JEE Atomic Structure Questions

For hydrogen atom, energy depends only on n (principal quantum number), not on l.

3p, 3d all have n=3 (same energy). 4s has n=4 (higher energy).

So 3px, 3d_{x²-y²}, 3d_{z²} all have the lowest energy among the options (n=3).

The correct answer is Option 1: (B), (C) and (D) only.

We need to find the radius of the first excited state of the Helium ion (He⁺).

The radius of the nth orbit for a hydrogen-like atom is:

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

Here, $$a_0$$ is the Bohr radius (the radius of the first orbit of hydrogen) and Z is the atomic number.

For He⁺, Z = 2, and the first excited state corresponds to n = 2. Substituting these values gives

$$r = \frac{(2)^2 \times a_0}{2} = \frac{4a_0}{2} = 2a_0$$

The correct answer is Option 2: $$r = 2a_0$$.

First recall the basic postulates of Bohr for the hydrogen atom:

(i) The electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of $$\hbar$$ and the energy of the atom is constant in each allowed orbit (stationary state).
(ii) While the electron stays in any one stationary state it does not radiate energy.
(iii) Energy is emitted or absorbed only when the electron jumps between two stationary states, the photon frequency being fixed by the energy difference of the two states.
(iv) Inside each stationary state the electron actually follows a definite circular path of radius $$r_n$$ around the nucleus.

Now compare these statements with the principles of the modern quantum-mechanical (wave-mechanical) model:

(a) Quantum mechanics also allows only discrete energies for bound electrons, hence statements about distinct energies and stationary states remain valid.
(b) The time-dependent Schrödinger equation shows that if an electron is in an energy eigen-state (stationary state) its probability density is time-independent, so no radiation is emitted — fully agreeing with Bohr’s “no radiation in a stationary state”.
(c) When an electron undergoes a transition between two energy eigen-states, quantum mechanics predicts emission or absorption of a photon whose energy equals the energy difference — matching Bohr’s third postulate.
(d) However, quantum mechanics does not assign a precise trajectory (circular or otherwise) to the electron. Instead, it gives only a probability distribution described by the wave function $$\psi(\mathbf{r})$$. Hence the idea of the electron moving in a well-defined circular orbit is incompatible with the uncertainty principle and the wave nature of the electron.

Examine each option in light of this comparison:

Option A: “An atom in a stationary state does not emit electromagnetic radiation …” — this is fully consistent with quantum mechanics.
Option B: “An atom can take only certain distinct energies $$E_1, E_2, E_3, …$$” — also consistent with quantum mechanics.
Option C: “When an electron makes a transition … it emits a photon …” — again agrees with quantum mechanics.
Option D: “The electron in a H atom’s stationary state moves in a circle around the nucleus” — this assumes a definite trajectory, which is ruled out by the quantum-mechanical picture.

Therefore, the postulate that contradicts the quantum mechanical model is Option D.

Final Answer: Option D

In Bohr’s model of the hydrogen atom the radius of the $$n^{\text{th}}$$ orbit is

$$r_n = r_1 n^{2}$$

where $$r_1$$ is the radius of the first orbit. Hence the ratio of radii of any two orbits is

$$\frac{r_n}{r_m} = \left(\frac{n}{m}\right)^{2}$$

Using this relation for every option:

Case A:
$$\frac{r_3}{r_1} = \left(\frac{3}{1}\right)^{2} = 9$$.
The radius of the 3rd orbit is nine times that of the 1st orbit.
Statement A is correct.

Case B:
$$\frac{r_8}{r_4} = \left(\frac{8}{4}\right)^{2} = 2^{2} = 4$$.
The radius of the 8th orbit is four times that of the 4th orbit.
Statement B is correct.

Case C:
$$\frac{r_6}{r_4} = \left(\frac{6}{4}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} = 2.25$$.
The radius of the 6th orbit is only 2.25 times that of the 4th orbit, not three times.
Statement C is incorrect.

Case D:
$$\frac{r_4}{r_2} = \left(\frac{4}{2}\right)^{2} = 2^{2} = 4$$.
The radius of the 4th orbit is four times that of the 2nd orbit.
Statement D is correct.

Only Statement C contradicts the Bohr radius relation. Therefore the incorrect statement is

Option C.

For any electron in an atomic orbital, the magnitude of the orbital angular momentum $$L$$ is given by
$$L = \sqrt{l(l+1)}\,\hbar$$ where $$l$$ is the azimuthal (orbital) quantum number and $$\hbar = \dfrac{h}{2\pi}$$.

Case 1: Electron in the $$2s$$ orbital
For an $$s$$-orbital, $$l = 0$$.
Therefore,
$$L_{2s} = \sqrt{0(0+1)}\,\hbar = 0$$.

Case 2: Electron in the $$2p$$ orbital
For a $$p$$-orbital, $$l = 1$$.
Therefore,
$$L_{2p} = \sqrt{1(1+1)}\,\hbar = \sqrt{2}\,\hbar = \sqrt{2}\,\dfrac{h}{2\pi}$$.

So the orbital angular momenta are 0 for $$2s$$ and $$\sqrt{2}\,\dfrac{h}{2\pi}$$ for $$2p$$.

Hence, the correct option is Option D.

Evaluate two statements about atomic orbitals.

Statement I: For a given shell, the total number of allowed orbitals is $$n^2$$.

For shell n, the subshells are l = 0, 1, 2, ..., (n-1). Each subshell has (2l+1) orbitals.

Total = $$\sum_{l=0}^{n-1}(2l+1) = n^2$$. This is TRUE.

Statement II: For any subshell, the spatial orientation is given by -l to +l values including zero.

The magnetic quantum number $$m_l$$ ranges from -l to +l. This gives (2l+1) values representing different spatial orientations. This is TRUE.

The correct answer is Option 3: Both Statement I and Statement II are true.

In a multi-electron atom (without external fields), we need to identify which orbitals have the same energy.

We start by recalling the energy ordering in multi-electron atoms. In multi-electron atoms, the energy of an orbital depends on both $$n$$ and $$l$$ (unlike hydrogen where it depends only on $$n$$). However, orbitals with the same $$n$$ and $$l$$ but different $$m_l$$ are degenerate (same energy) in the absence of external electric and magnetic fields.

Next, we identify the orbitals: A. $$n=1, l=0, m_l=0$$ — this is the $$1s$$ orbital; B. $$n=2, l=0, m_l=0$$ — this is the $$2s$$ orbital; C. $$n=2, l=1, m_l=1$$ — this is one of the $$2p$$ orbitals; D. $$n=3, l=2, m_l=1$$ — this is one of the $$3d$$ orbitals; E. $$n=3, l=2, m_l=0$$ — this is another $$3d$$ orbital.

Substituting these values, D and E both have $$n = 3$$ and $$l = 2$$ (both are $$3d$$ orbitals), differing only in $$m_l$$; they are degenerate. No other pair shares the same $$(n, l)$$ values.

The correct answer is Option D) D and E Only.

Evaluate two statements about X-ray spectra (Moseley's law).

Moseley's law: $$\sqrt{\nu} = a(Z - b)$$, where Z is atomic number, and a, b are constants.

Statement I: A plot of $$\sqrt{\nu}$$ vs atomic mass is a straight line.

Moseley's law relates $$\sqrt{\nu}$$ to atomic number Z, not atomic mass. Since atomic mass and Z don't have a perfectly linear relationship, this is FALSE.

Statement II: A plot of $$\nu$$ vs atomic number is a straight line.

From Moseley's law: $$\sqrt{\nu} = a(Z-b)$$, so $$\nu = a^2(Z-b)^2$$. This is a parabolic (quadratic) relationship, not linear. So this is FALSE.

The correct answer is Option 3: Both Statement I and Statement II are false.

We are given two statements about atomic orbitals and transitions. We need to evaluate their truth and choose the correct option.

First, consider Statement (II): $$2p_{x}$$ and $$2p_{y}$$ are degenerate orbitals.

Degenerate orbitals have the same energy. In hydrogen-like atoms (without external fields), orbitals with the same principal quantum number $$n$$ and azimuthal quantum number $$l$$ are degenerate. For $$n=2$$ and $$l=1$$ (p-orbitals), the orbitals $$2p_x$$, $$2p_y$$, and $$2p_z$$ all have identical energy. Therefore, $$2p_x$$ and $$2p_y$$ are degenerate.

Thus, Statement (II) is true.

Now, consider Statement (I): A spectral line will be observed for a $$2p_{x}\rightarrow 2p_{y}$$ transition.

Spectral lines arise from transitions between different energy levels, where a photon is emitted or absorbed due to a non-zero energy difference. Since $$2p_x$$ and $$2p_y$$ are degenerate (same energy), the energy difference $$\Delta E = 0$$. Hence, no photon is emitted or absorbed, and no spectral line is observed.

Additionally, electric dipole transitions follow selection rules. The change in azimuthal quantum number must be $$\Delta l = \pm 1$$. Here, both orbitals have $$l=1$$, so $$\Delta l = 0$$, which violates the selection rule. This confirms the transition is forbidden.

Therefore, Statement (I) is false.

In summary:

• Statement (I) is false.
• Statement (II) is true.

The correct option is B: Statement I is false but Statement II is true.

We need to find the de-Broglie wavelength of the electron in the second orbit of hydrogen atom in terms of the Bohr radius $$a_0$$.

We start by recalling the Bohr model relations. From angular momentum quantization: $$mv_n r_n = n\hbar$$, where $$\hbar = h/(2\pi)$$. The Bohr radius of the $$n$$th orbit is given by $$r_n = n^2 a_0$$.

Next, from $$m v_n r_n = n\hbar = \frac{nh}{2\pi}$$, we have $$ m v_n = \frac{nh}{2\pi r_n} = \frac{nh}{2\pi \cdot n^2 a_0} = \frac{h}{2\pi n a_0} $$. This gives the momentum of the electron in the $$n$$th orbit.

Substituting this into the de-Broglie relation $$\lambda = \frac{h}{mv_n}$$ leads to $$ \lambda = \frac{h}{\frac{h}{2\pi n a_0}} = 2\pi n a_0 $$.

Therefore, for the second orbit where $$n = 2$$, $$ \lambda = 2\pi \times 2 \times a_0 = 4\pi a_0 $$. This can also be written as $$\frac{8\pi a_0}{2} = \frac{8\pi a_0}{n}$$.

Looking at the options, Option A gives $$\frac{8\pi a_0}{n}$$, which for $$n = 2$$ becomes $$\frac{8\pi a_0}{2} = 4\pi a_0$$, matching the result.

The correct answer is Option A: $$\frac{8\pi a_0}{n}$$.

The azimuthal quantum number $$l$$ decides the subshell:
$$l = 0 \;(\text{s}), \; l = 1 \;(\text{p}), \; l = 2 \;(\text{d}), \; l = 3 \;(\text{f})$$

The ground-state electronic configuration of a chromium atom ($$Z = 24$$) is exceptional. Instead of the expected $$3d^4\,4s^2$$, chromium attains extra stability with the half-filled $$3d$$ subshell:

$$\text{Cr} : [\text{Ar}]\,3d^{5}\,4s^{1}$$

Expand the configuration to list all occupied subshells:

$$1s^{2}\;2s^{2}\;2p^{6}\;3s^{2}\;3p^{6}\;3d^{5}\;4s^{1}$$

The $$p$$ subshells present are $$2p^{6}$$ and $$3p^{6}$$.

Total $$p$$-electrons $$= 6 + 6 = 12$$.

The only $$d$$ subshell present is $$3d^{5}$$.

Total $$d$$-electrons $$= 5$$.

Hence, in the ground state of chromium, the numbers of electrons with $$l = 1$$ and $$l = 2$$ are $$12$$ and $$5$$ respectively.

Therefore the correct option is Option C.

The atomic number $$Z = 9$$ corresponds to fluorine. Its ground-state electronic configuration is
$$1s^2\, 2s^2\, 2p^5$$.

Useful facts:
(i) For every electron the set of quantum numbers $$\{n,\,l,\,m_l,\,m_s\}$$ must be unique (Pauli exclusion principle).
(ii) In a given subshell electrons first occupy different orbitals with parallel spins (Hund’s rule) and pairing begins only after every orbital is singly occupied.

Testing statement A
Fluorine has 9 electrons. Split them according to their spin quantum number $$m_s$$.
• $$1s^2$$ : one $$m_s = +\frac12$$ and one $$m_s = -\frac12$$ ⇒ 1 up, 1 down.
• $$2s^2$$ : one up, one down ⇒ +1 up, +1 down (cumulative 2 up, 2 down).
• $$2p^5$$ : first three electrons occupy the three $$2p$$ orbitals with $$m_s = +\frac12$$, next two pair with $$m_s = -\frac12$$.
⇒ 3 up + 2 down (cumulative 5 up, 4 down).
Possible distribution: 5 electrons with $$m_s = +\frac12$$ and 4 electrons with $$m_s = -\frac12$$. Hence statement A is correct.

Testing statement B
The five $$2p$$ electrons are placed as
$$\uparrow\downarrow \; (p_x),\; \uparrow \; (p_y),\; \uparrow \; (p_z).$$
Only one of the three $$2p$$ orbitals ends up doubly occupied; which one is arbitrary because the three orbitals are degenerate. Therefore it is not mandatory that exactly the $$p_z$$ orbital contains a single electron. Statement B is not universally correct and is rejected.

Testing statement C
For the last (ninth) electron the available orbitals in increasing energy order are
$$1s \lt 2s \lt 2p$$.
The $$1s$$ and $$2s$$ orbitals are already full; hence the ninth electron enters an orbital with $$n = 2,\, l = 1$$ (a $$2p$$ orbital). Statement C is correct.

Testing statement D
Angular nodes of an orbital $$= l$$.
List every occupied orbital:

1s : $$l = 0 \Rightarrow 0$$ angular nodes
2s : $$l = 0 \Rightarrow 0$$ angular nodes
2p_x, 2p_y, 2p_z : each has $$l = 1 \Rightarrow 1$$ angular node

Total angular nodes of all five distinct orbitals
$$0 + 0 + 1 + 1 + 1 = 3$$.
Statement D claims the sum is 1, which is incorrect.

Hence only statements A and C are correct.

Correct option: Option B (A and C Only).

Statement I: "It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle."

This is a direct statement of the Heisenberg Uncertainty Principle, which says:

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

where $$\Delta x$$ is the uncertainty in position and $$\Delta p$$ is the uncertainty in momentum. Since the product of the two uncertainties has a non-zero lower bound, we cannot make both arbitrarily small at the same time.

So Statement I is true.

Statement II: We are told that the uncertainty in position equals the uncertainty in momentum, i.e., $$\Delta x = \Delta p$$. We need to check whether $$\Delta v \ge \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}$$.

Starting from the Heisenberg Uncertainty Principle:

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

Since $$\Delta x = \Delta p$$, we substitute:

$$(\Delta p)^2 \ge \frac{h}{4\pi}$$

Taking the square root of both sides:

$$\Delta p \ge \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}$$

Now, momentum $$p = mv$$, so the uncertainty in momentum is related to the uncertainty in velocity by:

$$\Delta p = m \cdot \Delta v$$

Substituting:

$$m \cdot \Delta v \ge \frac{1}{2}\sqrt{\frac{h}{\pi}}$$

Dividing both sides by $$m$$:

$$\Delta v \ge \frac{1}{2m}\sqrt{\frac{h}{\pi}}$$

This can be rewritten as:

$$\Delta v \ge \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}$$

This is exactly what Statement II claims. So Statement II is also true.

Since both statements are true, the correct answer is Option C.

We need to find which spectral line of the hydrogen atom produces radiation of wavelength about 900 nm. The Rydberg formula is:

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

Given: $$\lambda = 900$$ nm $$= 9 \times 10^{-5}$$ cm, $$R_H = 10^5$$ cm$$^{-1}$$. Therefore,

$$ \frac{1}{\lambda} = \frac{1}{9 \times 10^{-5}} = \frac{10^5}{9} \text{ cm}^{-1} $$

For the Paschen series (where $$n_1 = 3$$) in the limit $$n_2 \to \infty$$, the formula gives

$$ \frac{1}{\lambda} = R_H \left(\frac{1}{9} - 0\right) = \frac{10^5}{9} \text{ cm}^{-1} $$

Since this matches the calculated value of $$\frac{1}{\lambda}$$, the wavelength 900 nm corresponds to the series limit of the Paschen series ($$\infty \to 3$$). The correct answer is Option 3: Paschen series, $$\infty \rightarrow 3$$.

The concept of “extra stability of half-filled subshells” is explained by Hund’s rule. According to this rule, when orbitals of equal energy (degenerate orbitals) are available, electrons occupy them singly with parallel spins before pairing begins. Three physical arguments support the added stability of a half-filled configuration.

1. Symmetrical distribution of charge
For a half-filled subshell, the probability cloud of electrons is distributed as uniformly as possible around the nucleus. This symmetry lowers the potential energy of the atom. Hence Statement (A) is correct.

2. Smaller coulombic repulsion and smaller shielding
When each degenerate orbital holds just one electron, there is no intra-orbital electron-electron repulsion. Inter-orbital repulsion is also reduced because the electrons tend to stay farther apart. Less repulsion means lower potential energy, and the mutual shielding of nuclear charge is also reduced. Thus both Statement (B) (“smaller coulombic repulsion energy”) and Statement (E) (“relatively smaller shielding of electrons by one another”) are correct.

3. Larger exchange energy
If there are $$n$$ electrons with parallel spins in a set of degenerate orbitals, the number of possible exchanges between them is $$\frac{n(n-1)}{2}$$. Each exchange lowers the energy slightly. A half-filled subshell maximises the number of such exchanges, giving the greatest lowering of energy. Therefore Statement (D) is correct.

4. Incorrect wording in Statement (C)
Statement (C) talks about “presence of electrons with the same spin in non-degenerate orbitals”. Hund’s rule applies to degenerate orbitals (same energy), not non-degenerate ones. So Statement (C) is incorrect.

Collecting the valid statements: (A), (B), (D) and (E).

Hence the correct option is Option B.

For any one-electron (hydrogen-like) species, the Bohr energy expression is
$$E_n = -13.6\,\text{eV}\,\frac{Z^{2}}{n^{2}}$$ where
$$Z$$ = atomic number  and  $$n$$ = principal quantum number.

Given: energy of the first Bohr orbit of hydrogen $$\bigl(Z = 1,\; n = 1\bigr)$$ is $$-13.6$$ eV, which matches the above formula and confirms its use.

For $$Be^{3+}$$ we have $$Z = 4$$ (beryllium’s atomic number).
The first excited state corresponds to the second orbit, so $$n = 2$$.

Substitute these values:
$$E_{2} = -13.6\,\text{eV}\,\frac{4^{2}}{2^{2}}$$

Compute numerator and denominator:
$$4^{2} = 16,\quad 2^{2} = 4$$

Therefore,
$$E_{2} = -13.6\,\text{eV}\,\frac{16}{4} = -13.6\,\text{eV}\,\times 4 = -54.4\,\text{eV}$$

The question asks for the magnitude (absolute value):
$$|E_{2}| = 54.4\,\text{eV}$$

To the nearest integer, the required energy magnitude is 54 eV.

We need to compare the energies of orbitals in a multi-electron system using the $$(n + l)$$ rule.

For multi-electron atoms, orbital energy increases with $$(n + l)$$. If two orbitals have the same $$(n + l)$$, the one with lower $$n$$ has lower energy.

Calculate $$(n + l)$$ for each:

(A) $$n = 4, l = 1$$: $$n + l = 5$$ (4p orbital)

(B) $$n = 4, l = 2$$: $$n + l = 6$$ (4d orbital)

(C) $$n = 3, l = 1$$: $$n + l = 4$$ (3p orbital)

(D) $$n = 3, l = 2$$: $$n + l = 5$$ (3d orbital)

(E) $$n = 4, l = 0$$: $$n + l = 4$$ (4s orbital)

Order by increasing energy:

$$n + l = 4$$: C (3p, $$n = 3$$) and E (4s, $$n = 4$$). Since both have $$n + l = 4$$, lower $$n$$ means lower energy: C < E

$$n + l = 5$$: D (3d, $$n = 3$$) and A (4p, $$n = 4$$). D < A

$$n + l = 6$$: B (4d)

So the increasing order of energy is: $$C < E < D < A < B$$

This matches Option D: (C) < (E) < (D) < (A) < (B).

The correct answer is Option D.

Statement I: "The orbitals having same energy are called as degenerate orbitals."

This is the correct definition of degenerate orbitals. Orbitals that have the same energy are called degenerate. Statement I is TRUE.

Statement II: "In hydrogen atom, $$3p$$ and $$3d$$ orbitals are not degenerate orbitals."

In a hydrogen atom (single-electron system), the energy of an orbital depends only on the principal quantum number $$n$$, not on the azimuthal quantum number $$l$$. Therefore, all orbitals with the same $$n$$ have the same energy.

Since $$3p$$ ($$n=3, l=1$$) and $$3d$$ ($$n=3, l=2$$) both have $$n = 3$$, they have the same energy in a hydrogen atom and are degenerate.

(Note: In multi-electron atoms, $$3p$$ and $$3d$$ would NOT be degenerate due to electron-electron repulsion and shielding effects. But in hydrogen atom specifically, they are degenerate.)

Statement II is FALSE.

The correct answer is: Statement-I is true but Statement-II is false.

We need to find the four quantum numbers for the valence electron of rubidium ($$Z = 37$$).

Determine the electronic configuration of Rb.

Rubidium has 37 electrons. Its electronic configuration is:

$$Rb: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 5s^1$$

Or equivalently: $$[Kr] \, 5s^1$$

The valence (outermost) electron is in the $$5s$$ orbital.

Assign the quantum numbers.

For an electron in the $$5s$$ orbital:

- Principal quantum number ($$n$$): $$n = 5$$ (the electron is in the 5th shell)

- Azimuthal quantum number ($$l$$): $$l = 0$$ (for an s-orbital, $$l$$ is always 0)

- Magnetic quantum number ($$m_l$$): $$m_l = 0$$ (for $$l = 0$$, the only possible value is 0)

- Spin quantum number ($$m_s$$): $$m_s = +\frac{1}{2}$$ (for the first electron in an orbital, by convention)

The set of quantum numbers is: $$5, 0, 0, +\frac{1}{2}$$.

The correct answer is Option (1): $$5, 0, 0, +\frac{1}{2}$$.

Neodymium (Nd) has atomic number 60. It is a lanthanide element. The noble gas core is Xenon (Xe, Z=54). After Xe, the electron filling order is: 6s, then 4f.

$$60 - 54 = 6$$ electrons to fill after Xe core.

First, 6s is filled: 2 electrons → $$6s^2$$ (4 remaining)

Then, 4f is filled: 4 electrons → $$4f^4$$

Electronic configuration: $$[Xe]\,4f^4\,6s^2$$.

The correct answer is Option 1: $$[Xe]\,4f^4\,6s^2$$.

Find the number of radial nodes for a 3p orbital.

The number of radial nodes (spherical nodes) in an orbital is given by:

$$ \text{Radial nodes} = n - l - 1 $$

where $$n$$ is the principal quantum number and $$l$$ is the azimuthal quantum number.

For the 3p orbital: $$n = 3$$ and $$l = 1$$ (since p corresponds to $$l = 1$$).

$$ \text{Radial nodes} = 3 - 1 - 1 = 1 $$

The correct answer is Option A: 1.

In isoelectronic species $$F^-$$, $$Ne$$, and $$Na^+$$, all three have 10 electrons, corresponding to the electron configuration 1s$$^2$$ 2s$$^2$$ 2p$$^6$$. They therefore share the same principal quantum number $$n = 2$$ for their outermost electrons and experience identical electron-electron repulsion.

Their distinguishing feature is the nuclear charge $$Z$$: $$Z_{F^-} = 9$$, $$Z_{Ne} = 10$$, and $$Z_{Na^+} = 11$$. A higher nuclear charge pulls the same number of electrons closer to the nucleus, reducing the ionic or atomic radius.

Consequently, $$F^-$$ (Z = 9) has the weakest attraction and the largest size, $$Ne$$ (Z = 10) is intermediate, and $$Na^+$$ (Z = 11) has the strongest attraction and the smallest size, giving the order: $$F^- > Ne > Na^+$$.

The factor that affects the size of these isoelectronic species is the nuclear charge $$Z$$. The correct answer is Option D: Nuclear charge $$Z$$.

Einsteinium has atomic number $$Z = 99$$. The closest noble gas preceding it is Radon (Rn) with $$Z = 86$$, so we represent the core as $$[Rn]$$ and calculate $$99 - 86 = 13$$ electrons remaining.

Following Radon, the orbitals fill in the order $$7s \rightarrow 5f \rightarrow 6d \rightarrow 7p$$. In the actinide series (elements 89-103), the general electronic configuration is $$[Rn] 5f^n 6d^0 7s^2$$, though one $$5f$$ electron may occasionally shift to $$6d$$.

Of the 13 electrons beyond the Radon core, two fill the $$7s$$ orbital to give $$7s^2$$, leaving eleven electrons to enter $$5f$$ as $$5f^{11}$$, with $$6d^0$$ remaining empty. Verifying that $$11 + 0 + 2 = 13$$ confirms that the total $$86 + 13 = 99$$ matches Einsteinium’s atomic number.

Therefore, the electronic configuration of Einsteinium is $$[Rn] 5f^{11} 6d^0 7s^2$$. The correct answer is Option 3: $$[Rn]5f^{11}6d^0 7s^2$$.

We need to find the four quantum numbers for the outermost electron of potassium (K, atomic number 19).

The four quantum numbers that describe an electron are the principal quantum number $$n$$, which determines the energy level or shell; the azimuthal quantum number $$l$$, which determines the subshell ($$l = 0$$ for s, $$l = 1$$ for p, $$l = 2$$ for d); the magnetic quantum number $$m$$, which ranges from $$-l$$ to $$+l$$; and the spin quantum number $$s$$, which can be either $$+\frac{1}{2}$$ or $$-\frac{1}{2}$$.

Following the Aufbau principle (filling order by increasing energy), the electron configuration of potassium (Z = 19) is $$\text{K}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1$$, which can also be written as $$[\text{Ar}] \, 4s^1$$ because the 4s subshell (n + l = 4 + 0 = 4) has a lower energy than the 3d subshell (n + l = 3 + 2 = 5).

Therefore, the outermost (valence) electron occupies the $$4s$$ orbital, for which the principal quantum number is $$n = 4$$, the azimuthal quantum number is $$l = 0$$, the magnetic quantum number is $$m = 0$$ (the only possible value when $$l = 0$$), and the spin quantum number is $$s = +\frac{1}{2}$$ (by convention, the first electron in an orbital has spin +1/2).

The correct answer is Option (2): $$n = 4, l = 0, m = 0, s = +\frac{1}{2}$$.

The question asks which statements about the behaviour of electrons in an atom are correct. We examine each option one by one, quoting the pertinent theoretical result before judging the statement.

Option A
Heisenberg’s Uncertainty Principle states $$\Delta x \,\Delta p \,\ge \,\frac{h}{4\pi}$$. Because the position $$x$$ and linear momentum $$p$$ of an electron cannot be known simultaneously with arbitrary accuracy, it is impossible to trace out a precise trajectory (definite path) for the electron as one does for macroscopic particles. Therefore the existence of well-defined orbits is ruled out.
Hence, Option A is correct.

Option B
In quantum‐mechanical (or even Bohr) treatment, the reference level of energy is taken to be zero for an electron at an infinite distance from the nucleus. Any electron bound to the nucleus has an energy that is negative (lower than zero). A 2s electron is certainly bound, so $$E_{2s} \lt 0$$, whereas $$E_{\infty}=0$$. Thus the energy of an electron in the 2s orbital is lower than that of an electron infinitely far away.
Hence, Option B is correct.

Option C
Bohr derived the energy of the electron in a hydrogenic atom as
$$E_n = -\frac{13.6\,\text{eV}\,Z^{2}}{n^{2}}$$.
Because the expression is inversely proportional to $$n^{2}$$ and carries a negative sign, the most negative (lowest) energy occurs at the smallest possible principal quantum number, $$n = 1$$. Lower (more negative) energy means greater stability, so the $$n = 1$$ orbit is the most stable.
Hence, Option C is correct.

Option D
Bohr also obtained the speed of the electron in the $$n^{\text{th}}$$ orbit:
$$v_n = \frac{2.18\times10^{6}\,\text{m s}^{-1}}{n}\;Z$$.
For a fixed nucleus ($$Z$$ constant), $$v_n$$ is inversely proportional to $$n$$, i.e. as $$n$$ increases, the magnitude of velocity decreases. The statement that velocity increases with $$n$$ is therefore wrong.
Hence, Option D is incorrect.

Therefore the correct statements are:
Option A, Option B, and Option C.

Final answer: Option A (Uncertainty principle rules out the existence of definite paths for electrons.), Option B (The energy of an electron in 2s orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.), Option C (According to Bohr's model, the most negative energy value for an electron is given by $$n = 1$$, which corresponds to the most stable orbit.)

1. Identify the Wavelength ($$\lambda$$)

Looking at the diagram, the distance of 1.5 pm represents the distance from a peak (crest) to a zero-crossing (node).

• A full wavelength ($$\lambda$$) consists of four such segments (peak to node, node to trough, trough to node, and node back to peak).
• Therefore, the wavelength is:
  $$\lambda = 4 \times 1.5 \text{ pm} = 6.0 \text{ pm}$$
• Converting picometers to meters:
  $$\lambda = 6.0 \times 10^{-12} \text{ m}$$

2. Relate Speed, Frequency, and Wavelength

Since this is an electromagnetic wave, it travels at the speed of light ($$c \approx 3 \times 10^8 \text{ m/s}$$). We use the standard wave equation:

$$c = \nu \lambda$$

Where:

• $$c$$ is the speed of light ($$3 \times 10^8 \text{ m/s}$$)
• $$\nu$$ is the frequency (Hz)
• $$\lambda$$ is the wavelength (m)

3. Calculate the Frequency ($$\nu$$)

Rearranging the formula to solve for frequency:

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{6 \times 10^{-12} \text{ m}}$$

Now, simplify the powers of ten:

$$\nu = 0.5 \times 10^{8 - (-12)} = 0.5 \times 10^{20} \text{ Hz}$$ 

$$\nu = 5 \times 10^{19} \text{ Hz}$$

By Heisenberg's uncertainty principle: $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$.

$$\Delta v = \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}$$

$$= \frac{6.626 \times 10^{-34}}{114.296 \times 10^{-46}} = \frac{6.626}{114.296} \times 10^{12} = 0.05798 \times 10^{12} = 5.798 \times 10^{10}$$

$$= 57.98 \times 10^9 \approx 58 \times 10^9$$ m/s.

The answer is $$\boxed{58}$$.

We wish to determine the frequency of the de-Broglie wave of the electron in Bohr’s first orbit of hydrogen. In de-Broglie’s theory, the wavelength is given by $$\lambda = \frac{h}{mv}$$, which yields a frequency $$\nu = \frac{v}{\lambda} = \frac{mv^2}{h}\,.$$ Noting that $$mv^2 = 2\times KE$$ and that the kinetic energy in the first orbit is $$|E_1| = R_H = 2.18\times10^{-18}$$ J, we have:

$$\nu = \frac{mv^2}{h} = \frac{2\times KE}{h} = \frac{2\times 2.18\times10^{-18}}{6.6\times10^{-34}}$$

$$= \frac{4.36\times10^{-18}}{6.6\times10^{-34}} = 6.606\times10^{15}\text{ Hz}$$

Expressed in the form $$\_\_\_ \times 10^{13}$$ Hz, this becomes $$\nu = 660.6\times10^{13}\approx661\times10^{13}$$ Hz. The answer is 661.

We need to find the total number of electrons with quantum numbers $$n = 4$$, $$|m_l| = 1$$, and $$m_s = -\frac{1}{2}$$.

For $$n = 4$$, the possible subshells are $$4s$$ ($$l = 0$$), $$4p$$ ($$l = 1$$), $$4d$$ ($$l = 2$$), and $$4f$$ ($$l = 3$$). We require $$|m_l| = 1$$, which corresponds to $$m_l = +1$$ or $$m_l = -1$$. In each subshell the allowed $$m_l$$ values range from $$-l$$ to $$+l$$.

In the $$4s$$ subshell ($$l = 0$$) the only value of $$m_l$$ is 0, so there are no orbitals with $$|m_l| = 1$$. In the $$4p$$ subshell ($$l = 1$$) the values of $$m_l$$ are $$-1, 0, +1$$, giving two orbitals with $$m_l = +1$$ and $$m_l = -1$$. Similarly, in the $$4d$$ subshell ($$l = 2$$) with $$m_l = -2, -1, 0, +1, +2$$ there are two orbitals with $$|m_l| = 1$$, and in the $$4f$$ subshell ($$l = 3$$) with $$m_l = -3, -2, -1, 0, +1, +2, +3$$ there are also two such orbitals.

The total number of orbitals satisfying $$|m_l| = 1$$ across all subshells is $$0 + 2 + 2 + 2 = 6$$. Since each orbital can accommodate one electron with $$m_s = -\frac{1}{2}$$, the total number of electrons is $$6$$.

The answer is 6.

In He$$^+$$, transition from 5th excited state (n = 6) to first excited state (n = 2).

Number of spectral lines from transitions between levels n = 6 to n = 2:

Number of levels involved = 6 - 2 + 1 = 5 (levels 2, 3, 4, 5, 6).

Number of spectral lines = $$\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$$.

Therefore, the answer is $$\boxed{10}$$.

For the nth orbit, the de Broglie condition states:

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

For n = 4: $$\lambda = \frac{2\pi r_4}{4}$$

The radius of the nth orbit: $$r_n = n^2 a_0$$, so $$r_4 = 16a_0$$

$$\lambda = \frac{2\pi \times 16a_0}{4} = 8\pi a_0$$

The answer is 8.

Find the maximum number of orbitals with $$n = 4$$ and $$m_l = 0$$.

For $$n = 4$$, the subshells correspond to $$l = 0, 1, 2, 3$$, that is, the 4s, 4p, 4d, and 4f subshells.

Next, check which subshells have $$m_l = 0$$. Every subshell has exactly one orbital with $$m_l = 0$$:

- $$l = 0$$: $$m_l = 0$$ → 1 orbital (4s)

- $$l = 1$$: $$m_l \in \{-1, 0, 1\}$$ → $$m_l = 0$$ exists → 1 orbital (4p)

- $$l = 2$$: $$m_l \in \{-2, -1, 0, 1, 2\}$$ → $$m_l = 0$$ exists → 1 orbital (4d)

- $$l = 3$$: $$m_l \in \{-3, ..., 3\}$$ → $$m_l = 0$$ exists → 1 orbital (4f)

In total, there are 4 orbitals.

The correct answer is 4.

We need to find the wavenumber for radiation with wavelength $$5800 \text{ \AA}$$. Since the wavenumber is defined as $$\bar{\nu} = \frac{1}{\lambda}$$, we first convert the wavelength to centimeters, giving $$\lambda = 5800 \text{ \AA} = 5800 \times 10^{-8} \text{ cm} = 5.8 \times 10^{-5} \text{ cm}$$.

Substituting this into the expression for the wavenumber yields $$\bar{\nu} = \frac{1}{5.8 \times 10^{-5}} = \frac{10^5}{5.8} = 17241.4 \text{ cm}^{-1}$$.

Since $$\bar{\nu} = x \times 10 \text{ cm}^{-1}$$, it follows that $$x = \frac{17241.4}{10} = 1724.14$$.

Therefore, rounding to the nearest integer gives $$x = \boxed{1724}$$.

We need to find the kinetic energy of the electron in the first excited state of hydrogen, expressed as $$x \times 10^{-1}$$ eV. For a hydrogen atom, the total energy, kinetic energy, and potential energy are related by $$E_n = -\frac{13.6}{n^2} \, \text{eV}$$, $$KE_n = -E_n = \frac{13.6}{n^2} \, \text{eV}$$, and $$PE_n = 2E_n = -\frac{27.2}{n^2} \, \text{eV}$$. The kinetic energy is always positive and equals the magnitude of the total energy, which follows from the virial theorem for a Coulomb potential: $$KE = -E$$ and $$PE = 2E$$.

Since the first excited state corresponds to $$n = 2$$, we have $$E_2 = -3.4 \, \text{eV}$$. Substituting into the expression for kinetic energy gives $$KE = -E_2 = -(-3.4) = 3.4 \, \text{eV}$$. Expressing this value in the form $$x \times 10^{-1}\, \text{eV}$$ yields $$3.4 \, \text{eV} = 34 \times 10^{-1} \, \text{eV}$$, so $$x = 34$$.

Option 34

We need to find the energy (in kJ/mol) corresponding to a wavelength of 242 nm.

Using $$E = \frac{hc}{\lambda}$$:

$$ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} $$

$$ E = \frac{19.878 \times 10^{-26}}{242 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{2.42 \times 10^{-7}} $$

$$ E = 8.214 \times 10^{-19} \text{ J} $$

Multiply by Avogadro's number ($$N_A = 6.022 \times 10^{23}$$):

$$ E_{mol} = 8.214 \times 10^{-19} \times 6.022 \times 10^{23} = 49.46 \times 10^{4} \text{ J/mol} $$

$$ E_{mol} = 494.6 \text{ kJ/mol} \approx 494 \text{ kJ/mol} $$

The answer is $$\boxed{494}$$ kJ/mol.

For $$n = 4$$, the possible subshells are: $$4s, 4p, 4d, 4f$$.

The completely filled subshells for $$n = 4$$ are:

- $$4s$$: 2 electrons (completely filled)

- $$4p$$: 6 electrons (completely filled)

- $$4d$$: 10 electrons (completely filled)

- $$4f$$: 14 electrons (completely filled)

Total electrons in all completely filled subshells with $$n = 4$$: $$2 + 6 + 10 + 14 = 32$$

Now, electrons with $$s = +\frac{1}{2}$$: In each completely filled subshell, exactly half the electrons have $$s = +\frac{1}{2}$$ and the other half have $$s = -\frac{1}{2}$$.

Number of electrons with $$s = +\frac{1}{2}$$: $$\frac{32}{2} = 16$$

The answer is $$\boxed{16}$$.

We need to find the velocity of the electron in Bohr's first orbit of the hydrogen atom, given the Rydberg constant and electron mass.

Recall the energy relation in Bohr's model.

In Bohr's model, for the hydrogen atom in the first orbit ($$n = 1$$), the total energy is:

$$ E_1 = -R_H $$

where $$R_H = 2.18 \times 10^{-18}$$ J is the Rydberg constant (energy).

The total energy is the sum of kinetic and potential energies. For a circular orbit under Coulomb's law:

$$ E = -KE = \frac{PE}{2} $$

This gives us: $$KE = -E = R_H$$

Set up the kinetic energy equation.

$$ KE = \frac{1}{2}mv^2 = R_H $$

$$ \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2 = 2.18 \times 10^{-18} $$

Solve for $$v^2$$.

$$ v^2 = \frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}} $$

$$ v^2 = \frac{4.36 \times 10^{-18}}{9.1 \times 10^{-31}} $$

$$ v^2 = \frac{4.36}{9.1} \times 10^{-18+31} = 0.4791 \times 10^{13} = 4.791 \times 10^{12} \text{ m}^2\text{s}^{-2} $$

Calculate $$v$$.

$$ v = \sqrt{4.791 \times 10^{12}} $$

$$ v = \sqrt{4.791} \times 10^6 $$

$$ v \approx 2.189 \times 10^6 \text{ m/s} $$

Express in the required form.

$$ v = 2.189 \times 10^6 = 21.89 \times 10^5 \text{ m/s} $$

Rounding to the nearest integer: $$v \approx 22 \times 10^5$$ m/s.

The answer is $$\boxed{22}$$.

We need to determine how many of the given ions have noble gas configuration.

A noble gas configuration means the ion's electron configuration matches that of a noble gas (He, Ne, Ar, Kr, Xe, Rn).

$$Sr^{2+}$$ (Z = 38): $$Sr$$ has configuration $$[Kr]5s^2$$. Removing 2 electrons gives $$[Kr]$$ — 36 electrons. This is the krypton noble gas configuration. Yes.

$$Cs^+$$ (Z = 55): $$Cs$$ has configuration $$[Xe]6s^1$$. Removing 1 electron gives $$[Xe]$$ — 54 electrons. This is the xenon noble gas configuration. Yes.

$$La^{2+}$$ (Z = 57): $$La$$ has configuration $$[Xe]5d^1 6s^2$$. Removing 2 electrons gives $$[Xe]5d^1$$ — 55 electrons. This is NOT a noble gas configuration (it has one extra d-electron beyond Xe). No.

$$Pb^{2+}$$ (Z = 82): $$Pb$$ has configuration $$[Xe]4f^{14}5d^{10}6s^2 6p^2$$. Removing 2 electrons (from 6p) gives $$[Xe]4f^{14}5d^{10}6s^2$$ — 80 electrons. This is NOT a noble gas configuration. No.

$$Yb^{2+}$$ (Z = 70): $$Yb$$ has configuration $$[Xe]4f^{14}6s^2$$. Removing 2 electrons gives $$[Xe]4f^{14}$$ — 68 electrons. This is NOT a noble gas configuration. No.

$$Fe^{2+}$$ (Z = 26): $$Fe$$ has configuration $$[Ar]3d^6 4s^2$$. Removing 2 electrons gives $$[Ar]3d^6$$ — 24 electrons. This is NOT a noble gas configuration. No.

Only $$Sr^{2+}$$ and $$Cs^+$$ have noble gas configurations.

The answer is 2.

The photon that strikes the metal comes from the electronic transition $$n = 4 \rightarrow n = 3$$ in a hydrogen-like ion of atomic number $$Z$$.

Step 1 : Energy of the photon
For a hydrogen-like ion, the energy of level $$n$$ is $$E_n = -\,\dfrac{Z^{2} \, 13.6\ \text{eV}}{n^{2}}$$. Hence for the transition $$4 \rightarrow 3$$

$$\Delta E \;=\; Z^{2}\,13.6\left(\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right) = Z^{2}\,13.6\Bigl(\dfrac{1}{9}-\dfrac{1}{16}\Bigr) = Z^{2}\,13.6\Bigl(\dfrac{16-9}{144}\Bigr) = Z^{2}\,13.6 \times \dfrac{7}{144}$$

Compute the numerical factor:

$$13.6 \times \dfrac{7}{144}= \dfrac{95.2}{144}=0.661111\;\text{eV}$$

Therefore the photon energy is

$$E_{\text{photon}} = 0.661111\,Z^{2}\ \text{eV} \qquad -(1)$$

Step 2 : Work function of the target metal
Threshold wavelength $$\lambda_0 = 310\ \text{nm}$$ gives the work function

$$\Phi = \dfrac{hc}{\lambda_0} = \dfrac{1240\ \text{eV·nm}}{310\ \text{nm}} = 4.00\ \text{eV}$$

Step 3 : Use photoelectric equation
Maximum kinetic energy of emitted electrons is $$K_{\max}=1.95\ \text{eV}$$. Hence

$$E_{\text{photon}} = \Phi + K_{\max} = 4.00 + 1.95 = 5.95\ \text{eV} \qquad -(2)$$

Step 4 : Equate $$E_{\text{photon}}$$ from (1) and (2)

$$0.661111\,Z^{2} = 5.95$$

$$Z^{2} = \dfrac{5.95}{0.661111} \approx 9.00$$

$$\Rightarrow\; Z \approx 3$$

Thus the atomic number of the hydrogen-like ion is 3.

Final Answer: $$Z = 3$$

Statement I: According to Bohr's model, the angular momentum is quantized: $$L = n\hbar$$. This is correct.

Statement II: In Bohr's model, the electron has a definite orbit with known position and momentum simultaneously. This violates the Heisenberg uncertainty principle (which states that position and momentum cannot both be precisely known). This is correct.

Both statements are correct.

This matches option 2.

The radius of the $$n$$th orbit of hydrogen atom is:

$$ r_n = n^2 a_0 $$

For the 3rd orbit: $$r_3 = 9a_0$$.

By Bohr's quantization condition, the circumference of the orbit equals $$n$$ times the de Broglie wavelength:

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

For $$n = 3$$:

$$ 2\pi(9a_0) = 3\lambda $$

$$ \lambda = \frac{18\pi a_0}{3} = 6\pi a_0 $$

The de Broglie wavelength of the electron in the 3rd orbit is $$6\pi a_0$$.

The radius of the nth orbit in a hydrogen-like atom is:

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

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

For Li²⁺ (Z = 3), 2nd orbit:

$$x = r_2 = \frac{4a_0}{3}$$

For Be³⁺ (Z = 4), 3rd orbit:

$$r_3 = \frac{9a_0}{4}$$

Ratio:

$$\frac{r_3}{x} = \frac{9a_0/4}{4a_0/3} = \frac{9a_0}{4} \times \frac{3}{4a_0} = \frac{27}{16}$$

So $$r_3 = \frac{27}{16}x$$

The correct answer is Option 3: $$\frac{27}{16}x$$.

We are given that $$\lambda$$ is the shortest wavelength in the Lyman series of hydrogen and we need the longest wavelength in the Balmer series of He$$^+$$.

The shortest wavelength in the Lyman series corresponds to the series limit (transition from $$n = \infty$$ to $$n = 1$$):

$$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R_H$$

So $$\lambda = \frac{1}{R_H}$$.

Now, the longest wavelength in the Balmer series of He$$^+$$ corresponds to the transition from $$n = 3$$ to $$n = 2$$. For He$$^+$$ with $$Z = 2$$:

$$\frac{1}{\lambda'} = R_H \times 4 \times \left(\frac{1}{4} - \frac{1}{9}\right) = R_H \times 4 \times \frac{5}{36} = \frac{20R_H}{36} = \frac{5R_H}{9}$$

Hence $$\lambda' = \frac{9}{5R_H} = \frac{9}{5} \times \frac{1}{R_H} = \frac{9\lambda}{5}$$.

Hence, the correct answer is Option 2: $$\frac{9\lambda}{5}$$.

We need to identify which set of ions represents a collection of isoelectronic species (species with the same number of electrons).

Given atomic numbers: F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21.

Count electrons for each option.

Option A: $$Li^+, Na^+, Mg^{2+}, Ca^{2+}$$

$$Li^+ : 3 - 1 = 2$$ electrons

$$Na^+ : 11 - 1 = 10$$ electrons

$$Mg^{2+} : 12 - 2 = 10$$ electrons

$$Ca^{2+} : 20 - 2 = 18$$ electrons

Electron counts: 2, 10, 10, 18 — Not isoelectronic.

Option B: $$Ba^{2+}, Sr^{2+}, K^+, Ca^{2+}$$

$$Ba^{2+} : 56 - 2 = 54$$ electrons

$$Sr^{2+} : 38 - 2 = 36$$ electrons

$$K^+ : 19 - 1 = 18$$ electrons

$$Ca^{2+} : 20 - 2 = 18$$ electrons

Electron counts: 54, 36, 18, 18 — Not isoelectronic.

Option C: $$N^{3-}, O^{2-}, F^-, S^{2-}$$

$$N^{3-} : 7 + 3 = 10$$ electrons

$$O^{2-} : 8 + 2 = 10$$ electrons

$$F^- : 9 + 1 = 10$$ electrons

$$S^{2-} : 16 + 2 = 18$$ electrons

Electron counts: 10, 10, 10, 18 — Not isoelectronic.

Option D: $$K^+, Cl^-, Ca^{2+}, Sc^{3+}$$

$$K^+ : 19 - 1 = 18$$ electrons

$$Cl^- : 17 + 1 = 18$$ electrons

$$Ca^{2+} : 20 - 2 = 18$$ electrons

$$Sc^{3+} : 21 - 3 = 18$$ electrons

Electron counts: 18, 18, 18, 18 — All isoelectronic!

Conclusion.

All four ions in Option D have 18 electrons each, making them isoelectronic with the noble gas Argon (Ar, atomic number 18). They all have the electron configuration $$1s^2 2s^2 2p^6 3s^2 3p^6$$.

The correct answer is Option D: $$K^+, Cl^-, Ca^{2+}, Sc^{3+}$$.

Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons — True.

Reason R: 5f orbitals are not as buried as 4f orbitals — True. Since 5f orbitals extend further from the nucleus, they can overlap more effectively with other orbitals, enabling bonding.

Both A and R are true and R is the correct explanation of A.

Assertion A: In the photoelectric effect, electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface.

This is correct. The photoelectric effect is instantaneous when the frequency of incident light exceeds the threshold frequency.

Reason R: When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place.

This is incorrect. Not any photon can cause energy transfer to eject electrons. The photon must have energy greater than or equal to the work function ($$h\nu \geq \phi$$) for the photoelectric effect to occur. Photons with energy less than the work function cannot eject electrons regardless of their intensity.

Therefore, A is correct but R is not correct.

as per Henry moseley $$\sqrt{\ V}$$ proportional to atomic number of elements 

We need to find how many atomic orbitals from the list $$7s, 7p, 6s, 8p, 8d$$ have exactly 5 radial nodes. The number of radial nodes in an atomic orbital is given by

where $$n$$ is the principal quantum number and $$l$$ is the azimuthal quantum number.

For 7s: $$n = 7$$, $$l = 0$$, so radial nodes $$= 7 - 0 - 1 = 6$$.

For 7p: $$n = 7$$, $$l = 1$$, so radial nodes $$= 7 - 1 - 1 = 5$$. This one qualifies.

For 6s: $$n = 6$$, $$l = 0$$, so radial nodes $$= 6 - 0 - 1 = 5$$. This one qualifies.

For 8p: $$n = 8$$, $$l = 1$$, so radial nodes $$= 8 - 1 - 1 = 6$$.

For 8d: $$n = 8$$, $$l = 2$$, so radial nodes $$= 8 - 2 - 1 = 5$$. This one qualifies.

The orbitals with exactly 5 radial nodes are 7p, 6s, and 8d. Hence, the answer is $$3$$.

Assertion A: Loss of electron from hydrogen atom results in nucleus of ~$$1.5 \times 10^{-3}$$ pm size.

When hydrogen loses its only electron, we are left with just a proton (H⁺). The size of a proton is approximately $$1.5 \times 10^{-3}$$ pm (= 1.5 femtometers). This is correct.

Reason R: Proton H⁺ always exists in combined form.

A bare proton (H⁺) is extremely small and has very high charge density. In aqueous solution, it immediately combines with water to form hydronium ion (H₃O⁺) or other hydrated species. It never exists as a free proton in solution. This is correct.

However, R does not explain A. The small size of the proton (A) is due to the nuclear dimensions, not because of its tendency to combine with other species (R). These are two independent facts.

The correct answer is Both A and R are correct but R is NOT the correct explanation of A.

The relative energies of the molecular orbitals (MOs) formed from the 2s- and 2p-orbitals depend on the atomic number of the two atoms.

For $$Z \ge 8$$ (that is, for $$O_2,\,F_2,\,Ne_2$$) the order is

$$\sigma_{2s}\;,\;\sigma^{*}_{2s}\;,\;\sigma_{2p_z}\;,\;\pi_{2p_x}=\pi_{2p_y}\;,\;\pi^{*}_{2p_x}=\pi^{*}_{2p_y}\;,\;\sigma^{*}_{2p_z} \quad -(1)$$

Fluorine has $$Z = 9$$, so $$F_2$$ must follow the sequence given in $$(1)$$.

Each fluorine atom supplies 7 valence electrons. Total valence electrons in $$F_2$$:

$$7 + 7 = 14$$

Placing these 14 electrons into the MOs of $$(1)$$:

$$\sigma_{2s}^2\,\sigma^{*}_{2s}^2\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi^{*}_{2p_x}^2\,\pi^{*}_{2p_y}^2\,\sigma^{*}_{2p_z}^0$$

Number of bonding electrons $$N_b = 8$$ (those in $$\sigma_{2s},\,\sigma_{2p_z},\,\pi_{2p_x},\,\pi_{2p_y}$$).
Number of antibonding electrons $$N_a = 6$$ (those in $$\sigma^{*}_{2s},\,\pi^{*}_{2p_x},\,\pi^{*}_{2p_y}$$).

Bond order

$$\text{B.O.} = \frac{N_b - N_a}{2} = \frac{8 - 6}{2} = 1$$

All electrons are paired, so $$F_2$$ is diamagnetic.

The diagram that shows the energy sequence of $$(1)$$, with $$\pi^{*}_{2p_x}$$ and $$\pi^{*}_{2p_y}$$ completely filled, $$\sigma^{*}_{2p_z}$$ empty, and no unpaired electrons, is Option C.

Therefore, the correct molecular orbital diagram for $$F_2$$ in its ground state is given by Option C.

For every cubic lattice the density, edge length and packing efficiency depend on three basic data: number of atoms per unit cell ($$Z$$), atomic radius ($$r$$) and molar mass ($$M$$).
We tabulate the standard relations first:

Face-centred cubic (fcc, metal $$x$$):
  • $$Z_x = 4$$     • contact along a face diagonal ⇒ $$\sqrt{2}\,L_x = 4\,r_x$$ ⇒ $$r_x = \dfrac{L_x}{2\sqrt{2}}$$.
Body-centred cubic (bcc, metal $$y$$):
  • $$Z_y = 2$$     • contact along a body diagonal ⇒ $$\sqrt{3}\,L_y = 4\,r_y$$ ⇒ $$r_y = \dfrac{\sqrt{3}}{4}\,L_y$$.
Simple cubic (sc, metal $$z$$):
  • $$Z_z = 1$$     • contact along an edge ⇒ $$L_z = 2\,r_z$$ ⇒ $$r_z = \dfrac{L_z}{2}$$.

The question gives the inter-relations

$$r_z = \frac{\sqrt{3}}{2}\,r_y,\qquad r_y = \frac{8}{\sqrt{3}}\,r_x,\qquad M_z = \frac{3}{2}\,M_y,\qquad M_z = 3\,M_x.$$

From the last two equalities

$$M_x = \frac{1}{3}M_z,\qquad M_y = \frac{2}{3}M_z \;\;\Longrightarrow\;\; \frac{M_x}{M_y} = \frac12.$$

1. Comparison of edge lengths

Using $$r_y = \dfrac{8}{\sqrt{3}}\,r_x$$ in $$L_y = \dfrac{4\,r_y}{\sqrt{3}}$$:
$$L_y = \frac{4}{\sqrt{3}}\left(\frac{8}{\sqrt{3}}\,r_x\right) = \frac{32}{3}\,r_x.$$
Using $$r_z = \frac{\sqrt{3}}{2}\,r_y = \frac{\sqrt{3}}{2}\left(\frac{8}{\sqrt{3}}\,r_x\right)=4\,r_x$$ in $$L_z=2\,r_z$$:
$$L_z = 8\,r_x.$$
For fcc, $$L_x = 2\sqrt{2}\,r_x.$$

Numerically (in units of $$r_x$$):
$$L_x = 2.828\,r_x,\qquad L_y = 10.667\,r_x,\qquad L_z = 8\,r_x.$$ Therefore $$L_y \gt L_z \gt L_x$$ and hence Option B ($$L_y \gt L_z$$) is correct, while Option C ($$L_x \gt L_y$$) is incorrect.

2. Packing efficiencies

Standard values are: fcc = 74 %, bcc = 68 %, sc = 52.4 %.
Thus$$\text{fcc (}x\text{)} \gt \text{bcc (}y\text{)} \gt \text{sc (}z\text{)},$$so Option A is correct.

3. Density comparison of $$x$$ and $$y$$

For any cubic lattice $$\rho = \dfrac{Z\,M}{N_A\,L^3}.$$ Taking the ratio $$ \frac{\rho_x}{\rho_y} = \frac{Z_x M_x L_y^{\,3}}{Z_y M_y L_x^{\,3}} = \left(\frac{4}{2}\right) \left(\frac{M_x}{M_y}\right) \left(\frac{L_y}{L_x}\right)^{3} = 2\left(\frac12\right)\left(\frac{L_y}{L_x}\right)^{3} = \left(\frac{L_y}{L_x}\right)^{3}. $$ We already have $$\dfrac{L_y}{L_x} = \dfrac{32/3}{2\sqrt{2}} = \dfrac{16}{3\sqrt{2}} \approx 3.772,$$ so $$\frac{\rho_x}{\rho_y} \approx 3.772^{3} \approx 53.7 \gt 1.$$ Hence $$\rho_x \gt \rho_y,$$ making Option D correct.

Final verdict:
Correct statements: Option A, Option B, Option D.

Neodymium (Nd) has atomic number 60.

Electronic configuration of Nd:

$$ \text{Nd} = [\text{Xe}] \; 4f^4 \; 6s^2 $$

For Nd$$^{2+}$$, two electrons are removed. In lanthanides, the $$6s$$ electrons are removed first (as they are in the outermost shell):

$$ \text{Nd}^{2+} = [\text{Xe}] \; 4f^4 $$

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.

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.

For Li²⁺ (Z=3), energy levels: $$E_n = -R_H \frac{Z^2}{n^2} = -\frac{9R_H}{n^2}$$

Energy for transition n → n+1:

$$\Delta E = 9R_H\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = 1.47 \times 10^{-17}$$

$$\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{1.47 \times 10^{-17}}{9 \times 2.18 \times 10^{-18}} = \frac{1.47}{1.962} = 0.75$$

Try n=1: $$1 - 1/4 = 0.75$$ ✓

The value of n is 1.

Energy of the incident photons is obtained from the Planck relation $$E = \dfrac{hc}{\lambda}$$.

Given: $$h = 6.6 \times 10^{-34}\,\text{J s}$$, $$c = 3.0 \times 10^{8}\,\text{m s}^{-1}$$, $$\lambda = 400\,\text{nm} = 400 \times 10^{-9}\,\text{m}$$.

Substituting, $$E = \dfrac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{400 \times 10^{-9}}$$.

$$E = \dfrac{19.8 \times 10^{-26}}{4.0 \times 10^{-7}} = 4.95 \times 10^{-19}\,\text{J}$$.

Convert this energy to electron-volts using $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$:

$$E = \dfrac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \,\text{eV} = 3.09\,\text{eV}\;(\text{approximately }3.1\,\text{eV}).$$

Photoelectric emission occurs when $$E \ge W_0$$. Compare $$3.1\,\text{eV}$$ with the given work functions:

Li : $$2.4\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
Na : $$2.3\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
K : $$2.25\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
Mg : $$3.7\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission
Cu : $$4.8\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission
Ag : $$4.3\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission

Therefore, only Li, Na, and K will show the photoelectric effect.

Total number of metals that emit electrons = $$3$$.

Answer : 3

We have velocity of electrons $$v = 1000$$ m/s, $$h = 6 \times 10^{-34}$$ Js, $$m_e = 9 \times 10^{-31}$$ kg.

(A) The de Broglie wavelength of the electron emitted is 666.67 nm.

$$\lambda = \frac{h}{m_e v} = \frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000} = \frac{6 \times 10^{-34}}{9 \times 10^{-28}} = 6.67 \times 10^{-7} \text{ m} = 666.7 \text{ nm}$$

This is TRUE.

(B) The characteristics of electrons emitted depend upon the material of the electrodes. This is FALSE — the properties of cathode rays (electrons) are independent of the electrode material, since electrons are fundamental particles with the same properties regardless of the source.

(C) The cathode rays start from cathode and move towards anode. This is TRUE — cathode rays are streams of electrons emitted from the cathode and accelerated towards the anode.

(D) The nature of the emitted electrons depends on the nature of the gas present. This is FALSE — the properties of electrons are independent of the gas used in the cathode ray tube.

So, the number of true statements is 2 (A and C). Hence, the answer is $$2$$.

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.

The de Broglie wavelength is given by:

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

where $$h = 6.6 \times 10^{-34}$$ Js, $$m = 9 \times 10^{-31}$$ kg, and $$K = 4.50 \times 10^{-29}$$ J.

We have $$2mK = 2 \times 9 \times 10^{-31} \times 4.50 \times 10^{-29} = 81 \times 10^{-60} = 8.1 \times 10^{-59}$$.

So $$\sqrt{2mK} = \sqrt{8.1 \times 10^{-59}} = 9 \times 10^{-30}$$ kg m/s.

Now the wavelength is:

$$\lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}} = 0.733 \times 10^{-4} = 7.33 \times 10^{-5}\;\text{m}$$

Rounding to the nearest integer, $$\lambda \approx 7 \times 10^{-5}$$ m. So, the answer is $$7$$.

Let us verify each statement about a black body:

(A) A black body emits and absorbs energy in the form of electromagnetic radiation. Correct.

(B) The frequency distribution of emitted radiation depends on temperature. Correct (Planck's law shows the spectral distribution changes with temperature).

(C) At a given temperature, the intensity vs frequency curve passes through a maximum value. Correct (Wien's displacement law).

(D) The maximum of the intensity vs frequency curve shifts to higher frequency at higher temperatures. Correct (Wien's displacement law: $$\nu_{max} \propto T$$).

All four statements are correct. The number of incorrect statements is 0.

We need to evaluate the Assertion-Reason statement about $$O^{2-}$$ and $$Mg^{2+}$$.

First, we check whether they are isoelectronic. $$O^{2-}$$ has 10 electrons (8 + 2), and $$Mg^{2+}$$ also has 10 electrons (12 - 2), so the Reason is true.

Next, we examine whether their ionic radii are the same. Since both species have 10 electrons but different nuclear charges (Z = 8 for $$O^{2-}$$ and Z = 12 for $$Mg^{2+}$$), the one with the greater nuclear charge pulls the electron cloud more tightly and has a smaller ionic radius. Therefore, $$r(O^{2-}) > r(Mg^{2+})$$, and the Assertion is false.

In conclusion, the Assertion is false but the Reason is true.

The correct answer is Option D.

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.

To determine the decreasing order of energy for the given orbitals, we apply the (n + l) rule and first identify each orbital by calculating its (n + l) value. For (A), $$n = 3, l = 0$$ corresponds to the 3s orbital, giving $$n + l = 3$$. For (B), $$n = 4, l = 0$$ corresponds to the 4s orbital, yielding $$n + l = 4$$. For (C), $$n = 3, l = 1$$ corresponds to the 3p orbital, yielding $$n + l = 4$$. Finally, for (D), $$n = 3, l = 2$$ corresponds to the 3d orbital, giving $$n + l = 5$$.

Since higher (n + l) values indicate higher energy, orbital D (with $$n + l = 5$$) has the highest energy, while orbital A (with $$n + l = 3$$) has the lowest. Orbitals B and C both have $$n + l = 4$$, so we compare their n values. Substituting these values shows that B has $$n = 4$$ whereas C has $$n = 3$$, and therefore orbital B lies above orbital C in energy.

From the above, the decreasing order of energy is (D) > (B) > (C) > (A). Thus, the correct answer is Option A: (D) > (B) > (C) > (A).

Identify the species and their isoelectronic nature.

All the given ions are isoelectronic with 10 electrons each:

$$N^{3-}$$ (7 + 3 = 10 e$$^-$$), $$O^{2-}$$ (8 + 2 = 10 e$$^-$$), $$F^-$$ (9 + 1 = 10 e$$^-$$), $$Na^+$$ (11 - 1 = 10 e$$^-$$), $$Mg^{2+}$$ (12 - 2 = 10 e$$^-$$).

Apply the rule for isoelectronic species.

For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. This is because a higher nuclear charge pulls the same number of electrons closer to the nucleus.

Arrange in order of increasing ionic size.

Nuclear charges: $$Mg^{2+}$$ (Z=12) > $$Na^+$$ (Z=11) > $$F^-$$ (Z=9) > $$O^{2-}$$ (Z=8) > $$N^{3-}$$ (Z=7)

Therefore, ionic size increases as:

$$Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$$

This matches Option A.

The answer is $$\boxed{\text{Option A}}$$.

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.

We need to find the number of radial and angular nodes in the $$4d$$ orbital.

For the $$4d$$ orbital: principal quantum number $$n = 4$$, azimuthal quantum number $$l = 2$$.

Number of angular nodes = $$l = 2$$

Number of radial nodes = $$n - l - 1 = 4 - 2 - 1 = 1$$

Total number of nodes = $$n - 1 = 3$$

Angular + Radial = $$2 + 1 = 3$$ ✓

The number of radial and angular nodes are $$1$$ and $$2$$, respectively.

The correct answer is Option A: $$1$$ & $$2$$.

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.

We need to find the energy of one mole of photons of radiation with wavelength $$\lambda = 300$$ nm; here $$h = 6.63 \times 10^{-34}$$ J s, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, and $$c = 3 \times 10^8$$ m/s.

The energy of a single photon is given by:

$$E = \frac{hc}{\lambda}$$

Since the wavelength is provided in nanometres, we convert:

$$\lambda = 300 \text{ nm} = 300 \times 10^{-9} \text{ m} = 3 \times 10^{-7} \text{ m}$$

Substituting into the formula leads to:

$$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}}$$

This simplifies to:

$$E = \frac{19.89 \times 10^{-26}}{3 \times 10^{-7}}$$

and so:

$$E = 6.63 \times 10^{-19} \text{ J}$$

Now, multiplying by Avogadro's number gives the energy per mole of photons:

$$E_{mole} = N_A \times E = 6.02 \times 10^{23} \times 6.63 \times 10^{-19} \text{ J}$$

From this we obtain:

$$E_{mole} = 6.02 \times 6.63 \times 10^{23-19} \text{ J}$$

which yields:

$$E_{mole} = 39.9126 \times 10^{4} \text{ J}$$

and therefore:

$$E_{mole} = 3.99 \times 10^{5} \text{ J} = 399 \text{ kJ/mol}$$

Therefore, the correct option is Option C: 399 kJ mol$$^{-1}$$.

The rules for quantum numbers are:

- Principal quantum number: $$n = 1, 2, 3, \ldots$$

- Azimuthal quantum number: $$l = 0, 1, 2, \ldots, (n-1)$$

- Magnetic quantum number: $$m_l = -l, -(l-1), \ldots, 0, \ldots, (l-1), l$$

- Spin quantum number: $$s = +\dfrac{1}{2} \text{ or } -\dfrac{1}{2}$$

Let us check each option:

Option A: $$n = 3, l = 2, m_l = 0, s = +\dfrac{1}{2}$$ — Valid. For $$n = 3$$, $$l$$ can be 0, 1, or 2. For $$l = 2$$, $$m_l$$ can range from $$-2$$ to $$+2$$.

Option B: $$n = 3, l = 2, m_l = -2, s = +\dfrac{1}{2}$$ — Valid. Same reasoning as above.

Option C: $$n = 3, l = 3, m_l = -3, s = -\dfrac{1}{2}$$ — Not allowed. For $$n = 3$$, the maximum value of $$l$$ is $$n - 1 = 2$$. So $$l = 3$$ is not permitted.

Option D: $$n = 3, l = 0, m_l = 0, s = -\dfrac{1}{2}$$ — Valid. For $$l = 0$$, $$m_l$$ must be 0.

Therefore, the correct answer is Option C.

We need to arrange the four electrons in order of increasing energy using the (n+l) rule. When two orbitals have the same (n+l) value, the one with the lower n has lower energy.

Electron A: $$n=3, l=2$$ — this is a 3d electron. We have $$n+l = 3+2 = 5$$.

Electron B: $$n=4, l=1$$ — this is a 4p electron. We have $$n+l = 4+1 = 5$$.

Electron C: $$n=4, l=2$$ — this is a 4d electron. We have $$n+l = 4+2 = 6$$.

Electron D: $$n=3, l=1$$ — this is a 3p electron. We have $$n+l = 3+1 = 4$$.

Now arranging by increasing (n+l): D has $$n+l = 4$$ (lowest), then A and B both have $$n+l = 5$$. Among A (n=3) and B (n=4), A has a lower n, so A has lower energy than B. Finally, C has $$n+l = 6$$ (highest).

The correct order of increasing energy is: $$D < A < B < C$$.

Hence, the correct answer is Option B.

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 are given an imaginary ion $$_{22}^{48}X^{3-}$$ and need to find the percentage ‘a’ by which neutrons exceed the number of electrons.

Since the atomic number (Z) = 22, the number of protons is 22. Furthermore, the mass number (A) = 48 implies that the number of neutrons is A − Z = 48 − 22 = 26. Because the ion carries a charge of 3−, the total number of electrons is 22 + 3 = 25.

Since the nucleus contains ‘a’ % more neutrons than the number of electrons, we set up $$\text{Number of neutrons} = \text{Number of electrons} + \frac{a}{100} \times \text{Number of electrons}$$.

Substituting the known values yields $$26 = 25 + \frac{a}{100} \times 25$$.

Rearranging gives $$26 - 25 = \frac{25a}{100}$$, which simplifies to $$1 = \frac{25a}{100}$$.

Solving for a, we find $$a = \frac{100}{25} = 4$$.

Therefore, the value of $$a$$ is 4.

We use the Heisenberg uncertainty principle to find the mass of the particle.

The Heisenberg uncertainty principle states:

$$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}$$

where $$\Delta x$$ is the uncertainty in position, $$\Delta v$$ is the uncertainty in velocity, $$m$$ is the mass, and $$h$$ is Planck's constant. The known values are $$\Delta v = 2.4 \times 10^{-26}$$ m/s, $$\Delta x = 10^{-7}$$ m, and $$h = 6.626 \times 10^{-34}$$ Js.

Rearranging the formula for mass and taking equality for minimum uncertainty gives:

$$m = \frac{h}{4\pi \cdot \Delta x \cdot \Delta v}$$

Substituting the given values, we have:

$$m = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 10^{-7} \times 2.4 \times 10^{-26}}$$

Calculating the denominator step by step, note that $$4 \times 3.14 = 12.56$$ and $$12.56 \times 2.4 = 30.144$$. The combined power of ten in the denominator is $$10^{-7} \times 10^{-26} = 10^{-33}$$, so the denominator becomes $$30.144 \times 10^{-33}$$.

Thus,

$$m = \frac{6.626 \times 10^{-34}}{30.144 \times 10^{-33}}$$

$$m = \frac{6.626}{30.144} \times 10^{-34+33}$$

$$m = 0.2198 \times 10^{-1}$$ kg, which simplifies to $$m = 0.02198$$ kg.

Converting this mass into grams yields $$m = 0.02198 \times 1000 = 21.98$$ g, and rounding to the nearest integer gives $$m \approx 22$$ g.

Hence, the mass of the particle is 22 g.

We need to find the maximum wavelength of a photon required to remove a photoelectron from a metal with work function $$\phi = 6.63 \times 10^{-19}$$ J. The photoelectric effect requires the photon energy to be at least equal to the work function of the metal; the minimum energy corresponds to the maximum wavelength since energy and wavelength are inversely related. At the threshold (minimum energy = work function), $$E_{photon} = \phi$$.

The energy of a photon is given by $$E = \frac{hc}{\lambda}$$, so at threshold $$\frac{hc}{\lambda_{max}} = \phi$$. Solving for $$\lambda_{max}$$ gives $$\lambda_{max} = \frac{hc}{\phi}$$.

Using $$h = 6.63 \times 10^{-34}$$ J s, $$c = 3 \times 10^8$$ m/s, and $$\phi = 6.63 \times 10^{-19}$$ J, we get $$\lambda_{max} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-19}}$$, which simplifies to $$\lambda_{max} = \frac{6.63 \times 3 \times 10^{-34+8}}{6.63 \times 10^{-19}}$$, then to $$\lambda_{max} = \frac{3 \times 10^{-26}}{10^{-19}}$$, and hence $$\lambda_{max} = 3 \times 10^{-7} \text{ m}$$.

Converting to nanometers: $$\lambda_{max} = 3 \times 10^{-7} \text{ m} = 300 \times 10^{-9} \text{ m} = 300 \text{ nm}$$. The correct answer is 300 nm.

We need to find the minimum uncertainty in the speed of an electron confined to a one-dimensional region of length $$2a_0$$, where $$a_0 = 52.9$$ pm is the Bohr radius.

We use the Heisenberg Uncertainty Principle in one dimension:

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Here, the uncertainty in position $$\Delta x = 2a_0$$. For the minimum uncertainty, we take the equality:

$$\Delta p = \frac{h}{4\pi \cdot \Delta x} = \frac{h}{4\pi \cdot 2a_0}$$

Since $$\Delta p = m \cdot \Delta v$$, we have:

$$\Delta v = \frac{h}{4\pi \cdot 2a_0 \cdot m}$$

Now we substitute the values. We have $$h = 6.63 \times 10^{-34}$$ J s, $$a_0 = 52.9$$ pm $$= 52.9 \times 10^{-12}$$ m, and $$m = 9.1 \times 10^{-31}$$ kg.

$$\Delta v = \frac{6.63 \times 10^{-34}}{4\pi \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31}}$$

Let us compute the denominator step by step. We have $$4\pi \approx 12.566$$, and:

$$12.566 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31}$$

$$= 12.566 \times 105.8 \times 10^{-12} \times 9.1 \times 10^{-31}$$

$$= 12.566 \times 962.78 \times 10^{-43}$$

$$= 12101.4 \times 10^{-43}$$

$$= 1.21014 \times 10^{-39}$$

Now dividing:

$$\Delta v = \frac{6.63 \times 10^{-34}}{1.21014 \times 10^{-39}} = \frac{6.63}{1.21014} \times 10^{5}$$

$$= 5.479 \times 10^{5} \text{ m s}^{-1}$$

Converting to km/s: $$\Delta v = 547.9 \approx 548$$ km s$$^{-1}$$.

Hence, the correct answer is 548.

We need to determine how many of the given sets of quantum numbers are valid.

Key Rules for Quantum Numbers: 1. Principal quantum number: $$n = 1, 2, 3, ...$$; 2. Azimuthal quantum number: $$l = 0, 1, 2, ..., (n-1)$$; 3. Magnetic quantum number: $$m_l = -l, -(l-1), ..., 0, ..., (l-1), l$$.

For set A ($$n = 3, l = 3, m_l = -3$$), for $$n = 3$$ the allowed values of $$l$$ are 0, 1, 2 (i.e., $$l$$ ranges from 0 to $$n - 1 = 2$$). Here $$l = 3$$ is NOT allowed since $$l$$ must be less than $$n$$. Hence, set A is invalid.

For set B ($$n = 3, l = 2, m_l = -2$$), $$l$$ can be 0, 1, 2 so $$l = 2$$ is valid. For $$l = 2$$, $$m_l$$ can be $$-2, -1, 0, +1, +2$$ so $$m_l = -2$$ is valid. Hence, set B is valid.

For set C ($$n = 2, l = 1, m_l = +1$$), $$l$$ can be 0, 1 so $$l = 1$$ is valid. For $$l = 1$$, $$m_l$$ can be $$-1, 0, +1$$ so $$m_l = +1$$ is valid. Hence, set C is valid.

For set D ($$n = 2, l = 2, m_l = +2$$), $$l$$ can be 0, 1 (i.e., $$l$$ ranges from 0 to $$n - 1 = 1$$). Here $$l = 2$$ is NOT allowed since $$l$$ must be less than $$n$$. Hence, set D is invalid.

Only sets B and C are valid. The number of correct sets of quantum numbers is 2.

We are given that an electron emitted from a hydrogen atom has a de Broglie wavelength of $$\lambda = 3.3 \times 10^{-10}$$ m. We need to find how many times the energy absorbed by the electron in the ground state is, compared to the minimum energy required for its escape (ionization energy).

The minimum energy required for the electron to escape from the hydrogen atom in the ground state is the ionization energy of hydrogen:

$$E_{min} = 13.6 \text{ eV} = 13.6 \times 1.6 \times 10^{-19} = 2.176 \times 10^{-18} \text{ J}$$

Now, let us find the kinetic energy of the emitted electron using its de Broglie wavelength. From the de Broglie relation $$\lambda = \frac{h}{mv}$$, the kinetic energy is:

$$KE = \frac{1}{2}mv^2 = \frac{h^2}{2m\lambda^2}$$

Substituting the given values:

$$KE = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (3.3 \times 10^{-10})^2}$$

Computing the numerator: $$(6.626)^2 = 43.90$$, so the numerator is $$43.90 \times 10^{-68}$$.

Computing the denominator: $$2 \times 9.1 \times (3.3)^2 = 2 \times 9.1 \times 10.89 = 198.2$$, with units $$10^{-31} \times 10^{-20} = 10^{-51}$$, so the denominator is $$198.2 \times 10^{-51}$$.

$$KE = \frac{43.90 \times 10^{-68}}{198.2 \times 10^{-51}} = 0.2215 \times 10^{-17} = 2.215 \times 10^{-18} \text{ J}$$

By the photoelectric effect principle, the total energy absorbed by the electron equals the minimum escape energy plus the kinetic energy of the emitted electron:

$$E_{absorbed} = E_{min} + KE = 2.176 \times 10^{-18} + 2.215 \times 10^{-18} = 4.391 \times 10^{-18} \text{ J}$$

Now we find the ratio of energy absorbed to the minimum escape energy:

$$\frac{E_{absorbed}}{E_{min}} = \frac{4.391 \times 10^{-18}}{2.176 \times 10^{-18}} \approx 2.02 \approx 2$$

So the energy absorbed by the electron is approximately 2 times the minimum energy required for its escape from the atom.

Hence, the correct answer is 2.

We need to find the maximum number of emission lines when an electron drops from $$n = 5$$ to the ground state ($$n = 1$$).

Use the formula for maximum emission lines

When an electron transitions from level $$n$$ to the ground state, the maximum number of spectral lines is given by:

$$\text{Number of lines} = \frac{n(n-1)}{2}$$

where $$n$$ is the initial energy level.

Substitute $$n = 5$$

$$\text{Number of lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = 10$$

List all possible transitions

The 10 possible transitions are:

$$5 \rightarrow 4, \; 5 \rightarrow 3, \; 5 \rightarrow 2, \; 5 \rightarrow 1$$

$$4 \rightarrow 3, \; 4 \rightarrow 2, \; 4 \rightarrow 1$$

$$3 \rightarrow 2, \; 3 \rightarrow 1$$

$$2 \rightarrow 1$$

The answer is 10.

We need to find the value of $$x$$ such that the velocity of the electron is $$x$$ times the velocity of the neutron when their de Broglie wavelengths are equal.

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{mv}$$

where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity of the particle.

For the electron and neutron to have the same wavelength:

$$\lambda_e = \lambda_n$$

$$\frac{h}{m_e v_e} = \frac{h}{m_n v_n}$$

Cancelling $$h$$ from both sides:

$$\frac{1}{m_e v_e} = \frac{1}{m_n v_n}$$

$$m_n v_n = m_e v_e$$

$$\frac{v_e}{v_n} = \frac{m_n}{m_e}$$

$$m_e = 9.1 \times 10^{-31} \text{ kg}$$

$$m_n = 1.6 \times 10^{-27} \text{ kg}$$

$$x = \frac{v_e}{v_n} = \frac{m_n}{m_e} = \frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$$

$$x = \frac{1.6}{9.1} \times 10^{-27-(-31)} = \frac{1.6}{9.1} \times 10^{4}$$

$$= 0.17582 \times 10^{4}$$

$$= 1758.2$$

Rounding to the nearest integer: $$x = 1758$$

Hence, the value of $$x$$ is 1758.

We need to find the longest wavelength of light that can ionise $$Li^{2+}$$ from its ground state. The longest wavelength corresponds to the minimum energy required for ionisation.

$$Li^{2+}$$ is a hydrogen-like ion with atomic number $$Z = 3$$ and a single electron. For such ions, the Bohr model applies directly. The energy of the electron in the $$n$$-th shell is $$E_n = E_1^{(H)} \times \frac{Z^2}{n^2}$$, where $$E_1^{(H)} = -2.2 \times 10^{-18}$$ J is the energy of the electron in the first shell of the hydrogen atom.

For the ground state of $$Li^{2+}$$, $$n = 1$$ and $$Z = 3$$: $$E_1 = -2.2 \times 10^{-18} \times \frac{9}{1} = -19.8 \times 10^{-18}$$ J.

The ionisation energy equals $$|E_1| = 19.8 \times 10^{-18}$$ J (energy to go from $$n = 1$$ to $$n = \infty$$).

Using the relation $$E = \frac{hc}{\lambda}$$, the longest wavelength of light that can ionise $$Li^{2+}$$ is $$\lambda = \frac{hc}{E}$$.

Substituting: $$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{19.8 \times 10^{-18}}$$.

Numerator: $$6.63 \times 3 = 19.89$$, so the numerator is $$19.89 \times 10^{-26}$$.

Therefore: $$\lambda = \frac{19.89 \times 10^{-26}}{19.8 \times 10^{-18}} = \frac{19.89}{19.8} \times 10^{-26+18} = 1.0045 \times 10^{-8}$$ m.

Thus $$x \approx 1$$ (nearest integer).

So, the answer is $$1$$.

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.

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.

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 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 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 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.

First, let us recall the Aufbau principle, which states that electrons occupy the available orbitals in the order of increasing $$(n+l)$$ value, and if two orbitals have the same $$(n+l)$$ value, then the one with the lower $$n$$ (principal quantum number) is filled first.

Now, we list all the orbitals in the increasing order of $$(n+l)$$ up to the region where the sixth period ends (i.e., up to the noble-gas element radon, $$Z = 86$$):

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

We see that the very first orbital whose principal quantum number is $$6$$ is $$6s$$. So, after finishing $$5p$$ (which completes the fifth period at xenon, $$Z = 54$$), the next orbital to receive electrons is $$6s$$.

After filling the $$6s$$ orbital with two electrons, we check the next entry in our ordered list of orbitals. That orbital is $$4f$$. Although its principal quantum number $$n = 4$$ is numerically less than 6, its $$(n+l)$$ value $$=4+3=7$$ is larger than that of $$6s$$ $$\big(n+l = 6+0=6\big)$$, so, according to the Aufbau ordering, $$4f$$ comes directly after $$6s$$. Hence, as we progress through the sixth period (elements $$Z = 57$$ to $$Z = 70$$, the lanthanoids), the $$4f$$ subshell gets filled.

Once $$4f$$ is completely filled with its $$14$$ electrons, the next orbital in the list is $$5d$$. Thus, elements $$Z = 71$$ to $$Z = 80$$ (the sixth-period transition elements) add electrons to $$5d$$.

When $$5d$$ reaches its capacity of $$10$$ electrons, the sequence finally moves on to $$6p$$. The $$6p$$ orbitals are filled across the last part of the sixth period, ending with radon, $$Z = 86$$.

Putting all these steps together, the orbitals that receive electrons in the sixth period are, in chronological order, $$6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$$.

Comparing this list with the supplied options, we observe that Option A lists the orbitals exactly as $$6s,\,4f,\,5d,\,6p$$, which matches our derived sequence.

Hence, the correct answer is Option A.

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.

To determine the atomic number of the element whose temporary IUPAC name is “Unnilunium”, we recall the systematic naming convention introduced by the International Union of Pure and Applied Chemistry (IUPAC) for elements whose permanent names were not yet assigned.

According to this convention, the name of an element is built from three-letter Latin/Greek numerical prefixes that denote each digit of its atomic number, read from left to right, and finished with the ending “-ium”. The officially adopted prefixes are:

$$\text{un} = 1, \qquad \text{bi} = 2, \qquad \text{tri} = 3,$$

$$\text{quad} = 4, \qquad \text{pent} = 5, \qquad \text{hex} = 6,$$

$$\text{sept} = 7, \qquad \text{oct} = 8, \qquad \text{enn} = 9,$$

$$\text{nil} = 0.$$

In the given name “Unnilunium” we can clearly identify three numerical prefixes followed by the suffix “-ium”. We write them one after another:

$$\text{un} \;+\; \text{nil} \;+\; \text{un} \;+\; \text{ium}$$

Now we translate each prefix into the corresponding digit:

$$\text{un} \rightarrow 1, \qquad \text{nil} \rightarrow 0, \qquad \text{un} \rightarrow 1.$$

Placing these digits in the same order, we obtain the three-digit sequence

$$101.$$

So, the temporary systematic name “Unnilunium” represents the element whose atomic number is $$101$$. Historically, once the element received its permanent name, it became known as Mendelevium (symbol $$\text{Md}$$).

Hence, the correct answer is Option C.

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

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

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

Let us now introduce the variable

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

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.

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.