final wavelength.)

According to the Planck-Einstein relation, the energy of a photon is connected to its frequency by

$$\Delta E = h\nu,$$

and the frequency and wavelength are related through the speed of light:

$$\nu = \frac{c}{\lambda}.$$

Combining the two relations gives the wavelength directly:

$$\Delta E = h\frac{c}{\lambda}\quad\Longrightarrow\quad \lambda = \frac{hc}{\Delta E}.$$

Now substitute the given constants $$h = 6.62 \times 10^{-34}\ \text{J\,s}$$ and $$c = 3.0 \times 10^{8}\ \text{m\,s}^{-1}$$ together with $$\Delta E = 1.6335 \times 10^{-18}\ \text{J}$$:

$$\lambda = \frac{\left(6.62 \times 10^{-34}\right)\left(3.0 \times 10^{8}\right)}{1.6335 \times 10^{-18}}.$$

Multiplying the numerator:

$$6.62 \times 3.0 = 19.86,$$ so $$h\,c = 19.86 \times 10^{-26}\ \text{J\,m} = 1.986 \times 10^{-25}\ \text{J\,m}.$$

Now divide:

$$\lambda = \frac{1.986 \times 10^{-25}}{1.6335 \times 10^{-18}} = 1.216 \times 10^{-7}\ \text{m}.$$

This is essentially the same as $$1.214 \times 10^{-7}\ \text{m}$$ when rounded to three significant figures, which matches one of the options.

Hence, the correct answer is Option C.

For any particle of mass $$m$$ moving with momentum $$p$$, the de-Broglie wavelength is

$$\lambda = \frac{h}{p}$$

For an electron whose speed is much less than the speed of light, the classical relation between kinetic energy $$K$$ and momentum $$p$$ is

$$K = \frac{p^{2}}{2m} \; \; \Longrightarrow \; \; p = \sqrt{2mK}$$

Substituting this $$p$$ in the de-Broglie formula,

$$\lambda = \frac{h}{\sqrt{2mK}}$$

Thus, the wavelength is inversely proportional to the square root of the kinetic energy:

$$\lambda \propto \frac{1}{\sqrt{K}}$$

If the kinetic energy is increased four times, i.e. $$K' = 4K$$, then

$$\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}$$

Therefore, the de-Broglie wavelength becomes one-half of its original value.

Option B which is: half

For an electron in the $$n^{\text{th}}$$ Bohr orbit, the de Broglie wavelength $$\lambda$$ satisfies the condition that an integral number of wavelengths fits exactly on the circumference:

$$2\pi r_n = n\,\lambda \quad -(1)$$

The radius of the $$n^{\text{th}}$$ orbit for a hydrogen atom is given by Bohr’s formula:

$$r_n = n^{2} a_0 \quad -(2)$$

Substitute $$r_n$$ from $$-(2)$$ into the quantization condition $$-(1)$$ and solve for $$\lambda$$:

$$2\pi (n^{2} a_0) = n\,\lambda$$

$$\lambda = \frac{2\pi n^{2} a_0}{n} = 2\pi n a_0 \quad -(3)$$

For the third orbit, $$n = 3$$. Using $$-(3)$$:

$$\lambda_3 = 2\pi \times 3 \times a_0 = 6\pi a_0$$

Thus, the de Broglie wavelength of an electron in the third orbit is $$6\pi a_0$$.

Option C which is: $$6\pi a_0$$

The Balmer series corresponds to electronic transitions that terminate at the level $$n_1 = 2$$ in the hydrogen atom.

For any line of a hydrogen series, the Rydberg formula gives the frequency $$\nu$$ as:
$$\nu = R_\infty\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right)$$ where $$n_1$$ is the lower (final) level and $$n_2$$ (>$$n_1$$) is the upper (initial) level.

The limiting (or series) line is obtained when the electron falls from $$n_2 = \infty$$ to $$n_1 = 2$$. At $$n_2 = \infty$$, the term $$\dfrac{1}{n_2^{2}}$$ becomes zero. Hence
$$\nu_{\text{limit}} = R_\infty\left(\frac{1}{2^{2}} - 0\right) = R_\infty \times \frac{1}{4}$$

Insert the given value $$R_\infty = 3.29 \times 10^{15}\ \text{cycles s}^{-1}$$:
$$\nu_{\text{limit}} = \frac{3.29 \times 10^{15}}{4} = 0.8225 \times 10^{15}\ \text{s}^{-1} = 8.225 \times 10^{14}\ \text{s}^{-1}$$

Rounded to three significant figures, $$\nu_{\text{limit}} \approx 8.22 \times 10^{14}\ \text{s}^{-1}$$.

Option A which is: $$8.22 \times 10^{14}\ \text{s}^{-1}$$

When a gas atom or molecule first absorbs a photon and later de-excites in two separate steps, the total energy of the two emitted photons must equal the energy of the single absorbed photon.

Photon energy is given by $$E = \frac{hc}{\lambda}$$, so energy conservation gives
$$\frac{hc}{\lambda_{\text{abs}}}= \frac{hc}{\lambda_1}+ \frac{hc}{\lambda_2}$$

The constants $$h$$ and $$c$$ cancel out, leaving
$$\frac{1}{\lambda_{\text{abs}}}= \frac{1}{\lambda_1}+ \frac{1}{\lambda_2} \quad -(1)$$

Data from the question:
$$\lambda_{\text{abs}} = 355 \,\text{nm}, \quad \lambda_1 = 680 \,\text{nm}, \quad \lambda_2 = ?$$

Substitute in equation $$-(1)$$:
$$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_2}$$

Solve for $$\lambda_2$$:
$$\frac{1}{\lambda_2} = \frac{1}{355} - \frac{1}{680}$$
$$\frac{1}{\lambda_2} = 0.0028169\;\text{nm}^{-1} - 0.0014706\;\text{nm}^{-1}$$
$$\frac{1}{\lambda_2} = 0.0013463\;\text{nm}^{-1}$$

Taking the reciprocal:
$$\lambda_2 = \frac{1}{0.0013463}\;\text{nm} \approx 742.8\;\text{nm}$$

Rounded to the nearest nanometre, $$\lambda_2 \approx 743 \,\text{nm}$$.

Therefore, the other emitted wavelength is $$743 \,\text{nm}$$.

Option B which is: 743 nm

For simultaneous measurement of position and momentum, Heisenberg’s uncertainty principle states
$$\Delta x\,\Delta p \;\ge\; \dfrac{h}{4\pi}$$

The electron’s momentum uncertainty is related to its speed uncertainty by $$\Delta p = m_e\,\Delta v$$.

Step 1: Determine the speed uncertainty.
Given speed $$v = 600\,$$m s$$^{-1}$$ with an accuracy of $$0.005\% = 0.005/100 = 5.0\times10^{-5}$$, therefore
$$\Delta v = v \times 5.0\times10^{-5} = 600 \times 5.0\times10^{-5} = 0.03\,$$m s$$^{-1}$$.

Step 2: Calculate the momentum uncertainty.
Electron mass $$m_e = 9.1\times10^{-31}\,$$kg, hence
$$\Delta p = m_e \,\Delta v = 9.1\times10^{-31}\times0.03 = 2.73\times10^{-32}\,\text{kg m s}^{-1}.$$

Step 3: Apply the uncertainty relation to find positional uncertainty.
$$\Delta x \;\ge\; \dfrac{h}{4\pi\,\Delta p} = \dfrac{6.6\times10^{-34}}{4\pi \times 2.73\times10^{-32}}.$$

Compute the denominator first:
$$4\pi \times 2.73\times10^{-32} \approx 12.566 \times 2.73\times10^{-32} = 3.43\times10^{-31}.$$

Hence
$$\Delta x \;\ge\; \dfrac{6.6\times10^{-34}}{3.43\times10^{-31}} \approx 1.92\times10^{-3}\,\text{m}.$$

The minimum uncertainty in the electron’s position is therefore $$1.92 \times 10^{-3}\,$$m.

Option C which is: $$1.92 \times 10^{-3}\,$$m

For a moving particle, the de Broglie relation gives its wavelength: $$\lambda = \frac{h}{mv}$$ where
$$h$$ is Planck’s constant, $$m$$ is the particle’s mass and $$v$$ is its speed.

Substitute the given values:
$$h = 6.63 \times 10^{-34}\; \text{J s}$$,  $$m = 1.67 \times 10^{-27}\; \text{kg}$$,  $$v = 1.0 \times 10^{3}\; \text{m s}^{-1}$$

The momentum of the proton is
$$mv = \left(1.67 \times 10^{-27}\right)\left(1.0 \times 10^{3}\right) = 1.67 \times 10^{-24}\; \text{kg m s}^{-1}$$

Hence the wavelength is
$$\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-24}} = 3.97 \times 10^{-10}\; \text{m}$$

Convert metres to nanometres (1 nm = $$10^{-9}$$ m):
$$3.97 \times 10^{-10}\; \text{m} = \frac{3.97 \times 10^{-10}}{10^{-9}}\; \text{nm} = 0.397\; \text{nm} \approx 0.40\; \text{nm}$$

Therefore, the wavelength associated with the proton is $$0.40$$ nm.

Option B which is: $$0.40\;\text{nm}$$

For a H-atom the energy of the electron in the $$n^{\text{th}}$$ orbit is given by Bohr’s expression
$$E_n = -\dfrac{R_H}{n^2}$$, where $$R_H$$ (in energy units per mole) is the Rydberg constant.

The given ionization enthalpy equals the energy needed to lift the electron from the ground state $$n = 1$$ to $$n = \infty$$, i.e.
$$\lvert E_1 \rvert = R_H = 1.312 \times 10^6\; \text{J mol}^{-1}$$.

Energy in the first excited state $$n = 2$$ is
$$E_2 = -\dfrac{R_H}{2^2} = -\dfrac{R_H}{4}$$.

The energy required to excite the electron from $$n = 1$$ to $$n = 2$$ is the difference
$$\Delta E = E_2 - E_1 = \left(-\dfrac{R_H}{4}\right) - \left(-R_H\right) = R_H\!\left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4}R_H.$$

Substituting $$R_H = 1.312 \times 10^6\; \text{J mol}^{-1}$$:
$$\Delta E = \dfrac{3}{4} \times 1.312 \times 10^6 = 0.984 \times 10^6\; \text{J mol}^{-1} = 9.84 \times 10^5\; \text{J mol}^{-1}.$$

Hence, the required excitation energy is $$9.84 \times 10^5\ \text{J mol}^{-1}$$.

Option D which is: $$9.84 \times 10^5\ J\ mol^{-1}$$

An isotope is defined as a nuclide having the same atomic number $$Z$$ (same element) but a different mass number $$A$$. Therefore, in any nuclear reaction that produces an isotope, $$Z$$ must remain unchanged while $$A$$ must change.

Case A: Neutron particle emission
A free neutron leaves the nucleus:
$$^{A}_{Z}X \;\rightarrow\; ^{A-1}_{Z}X \;+\; ^{1}_{0}n$$
• Atomic number: $$Z$$ remains the same.
• Mass number: decreases by 1 (from $$A$$ to $$A-1$$).
Hence the daughter nuclide is the isotope of the parent element.

Case B: Positron ( $$\beta^{+}$$ ) emission
A proton converts into a neutron and emits a positron:
$$^{A}_{Z}X \;\rightarrow\; ^{A}_{Z-1}Y \;+\; ^{0}_{+1}\beta$$
• Atomic number: decreases by 1.
• Mass number: unchanged.
Different element → not an isotope.

Case C: $$\alpha$$-particle emission
An $$\alpha$$ particle ($$^{4}_{2}\mathrm{He}$$) is emitted:
$$^{A}_{Z}X \;\rightarrow\; ^{A-4}_{Z-2}Y \;+\; ^{4}_{2}\alpha$$
• Atomic number: decreases by 2.
• Mass number: decreases by 4.
Different element → not an isotope.

Case D: $$\beta^{-}$$ (electron) emission
A neutron converts into a proton and emits an electron:
$$^{A}_{Z}X \;\rightarrow\; ^{A}_{Z+1}Y \;+\; ^{0}_{-1}\beta$$
• Atomic number: increases by 1.
• Mass number: unchanged.
Different element → not an isotope.

Only neutron emission keeps $$Z$$ unchanged while altering $$A$$, thus producing an isotope of the same element.

Final Answer: Option A which is: neutron particle emission

The electron is falling from $$n_2=\infty$$ to the first stationary state $$n_1=1$$ in a hydrogen atom. Such transitions belong to the Lyman series.

Using the Rydberg formula,

$$\frac{1}{\lambda}=R_HZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

For hydrogen, $$Z=1$$, $$n_1=1$$ and $$n_2=\infty$$.

Substituting these values,

$$\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)$$

Since,

$$\frac{1}{\infty^2}=0$$

the equation becomes

$$\frac{1}{\lambda}=1.097\times10^7\ \text{m}^{-1}$$

Therefore,

$$\lambda=\frac{1}{1.097\times10^7}\ \text{m}$$

$$\lambda=0.9115\times10^{-7}\ \text{m}$$

Converting into nanometers,

$$\lambda=0.9115\times10^{-7}\times10^9\ \text{nm}$$

$$\lambda=91.15\ \text{nm}$$

Hence, the wavelength of the emitted radiation is

$$\boxed{91\ \text{nm}}$$

Thus, the correct answer is Option (A).

In every nuclear decay, both the mass number $$A$$ and the atomic number $$Z$$ must balance on the two sides of the reaction.
  • $$\alpha$$-particle : $$\ ^4_2He \; ( \Delta A = -4,\; \Delta Z = -2)$$
  • $$\beta^-$$-particle : $$\ ^0_{-1}e \; ( \Delta A = 0,\; \Delta Z = +1)$$
  • $$\beta^+$$-particle : $$\ ^0_{+1}e \; ( \Delta A = 0,\; \Delta Z = -1)$$

$$^{238}_{92}U \;\longrightarrow\; {}^{234}_{91}Pa$$
Required changes: $$\Delta A = 238-234 = 4,\; \Delta Z = 92-91 = 1$$ (mass ↓4, atomic ↓1).
One $$\alpha$$ gives $$\Delta A = 4, \Delta Z = 2$$ too much fall in $$Z$$. Adding one $$\beta^-$$ raises $$Z$$ by 1, giving net $$\Delta A = 4, \Delta Z = 1$$ as needed.
Emission: one $$\alpha$$ + one $$\beta^-$$ ⇒ List-II (4).

$$^{214}_{82}Pb \;\longrightarrow\; {}^{210}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = 0$$.
One $$\alpha$$ accounts for $$\Delta A = 4, \Delta Z = 2$$. To restore $$Z$$, two $$\beta^-$$ (each +1) add $$\Delta Z = -2+2 = 0$$.
Emission: one $$\alpha$$ + two $$\beta^-$$ ⇒ List-II (3).

$$^{210}_{81}Tl \;\longrightarrow\; {}^{206}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = -1$$ (mass ↓4, atomic ↑1).
One $$\alpha$$ gives $$\Delta A = 4,\; \Delta Z = 2$$ too low atomic number. Three $$\beta^-$$ (total +3) correct it: $$-2 + 3 = +1$$.
Emission: one $$\alpha$$ + three $$\beta^-$$ ⇒ List-II (2).

$$^{228}_{91}Pa \;\longrightarrow\; {}^{224}_{88}Ra$$
Required changes: $$\Delta A = 4,\; \Delta Z = 3$$ (mass ↓4, atomic ↓3).
One $$\alpha$$ produces $$\Delta Z = 2$$; an extra drop of 1 is still needed. One $$\beta^+$$ (−1) gives the balance: $$-2 -1 = -3$$.
Emission: one $$\alpha$$ + one $$\beta^+$$ ⇒ List-II (1).

Thus the matches are:
P → 4,  Q → 3,  R → 2,  S → 1.

Option A which is: P → 4, Q → 3, R → 2, S → 1

Explanation

The ion contains 55 protons.

An element with atomic number

$$Z=55$$

is cesium, $$Cs$$.

The electronic configuration of neutral cesium is

$$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\ 6s^1$$

or

$$[Xe],6s^1$$

The ion is in the unipositive state, which means it has a charge of $$+1$$.

To form $$Cs^+$$, the outermost electron is removed from the $$6s$$ orbital.

Therefore, the electronic configuration of $$Cs^+$$ becomes

$$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6$$

Now count the electrons present in all the $$s$$ subshells.

From $$1s$$

$$2$$ electrons

From $$2s$$

$$2$$ electrons

From $$3s$$

$$2$$ electrons

From $$4s$$

$$2$$ electrons

From $$5s$$

$$2$$ electrons

The $$6s$$ orbital is empty in $$Cs^+$$.

Hence, the total number of $$s$$ electrons is

$$2+2+2+2+2=10$$

Therefore, the number of $$s$$ electrons present in the unipositive ion is

$$10$$

Hence, the correct answer is

$$10$$

For a hydrogen-like ion, the radius of the $$n^{\text{th}}$$ Bohr orbit is given by
$$r_n = \frac{n^2 a_0}{Z}$$ where $$a_0 = 52.9 \text{ pm}$$ is the Bohr radius and $$Z$$ is the atomic number.

For $$\text{He}^+$$ the nuclear charge is $$Z = 2$$.
First identify the quantum numbers corresponding to the two given radii.

Initial orbit (larger radius):
$$r_i = 105.8 \text{ pm}$$

$$n_i^2 = \frac{r_i Z}{a_0} = \frac{105.8 \times 2}{52.9} = 4$$
$$\Rightarrow \; n_i = 2$$

Final orbit (smaller radius):
$$r_f = 26.45 \text{ pm}$$

$$n_f^2 = \frac{r_f Z}{a_0} = \frac{26.45 \times 2}{52.9} = 1$$
$$\Rightarrow \; n_f = 1$$

The transition is therefore $$n = 2 \rightarrow n = 1$$.

For a hydrogen-like species, the energy of the $$n^{\text{th}}$$ orbit is
$$E_n = -\frac{Z^2 R_H}{n^2}$$ where the Rydberg constant in energy units is $$R_H = 2.2 \times 10^{-18}\,\text{J}$$.

Compute the energy difference:

$$E_2 = -\frac{(2)^2 R_H}{2^2} = -R_H$$
$$E_1 = -\frac{(2)^2 R_H}{1^2} = -4R_H$$

Energy of the emitted photon:
$$\Delta E = E_1 - E_2 = (-4R_H) - (-R_H) = -3R_H$$
Magnitude: $$|\Delta E| = 3R_H = 3 \times 2.2 \times 10^{-18} = 6.6 \times 10^{-18}\,\text{J}$$

Relate the photon energy to its wavelength $$\lambda$$ using $$E = \frac{hc}{\lambda}$$:

$$\lambda = \frac{hc}{E}$$

With $$h = 6.6 \times 10^{-34}\,\text{J s}$$ and $$c = 3.0 \times 10^{8}\,\text{m s}^{-1}$$,
$$hc = 6.6 \times 10^{-34} \times 3.0 \times 10^{8} = 1.98 \times 10^{-25}\,\text{J m}$$

$$\lambda = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-18}} = 3.0 \times 10^{-8}\,\text{m}$$

Convert metres to nanometres (1 nm = $$10^{-9}$$ m):
$$\lambda = 30\,\text{nm}$$

Hence, the wavelength of the emitted photon is 30 nm.

The correct statements are C and E.

Final Answer: 2

We need to find the radius of the third Bohr orbit of $$\text{He}^+$$ given that the radius of the first Bohr orbit of hydrogen is $$0.6 \, \text{\AA}$$.

The Bohr radius formula is $$r_n = \dfrac{n^2}{Z} \cdot a_0$$.

In this formula, $$n$$ is the orbit number, $$Z$$ is the atomic number, and $$a_0$$ is the first Bohr radius of hydrogen.

For $$\text{He}^+$$, we have $$Z = 2$$, $$n = 3$$, and $$a_0 = 0.6 \, \text{\AA}$$.

Substituting these values into the formula gives $$r_3 = \dfrac{3^2}{2} \times 0.6 = \dfrac{9}{2} \times 0.6 = 4.5 \times 0.6 = 2.7 \, \text{\AA}$$.

Since $$1 \, \text{\AA} = 100 \, \text{pm}$$, we find $$r_3 = 2.7 \times 100 = 270 \, \text{pm}$$.

The radius of the third Bohr orbit of $$\text{He}^+$$ is $$\boxed{270}$$ picometers.

We begin by considering the Paschen series, which involves transitions to $$n = 3$$.

First, for the 1st line (4→3), we write $$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R \times \frac{7}{144}$$, and hence $$\lambda_1 = 720$$ nm.

Next, for the 2nd line (5→3), we have $$\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{25}\right) = R \times \frac{16}{225}$$.

Then, substituting these into the ratio, we find $$\frac{\lambda_2}{\lambda_1} = \frac{7/144}{16/225} = \frac{7 \times 225}{144 \times 16} = \frac{1575}{2304}$$,
and therefore $$\lambda_2 = 720 \times \frac{1575}{2304} = 492 \text{ nm}$$.

Analyzing each statement:

A. For 1s orbital, probability density is maximum at the nucleus — True. The radial probability density $$|\psi|^2$$ for 1s is maximum at r = 0.

B. For 2s orbital, probability density first increases then decreases sharply to zero — False. The 2s orbital has a radial node where probability density goes to zero, then increases again.

C. Boundary surface diagrams enclose 100% probability — False. They typically enclose about 90-95% probability.

D. p and d orbitals have 1 and 2 angular nodes respectively — True. Angular nodes = l (l=1 for p, l=2 for d).

E. Probability density of p orbital is zero at the nucleus — True. The p orbital has a node at the nucleus.

Correct statements: A, D, E = 3

Analyzing each statement about atomic spectra:

A) Line emission spectra are used to study the electronic structure.

This is correct. Line spectra reveal the energy levels and electronic structure of atoms.

B) The emission spectra of atoms in the gas phase show a continuous spread of wavelength from red to violet.

This is incorrect. Atoms in the gas phase produce line spectra (discrete wavelengths), not continuous spectra. Continuous spectra are produced by hot solids or dense gases.

C) An absorption spectrum is like the photographic negative of an emission spectrum.

This is correct. Where emission spectra show bright lines against a dark background, absorption spectra show dark lines at the same wavelengths against a bright background.

D) The element helium was discovered in the sun by spectroscopic method.

This is correct. Helium was first identified in the solar spectrum by Lockyer and Janssen in 1868, before being found on Earth.

Only statement B is incorrect.

The number of incorrect statements is 1.

Let us analyze each statement:

Statement A: "A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit."

This is correct. In Bohr's model, electrons move in fixed circular orbits around the nucleus.

Statement B: "An orbital is the one electron wave function."

This is correct. In quantum mechanics, an atomic orbital is described by the wave function \(\psi\), which is a one-electron wave function.

Statement C: "The existence of Bohr's orbits is supported by hydrogen spectrum."

This is correct. Bohr's model successfully explained the hydrogen emission spectrum and the energy levels matched the observed spectral lines.

Statement D: "Atomic orbital is characterised by the quantum numbers n and l only."

This is incorrect. An atomic orbital is characterised by three quantum numbers: the principal quantum number \(n\), the azimuthal quantum number \(l\), and the magnetic quantum number \(m_l\). The magnetic quantum number \(m_l\) determines the orientation of the orbital in space, and different values of \(m_l\) correspond to different orbitals (e.g., \(p_x, p_y, p_z\)).

Therefore, the correct answer is Option D.

Threshold frequency of platinum: $$\nu_0 = 1.3 \times 10^{15}$$ s$$^{-1}$$

Planck's constant: $$h = 6.6 \times 10^{-34}$$ Js

The minimum energy required equals the work function, which is the energy corresponding to the threshold frequency:

$$E = h\nu_0$$

$$E = 6.6 \times 10^{-34} \times 1.3 \times 10^{15}$$

$$E = 6.6 \times 1.3 \times 10^{-34+15}$$

$$E = 8.58 \times 10^{-19} \text{ J}$$

Hence, the correct answer is Option A.

To find the pair of ions isoelectronic with $$Al^{3+}$$, note that aluminium (Al) has atomic number 13, so $$Al^{3+}$$ has $$13 - 3 = 10$$ electrons.

In Option A, $$Br^-$$ has $$35 + 1 = 36$$ electrons and $$Be^{2+}$$ has $$4 - 2 = 2$$ electrons, so neither has 10.
In Option B, $$Cl^-$$ has $$17 + 1 = 18$$ electrons and $$Li^+$$ has $$3 - 1 = 2$$ electrons, so neither has 10.
In Option C, $$S^{2-}$$ has $$16 + 2 = 18$$ electrons and $$K^+$$ has $$19 - 1 = 18$$ electrons, so neither has 10.
In Option D, $$O^{2-}$$ has $$8 + 2 = 10$$ electrons and $$Mg^{2+}$$ has $$12 - 2 = 10$$ electrons, so both have 10 electrons.

Since both $$O^{2-}$$ and $$Mg^{2+}$$ have the same number of electrons as $$Al^{3+}$$, they are isoelectronic. The correct answer is Option D.

We need to identify which pairs of electrons are in degenerate orbitals. Degenerate orbitals are orbitals that have the same energy. In a hydrogen-like atom, orbitals with the same $$n$$ are degenerate, but in multi-electron atoms, orbitals with the same $$n$$ and $$l$$ are degenerate (due to shielding effects).

Pair (A):

(a) $$n = 3, l = 1$$ → 3p orbital

(b) $$n = 3, l = 2$$ → 3d orbital

These are in different subshells (3p vs 3d). In a multi-electron atom, 3p and 3d have different energies. Not degenerate.

Pair (B):

(a) $$n = 3, l = 2, m_l = -2$$ → one of the 3d orbitals

(b) $$n = 3, l = 2, m_l = -1$$ → another of the 3d orbitals

Both electrons are in 3d orbitals (same $$n$$ and same $$l$$). Orbitals with the same $$n$$ and $$l$$ but different $$m_l$$ are degenerate. Degenerate.

Pair (C):

(a) $$n = 4, l = 2$$ → 4d orbital

(b) $$n = 3, l = 2$$ → 3d orbital

These have different principal quantum numbers (4d vs 3d). They have different energies. Not degenerate.

Only pair (B) has electrons in degenerate orbitals.

The correct answer is Option B: Only (B).

Let us analyze each statement.

Statement (A): "The principal quantum number 'n' is a positive integer with values of $$n = 1, 2, 3, \dots$$." This is correct. The principal quantum number is always a positive integer.

Statement (B): "The azimuthal quantum number 'l' for a given 'n' can have values as $$l = 0, 1, 2, \dots, n$$ and has $$(2n + 1)$$ values." This is incorrect because the azimuthal quantum number $$l$$ ranges from $$0$$ to $$(n - 1)$$, giving $$n$$ values, not $$(2n + 1)$$ values, and $$l$$ goes up to $$(n - 1)$$, not $$n$$.

Statement (C): "Magnetic orbital quantum number 'm' for a particular 'l' has $$(2l + 1)$$ values." This is correct. The magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$, giving $$(2l + 1)$$ values.

Statement (D): "$$\pm \frac{1}{2}$$ are the two possible orientations of electron spin." This is correct. The spin quantum number $$m_s$$ can be $$+\frac{1}{2}$$ or $$-\frac{1}{2}$$.

Statement (E): "For $$l = 5$$, there will be a total of 9 orbitals." This is incorrect because for $$l = 5$$, the number of orbitals is $$(2l + 1) = 2(5) + 1 = 11$$, not $$9$$.

The correct statements are (A), (C) and (D).

Hence, the correct answer is Option C.

We need to evaluate the assertion and reason about orbital energies of hydrogen and lithium.

The assertion (A) states that the energy of the $$2s$$ orbital of hydrogen atom is greater than that of the $$2s$$ orbital of lithium. The reason (R) given is that energies of the orbitals in the same subshell decrease with increase in the atomic number.

For hydrogen ($$Z = 1$$), the energy of the $$2s$$ orbital is:

$$E = -\frac{13.6}{n^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$$

For lithium ($$Z = 3$$), the effective nuclear charge experienced by the $$2s$$ electron is greater than 1 due to incomplete shielding. The $$2s$$ electron in lithium is more tightly bound (has more negative energy) than the $$2s$$ electron in hydrogen. Since the energy of the $$2s$$ orbital in hydrogen is less negative (higher/greater) than that in lithium, the assertion is true.

As atomic number increases, the nuclear charge increases. For the same subshell, the increased nuclear charge (even after accounting for shielding) pulls the electrons closer to the nucleus, making the energy more negative (lower). So the energy of orbitals in the same subshell decreases (becomes more negative) with increasing atomic number, and the reason is true.

The increased nuclear attraction in lithium directly explains why the $$2s$$ energy in hydrogen (lower $$Z$$) is higher (less negative) than in lithium (higher $$Z$$), so the reason correctly explains the assertion. The correct answer is Option A: Both A and R are true and R is the correct explanation of A.

Let us evaluate each statement:

(A) The electronic configuration of Cr is [Ar] 3d$$^5$$4s$$^1$$.

This is correct. Chromium has an exceptional configuration due to the extra stability of a half-filled 3d subshell.

(B) The magnetic quantum number may have a negative value.

This is correct. The magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$, so it can take negative values.

(C) In the ground state of an atom, the orbitals are filled in order of their increasing energy order.

This is correct. This is the Aufbau principle, which states that electrons fill orbitals starting from the lowest energy level.

(D) The total number of nodes are given by n - 2.

This is incorrect. The total number of nodes (radial + angular) is given by $$n - 1$$, not $$n - 2$$.

Therefore, statements (A), (B), and (C) are correct.

Hence, the correct answer is Option D ((A), (B) & (C) only).

Hydrogen has three isotopes: Protium ($$^1$$H) with mass number 1 (1 proton, 0 neutrons), Deuterium ($$^2$$H or D) with mass number 2 (1 proton, 1 neutron), and Tritium ($$^3$$H or T) with mass number 3 (1 proton, 2 neutrons).

Since they are isotopes of the same element (hydrogen), they have the same number of protons (1 each) — so Option A is incorrect; the same atomic number ($$Z = 1$$) — so Option B is incorrect; and the same electronic configuration ($$1s^1$$) — so Option C is incorrect.

They differ in the number of neutrons, which means they have different mass numbers and therefore different atomic masses.

This is why isotopes have nearly the same chemical properties (determined by electronic configuration) but different physical properties (affected by mass).

Hence, the correct answer is Option D (Atomic mass).

Let the edge length of the fcc unit cell be $$a$$ and the radius of every atom (both on lattice points and in the voids) be $$r$$.

Step 1 - Find the shortest centre-to-centre distance.
In an fcc lattice a tetrahedral void is situated, for example, at $$\left(\tfrac14,\tfrac14,\tfrac14\right)$$ while the nearest lattice atom is at the corner $$\left(0,0,0\right)$$. Their separation is $$d_{\text{tv-fcc}} = a\sqrt{\left(\tfrac14\right)^2+\left(\tfrac14\right)^2+\left(\tfrac14\right)^2} = a\frac{\sqrt3}{4} \;$$

The next smaller distance is between two lattice atoms (corner to face-centre): $$d_{\text{fcc-fcc}} = \frac{a}{\sqrt2}$$ Since $$\dfrac{a}{\sqrt2}\gt\dfrac{a\sqrt3}{4}$$, the shortest distance in the occupied arrangement is $$d_{\text{tv-fcc}} = a\sqrt3/4$$.

Step 2 - Relate $$r$$ to $$a$$.
For closest packing we let the atoms in a tetrahedral void just touch the nearest lattice atom: $$2r = \frac{a\sqrt3}{4}\;\Rightarrow\; r = \frac{a\sqrt3}{8}$$

Step 3 - Count the number of atoms per unit cell.
• fcc lattice sites: 8 corners $$\times$$ $$\frac18$$ + 6 faces $$\times$$ $$\frac12$$ = 4 atoms.
• Tetrahedral voids in one unit cell = 8; only alternate (half) are filled ⇒ 4 atoms.
Total atoms in the cell $$n = 4 + 4 = 8$$.

Step 4 - Volume occupied by all atoms.
Volume of one sphere $$= \frac43\pi r^{3}$$, therefore $$V_{\text{atoms}} = 8 \times \frac43\pi r^{3} = \frac{32}{3}\pi r^{3}$$ Substituting $$r = \dfrac{a\sqrt3}{8}$$: $$V_{\text{atoms}} = \frac{32}{3}\pi\left(\frac{a\sqrt3}{8}\right)^{3} = \frac{32}{3}\pi a^{3}\frac{(\sqrt3)^{3}}{8^{3}} = \frac{32}{3}\pi a^{3}\frac{3\sqrt3}{512} = \frac{\pi\sqrt3}{16}\,a^{3}$$

Step 5 - Packing efficiency.
Unit-cell volume $$= a^{3}$$. Packing efficiency $$\eta = \frac{V_{\text{atoms}}}{a^{3}}\times100\% = \frac{\pi\sqrt3}{16}\times100\% \approx \frac{3.1416\times1.732}{16}\times100\% \approx 34\%$$

The value calculated (≈34 %) is closest to 35 %.

Hence, Option B which is: 35 %

We have 1 $$\mu$$g of radioactive element X with a half-life of 30 years, and we need to find the amount remaining after 100 years, expressed as $$\_\_\_ \times 10^{-1}$$ $$\mu$$g.

We use the radioactive decay formula:

$$N = N_0 \times e^{-\lambda t}$$

where $$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{30}$$ per year. After $$t = 100$$ years:

$$N = 1 \times e^{-\frac{0.693}{30} \times 100} = e^{-\frac{69.3}{30}} = e^{-2.31}$$

Now we evaluate $$e^{-2.31}$$. We know that $$\ln 10 = 2.303$$, so $$e^{-2.303} = 10^{-1} = 0.1$$. Since $$2.31$$ is very close to $$2.303$$, we can use the given values more precisely.

Using the alternative form: $$N = N_0 \times \left(\frac{1}{2}\right)^{t/t_{1/2}} = 1 \times \left(\frac{1}{2}\right)^{100/30} = \left(\frac{1}{2}\right)^{10/3}$$

Taking logarithm: $$\log N = -\frac{10}{3} \times \log 2 = -\frac{10}{3} \times 0.30 = -1.0$$

Therefore $$N = 10^{-1.0} = 0.1$$ $$\mu$$g.

Expressing this as $$\_\_\_ \times 10^{-1}$$ $$\mu$$g, we get $$0.1 = 1 \times 10^{-1}$$ $$\mu$$g.

Hence, the correct answer is 1.

To find the total energy of the electron revolving in a hydrogen atom, we consider the Bohr model where the electron moves in a circular orbit of radius $$r$$ around the proton. The electron has mass $$m$$ and charge $$e$$, while the proton has charge $$+e$$. The electrostatic force of attraction provides the necessary centripetal force for circular motion.

The electrostatic force between the electron and proton is given by Coulomb's law. Since the charges are opposite, the magnitude is $$\frac{1}{4\pi\epsilon_0} \frac{e \cdot e}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$$. This force acts as the centripetal force, which is $$\frac{mv^2}{r}$$, where $$v$$ is the orbital speed of the electron.

Setting these equal:

$$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$$

Multiply both sides by $$r$$ to simplify:

$$mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

Solve for $$v^2$$:

$$v^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r}$$

The kinetic energy $$K$$ of the electron is $$\frac{1}{2} m v^2$$. Substitute the expression for $$v^2$$:

$$K = \frac{1}{2} m \left( \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r} \right)$$

The $$m$$ cancels out:

$$K = \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

The potential energy $$U$$ for the electrostatic attraction is negative and given by $$-\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$.

The total energy $$E$$ is the sum of kinetic and potential energy:

$$E = K + U = \left( \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right) + \left( -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right)$$

Factor out $$\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$:

$$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \left( \frac{1}{2} - 1 \right)$$

Simplify the term in parentheses:

$$\frac{1}{2} - 1 = -\frac{1}{2}$$

So:

$$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \times \left( -\frac{1}{2} \right) = -\frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

In atomic physics, it is common to use units where the constant $$\frac{1}{4\pi\epsilon_0}$$ is set to 1 (as in cgs units). Thus, the expression simplifies to:

$$E = -\frac{1}{2} \frac{e^2}{r}$$

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.

For a $$d$$-subshell ($$l = 2$$), there are $$2l + 1 = 5$$ possible orbitals. These are represented by the magnetic quantum numbers:

$$m_l = -2, -1, 0, +1, +2$$

Each unique value of $$m_l$$ corresponds to exactly one specific orbital. Since the question specifies a unique value for $$m_l$$ ($$+2$$), it refers to only one orbital within the $$3d$$ subshell.

The de Broglie wavelength $$\lambda$$ is given by the formula $$\lambda = \frac{h}{m v}$$, where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity.

Given: mass $$m = 1000 \text{kg}$$, velocity $$v = 36 \text{km/hr}$$.

First, convert velocity from $$\text{km/hr}$$ to $$\text{m/s}$$ because SI units are required. We know that $$1 \text{km} = 1000 \text{m}$$ and $$1 \text{hour} = 3600 \text{seconds}$$. So,

$$v = 36 \frac{\text{km}}{\text{hr}} \times \frac{1000 \text{m}}{1 \text{km}} \times \frac{1 \text{hr}}{3600 \text{s}}$$

Simplify the conversion:

$$v = 36 \times \frac{1000}{3600} \text{m/s}$$

Reduce the fraction $$\frac{1000}{3600}$$ by dividing numerator and denominator by 200:

$$\frac{1000 \div 200}{3600 \div 200} = \frac{5}{18}$$

So,

$$v = 36 \times \frac{5}{18} \text{m/s}$$

Now, $$36 \div 18 = 2$$, so:

$$v = 2 \times 5 = 10 \text{m/s}$$

Now, Planck's constant $$h = 6.626 \times 10^{-34} \text{J s}$$, and $$1 \text{J} = 1 \text{kg m}^2 \text{s}^{-2}$$, so $$h = 6.626 \times 10^{-34} \text{kg m}^2 \text{s}^{-1}$$.

Substitute the values into the de Broglie formula:

$$\lambda = \frac{h}{m v} = \frac{6.626 \times 10^{-34}}{1000 \times 10}$$

First, compute the denominator:

$$m v = 1000 \times 10 = 10000 = 10^4$$

So,

$$\lambda = \frac{6.626 \times 10^{-34}}{10^4}$$

Dividing by $$10^4$$ is equivalent to multiplying by $$10^{-4}$$:

$$\lambda = 6.626 \times 10^{-34} \times 10^{-4}$$

When multiplying powers of 10, add the exponents:

$$10^{-34} \times 10^{-4} = 10^{-34 + (-4)} = 10^{-38}$$

Therefore,

$$\lambda = 6.626 \times 10^{-38} \text{m}$$

Comparing with the options:

• A. $$6.626 \times 10^{-34} \text{m}$$
• B. $$6.626 \times 10^{-38} \text{m}$$
• C. $$6.626 \times 10^{-31} \text{m}$$
• D. $$6.626 \times 10^{-30} \text{m}$$

The calculated wavelength matches option B.

Hence, the correct answer is Option B.

The wave number for a spectral line in the hydrogen spectrum is given by the Rydberg formula:

$$\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

where $$\bar{\nu}$$ is the wave number, $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the higher energy level.

For the Balmer series, the electron transitions end at the $$n = 2$$ level. Therefore, $$n_1 = 2$$. The first emission line corresponds to the smallest energy transition, which is from the next higher level, $$n_2 = 3$$, down to $$n_1 = 2$$.

Substituting $$n_1 = 2$$ and $$n_2 = 3$$ into the formula:

$$\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$

Calculate the squares: $$2^2 = 4$$ and $$3^2 = 9$$, so:

$$\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{9} \right)$$

To subtract these fractions, find a common denominator. The least common multiple of 4 and 9 is 36. Rewrite each fraction:

$$\frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36}$$

Now subtract:

$$\frac{9}{36} - \frac{4}{36} = \frac{5}{36}$$

Therefore:

$$\bar{\nu} = R \times \frac{5}{36} = \frac{5}{36}R$$

Comparing with the options, $$\frac{5}{36}R$$ corresponds to option A.

Hence, the correct answer is Option A.

To solve this problem, we need to determine the maximum number of electrons in an atom that can have the quantum numbers specified in parts (A) and (B). Quantum numbers describe the properties of atomic orbitals and the electrons within them. The principal quantum number $$n$$ indicates the shell, the azimuthal quantum number $$l$$ indicates the subshell (where $$l$$ ranges from 0 to $$n-1$$), the magnetic quantum number $$m_l$$ indicates the orbital (where $$m_l$$ ranges from $$-l$$ to $$+l$$), and the spin quantum number $$m_s$$ indicates the electron spin (either $$+\frac{1}{2}$$ or $$-\frac{1}{2}$$).

Starting with part (A): given $$n = 5$$ and $$m_l = +1$$. We are not given $$l$$ or $$m_s$$, so we must consider all possible values of $$l$$ for $$n = 5$$. Since $$n = 5$$, $$l$$ can be 0, 1, 2, 3, or 4.

For each $$l$$, we check if $$m_l = +1$$ is possible. The value of $$m_l$$ must satisfy $$-l \leq m_l \leq +l$$:

• If $$l = 0$$, $$m_l$$ can only be 0. So $$m_l = +1$$ is not possible.
• If $$l = 1$$, $$m_l$$ can be $$-1, 0, +1$$. So $$m_l = +1$$ is possible.
• If $$l = 2$$, $$m_l$$ can be $$-2, -1, 0, +1, +2$$. So $$m_l = +1$$ is possible.
• If $$l = 3$$, $$m_l$$ can be $$-3, -2, -1, 0, +1, +2, +3$$. So $$m_l = +1$$ is possible.
• If $$l = 4$$, $$m_l$$ can be $$-4, -3, -2, -1, 0, +1, +2, +3, +4$$. So $$m_l = +1$$ is possible.

Thus, $$m_l = +1$$ is possible for $$l = 1, 2, 3, 4$$. For each of these $$l$$ values, there is exactly one orbital with $$m_l = +1$$. Each orbital can hold up to 2 electrons (one with $$m_s = +\frac{1}{2}$$ and one with $$m_s = -\frac{1}{2}$$). Therefore:

• For $$l = 1$$, the orbital with $$m_l = +1$$ can hold 2 electrons.
• For $$l = 2$$, the orbital with $$m_l = +1$$ can hold 2 electrons.
• For $$l = 3$$, the orbital with $$m_l = +1$$ can hold 2 electrons.
• For $$l = 4$$, the orbital with $$m_l = +1$$ can hold 2 electrons.

Adding these together: $$2 + 2 + 2 + 2 = 8$$ electrons. Hence, the maximum number of electrons with $$n = 5$$ and $$m_l = +1$$ is 8.

Now for part (B): given $$n = 2$$, $$l = 1$$, $$m_l = -1$$, and $$m_s = -\frac{1}{2}$$. Here, all four quantum numbers are specified. This uniquely identifies a specific orbital (defined by $$n$$, $$l$$, and $$m_l$$) and a specific spin state ($$m_s$$).

For $$n = 2$$ and $$l = 1$$, the possible $$m_l$$ values are $$-1, 0, +1$$. The orbital with $$m_l = -1$$ is one specific orbital. Each orbital can hold a maximum of 2 electrons, but they must have different spins: one with $$m_s = +\frac{1}{2}$$ and one with $$m_s = -\frac{1}{2}$$. Since we are specifying $$m_s = -\frac{1}{2}$$, there can be only one electron in this atom that has exactly these quantum numbers: $$n = 2$$, $$l = 1$$, $$m_l = -1$$, and $$m_s = -\frac{1}{2}$$.

Therefore, the maximum number of electrons for part (B) is 1.

Combining the results, the maximum number of electrons for (A) is 8 and for (B) is 1. Comparing with the options:

• A. 25 and 1
• B. 8 and 1
• C. 2 and 4
• D. 4 and 1

Hence, the correct answer is Option B.

First, we need to understand what heavy water is. Heavy water is deuterium oxide, which is represented as $$ \text{D}_2\text{O} $$. This means one molecule of heavy water consists of two deuterium atoms and one oxygen atom.

Deuterium is an isotope of hydrogen. The atomic number of hydrogen is $$ 1 $$, which means it has $$ 1 $$ proton in its nucleus. Deuterium has a mass number of $$ 2 $$, so it has $$ 1 $$ proton and $$ 1 $$ neutron (since mass number = protons + neutrons, so neutrons = mass number - protons = $$ 2 - 1 = 1 $$). As a neutral atom, deuterium also has $$ 1 $$ electron.

Oxygen has an atomic number of $$ 8 $$, so it has $$ 8 $$ protons in its nucleus. The most common isotope of oxygen in heavy water is oxygen-16, with a mass number of $$ 16 $$. Therefore, it has $$ 16 - 8 = 8 $$ neutrons. As a neutral atom, oxygen has $$ 8 $$ electrons.

Now, for a molecule of heavy water, $$ \text{D}_2\text{O} $$, we calculate the total number of protons, electrons, and neutrons by adding the contributions from the two deuterium atoms and the one oxygen atom.

Total protons: Protons are in the nucleus and do not change when atoms form a molecule. Each deuterium atom has $$ 1 $$ proton, so two deuterium atoms contribute $$ 2 \times 1 = 2 $$ protons. The oxygen atom has $$ 8 $$ protons. Therefore, total protons = $$ 2 + 8 = 10 $$.

Total neutrons: Neutrons are also in the nucleus and do not change. Each deuterium atom has $$ 1 $$ neutron, so two deuterium atoms contribute $$ 2 \times 1 = 2 $$ neutrons. The oxygen atom has $$ 8 $$ neutrons. Therefore, total neutrons = $$ 2 + 8 = 10 $$.

Total electrons: In a neutral molecule, the total number of electrons equals the total number of protons, which is $$ 10 $$. We can also calculate it by adding the electrons from each atom. Each deuterium atom contributes $$ 1 $$ electron, so two deuterium atoms contribute $$ 2 \times 1 = 2 $$ electrons. The oxygen atom contributes $$ 8 $$ electrons. Therefore, total electrons = $$ 2 + 8 = 10 $$.

So, the numbers of protons, electrons, and neutrons in a molecule of heavy water are $$ 10 $$, $$ 10 $$, and $$ 10 $$ respectively.

Now, comparing with the options:

A. 8, 10, 11

B. 10, 10, 10

C. 10, 11, 10

D. 11, 10, 10

Option B matches our calculation.

Hence, the correct answer is Option B.

For one-electron atoms and for multishell electrons in multielectron atoms, the relative energy of an orbital is decided by the $$n+l$$ rule (also called the Aufbau principle).

Rule:
1. Orbitals having the smaller value of $$(n+l)$$ possess lower energy.
2. If two orbitals have the same $$(n+l)$$ value, the orbital with the smaller principal quantum number $$n$$ is lower in energy.

Compute $$(n+l)$$ for each given set of quantum numbers:

(a) $$n=4,\;l=1 \;(\text{4p}) \;\Rightarrow\; n+l = 4+1 = 5$$
(b) $$n=4,\;l=0 \;(\text{4s}) \;\Rightarrow\; n+l = 4+0 = 4$$
(c) $$n=3,\;l=2 \;(\text{3d}) \;\Rightarrow\; n+l = 3+2 = 5$$
(d) $$n=3,\;l=1 \;(\text{3p}) \;\Rightarrow\; n+l = 3+1 = 4$$

Step 1 - Compare $$(n+l)$$ values:
• The smallest value is 4 → orbitals (b) and (d).
• The next value is 5 → orbitals (a) and (c).

Step 2 - Tie-break when $$(n+l)$$ is equal:
For $$(n+l)=4$$: choose the smaller $$n$$.
  • (d) has $$n=3$$, (b) has $$n=4$$, so (d) lies lower than (b).
For $$(n+l)=5$$: again choose the smaller $$n$$.
  • (c) has $$n=3$$, (a) has $$n=4$$, so (c) lies lower than (a).

Overall increasing energy order:
$$\text{(d)} \lt \text{(b)} \lt \text{(c)} \lt \text{(a)}$$

Thus the correct option is:
Option B which is: (d) < (b) < (c) < (a)

For multi-electron atoms the energy of a subshell is decided by the $$\,(n+l)\,$ rule (also called the Madelung rule):

1. Subshells with lower $$n+l$$ have lower energy.
2. If two subshells have the same $$n+l$$, the subshell with lower $$n$$ is lower in energy.

List each set of quantum numbers, identify the subshell and calculate $$n+l$$:

• (i) $$n = 4,\; l = 1 \;\;(4p)$$ → $$n+l = 4+1 = 5$$

• (ii) $$n = 4,\; l = 0 \;\;(4s)$$ → $$n+l = 4+0 = 4$$

• (iii) $$n = 3,\; l = 2 \;\;(3d)$$ → $$n+l = 3+2 = 5$$

• (iv) $$n = 3,\; l = 1 \;\;(3p)$$ → $$n+l = 3+1 = 4$$

Arrange by increasing $$n+l$$ first:

• $$n+l = 4$$: 4s (ii) and 3p (iv)
• $$n+l = 5$$: 3d (iii) and 4p (i)

For ties, compare $$n$$:

• Between 3p (iv) and 4s (ii): lower $$n$$ (3) means 3p is lower in energy than 4s.

• Between 3d (iii) and 4p (i): lower $$n$$ (3) means 3d is lower in energy than 4p.

Hence the overall increasing energy order is

$$3p \lt 4s \lt 3d \lt 4p$$

Writing this sequence with the given Roman numerals:

$$(iv) \lt (ii) \lt (iii) \lt (i)$$

Therefore, the correct choice is:
Option B which is: (iv) < (ii) < (iii) < (i)

For every hydrogen-like species (one-electron ion) the Bohr energy expression is

$$E_n = -\frac{Z^{2}}{n^{2}}\;E_H$$

Here $$E_H$$ is the magnitude of the ground-state energy of the hydrogen atom (which equals the ionisation energy of hydrogen), $$Z$$ is the nuclear charge and $$n$$ is the principal quantum number.

Step 1 Extract $$E_H$$ from the given data
For $$\text{He}^+$$, $$Z = 2$$ and for the ground state $$n = 1$$. The ionisation energy given in the question is

$$|\;E_1(\text{He}^+)| = 19.6 \times 10^{-18}\;\text{J atom}^{-1}$$

But $$|\;E_1(\text{He}^+)| = \frac{Z^{2}}{n^{2}}\,E_H = 4\,E_H$$ (because $$Z=2,\;n=1$$).

Therefore

$$E_H = \frac{19.6 \times 10^{-18}}{4} = 4.9 \times 10^{-18}\;\text{J atom}^{-1}$$

Step 2 Energy of the first stationary state of $$\text{Li}^{2+}$$
For $$\text{Li}^{2+}$$, $$Z = 3$$ and in the first Bohr orbit $$n = 1$$. Applying the formula,

$$E_1(\text{Li}^{2+}) = -\frac{Z^{2}}{n^{2}}\,E_H = -9\,E_H$$

$$\Rightarrow\;E_1(\text{Li}^{2+}) = -9 \times 4.9 \times 10^{-18}\; \text{J}$$

$$E_1(\text{Li}^{2+}) = -44.1 \times 10^{-18}\;\text{J} = -4.41 \times 10^{-17}\;\text{J atom}^{-1}$$

Result: the energy of the first stationary state of $$\text{Li}^{2+}$$ is $$-4.41 \times 10^{-17}\;\text{J atom}^{-1}$$.

Hence, the correct choice is
Option B which is: $$-4.41 \times 10^{-17}\;\text{J atom}^{-1}$$

Energy of dissociation is given for one mole of $$Cl-Cl$$ bonds: $$D = 242\text{ kJ mol}^{-1} = 242 \times 10^{3}\text{ J mol}^{-1}$$.

Step 1 - convert this to energy per single bond.
Energy per bond $$E = \dfrac{D}{N_A} = \dfrac{242 \times 10^{3}\ \text{J mol}^{-1}}{6.02 \times 10^{23}\ \text{mol}^{-1}}$$.

Evaluating, $$E \approx 4.02 \times 10^{-19}\text{ J}$$.

Step 2 - relate photon energy to its wavelength.
For one photon, $$E = h c / \lambda$$, where $$h = 6.626 \times 10^{-34}\text{ J s}$$ and $$c = 3.00 \times 10^{8}\text{ m s}^{-1}$$.

Hence $$\lambda = \dfrac{h c}{E} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.02 \times 10^{-19}} \text{ m}$$.

Calculating, $$\lambda \approx 4.94 \times 10^{-7}\text{ m}$$.

Since $$1\text{ nm} = 10^{-9}\text{ m}$$, $$\lambda \approx 494\text{ nm}$$.

The longest (minimum-energy) wavelength capable of breaking a single $$Cl-Cl$$ bond is therefore $$\mathbf{494\ nm}$$.

Option D which is: $$494\ \text{nm}$$

For multi-electron atoms, the relative energy of subshells is predicted by the $$n + \ell$$ rule.

1. Arrange subshells in increasing order of $$n + \ell$$ (principal quantum number $$n$$ plus azimuthal quantum number $$\ell$$).
2. If two subshells have the same $$n + \ell$$ value, the subshell with the larger $$n$$ lies higher in energy.

Compute $$n + \ell$$ for each option:

Option A $$n = 3,\; \ell = 2$$ ⟹ $$n + \ell = 3 + 2 = 5$$

Option B $$n = 4,\; \ell = 2$$ ⟹ $$n + \ell = 4 + 2 = 6$$

Option C $$n = 4,\; \ell = 0$$ ⟹ $$n + \ell = 4 + 0 = 4$$

Option D $$n = 3,\; \ell = 0$$ ⟹ $$n + \ell = 3 + 0 = 3$$

• The largest $$n + \ell$$ value is $$6$$ for Option B.
• No other option ties with this value, so no further comparison is needed.

Hence the subshell with the highest energy among the given choices is described by Option B.

Option B which is: $$n = 4, \ell = 2, m = 1, s = +1/2$$

Bohr postulated that the angular momentum $$L$$ of an electron in a stationary orbit is quantised according to

$$L = n \dfrac{h}{2\pi}$$

where

• $$n = 1,2,3,\dots$$ is the principal quantum number (orbit number).
• $$h$$ is Planck’s constant.

For the $$5^{\text{th}}$$ orbit we have $$n = 5$$, therefore

$$L = 5 \dfrac{h}{2\pi} = \dfrac{5}{2}\dfrac{h}{\pi} = 2.5\dfrac{h}{\pi}$$

Hence the angular momentum of the electron in the fifth Bohr orbit is $$2.5\dfrac{h}{\pi}$$.

Option D which is: $$2.5\dfrac{h}{\pi}$$

Heisenberg uncertainty principle: $$\Delta x\,\Delta p \ge \frac{h}{4\pi}$$.

Given velocity $$v = 300\,\text{m s}^{-1}$$ and accuracy $$0.001\%$$.
Percentage $$0.001\% = 0.001/100 = 1 \times 10^{-5}$$.

Uncertainty in velocity:
$$\Delta v = v \times 10^{-5} = 300 \times 10^{-5}\,\text{m s}^{-1} = 3.0 \times 10^{-3}\,\text{m s}^{-1}.$$

Electron mass: $$m = 9.1 \times 10^{-31}\,\text{kg}$$.
Uncertainty in momentum:
$$\Delta p = m\,\Delta v = (9.1 \times 10^{-31})(3.0 \times 10^{-3})$$
$$\;\;\; = 2.73 \times 10^{-33}\,\text{kg m s}^{-1}.$$

Hence uncertainty in position:
$$\Delta x = \frac{h}{4\pi\,\Delta p} = \frac{6.63 \times 10^{-34}}{4\pi \times 2.73 \times 10^{-33}}$$
$$= \frac{6.63 \times 10^{-34}}{12.56 \times 2.73 \times 10^{-33}}$$
$$= \frac{6.63}{34.3} \times 10^{-34+33} \approx 1.93 \times 10^{-2}\,\text{m}.$$

Therefore the uncertainty in position is $$\boxed{1.92 \times 10^{-2}\,\text{m}}$$.

Option C which is: $$1.92 \times 10^{-2}\,m$$.