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.

The subsidiary (azimuthal) quantum number $$l$$ determines the subshell in which an electron is present. Different values of $$l$$ correspond to different orbital shapes such as $$s$$, $$p$$, $$d$$, and $$f$$ orbitals. Therefore, $$l$$ determines the shape of the orbital.

Hence, Statement (A) is true.

The given boundary surface diagram represents a dumbbell-shaped orbital with its two lobes aligned along the $$x$$-axis. Such an orientation is characteristic of the $$2p_x$$ orbital.

Hence, Statement (B) is true.

The $$+$$ and $$-$$ signs shown on the two lobes of a $$p$$ orbital represent the phase (sign) of the wave function $$\psi$$. They do not indicate positive or negative electrical charge, since the entire electron cloud carries the same negative charge.

Hence, Statement (C) is false.

A $$2p_x$$ orbital has its nodal plane perpendicular to the $$x$$-axis, which is the $$yz$$ plane ($$x=0$$). The probability density is zero only on this plane and not on the $$xy$$ plane.

Hence, Statement (D) is false.

Therefore, only Statements (A) and (B) are correct.

Hence, the correct answer is Option (D): (A) and (B) only.

For a one-electron (hydrogen-like) species the energy of the stationary state with principal quantum number $$n$$ is given by

$$E=-\frac{13.6\;{\rm eV}\;Z^{2}}{n^{2}} \qquad -(1)$$

where $$Z$$ is the atomic number of the nucleus.

For the present question $$n$$ is kept constant. Therefore from $$-(1)$$ we get the functional dependence

$$E \propto -Z^{2} \qquad -(2)$$

Key features of the relation $$-(2)$$:

1. The energy is always negative (bound state), so every point of the curve lies below the $$E=0$$ axis.
2. As $$Z$$ increases, $$|E|$$ increases quadratically; hence the curve falls more and more steeply—in other words, it is a downward-opening parabola in the fourth quadrant (positive $$Z$$, negative $$E$$).
3. If one extrapolates mathematically to $$Z=0$$ the energy tends to $$0$$, so the parabola’s vertex touches the origin, but for real atoms the physically relevant domain is $$Z\ge 1$$ where the curve starts at $$E=-13.6/n^{2}$$ eV and descends.

Among the given sketches, the only graph that shows a parabolic fall completely in the negative $$E$$ region, becoming more negative as $$Z$$ increases, is Option A.

Hence the correct choice is:
Option A which is: the downward-opening parabolic $$E$$ vs $$Z$$ plot.

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:

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

Case 1: Electrons with $$l = 1$$ (p-electrons)

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

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

Case 2: Electrons with $$l = 2$$ (d-electrons)

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.

To determine the incorrect statement, we evaluate each option using the quantum mechanical model of the hydrogen atom.

(A) The probability density of finding the electron is maximum at the nucleus.

This statement is correct. The probability density is given by $$|\psi|^2.$$

For an $$s$$-orbital, where the angular momentum quantum number is $$l=0,$$ the wavefunction does not vanish at the origin. Therefore, $$|\psi|^2$$ attains its maximum value at the nucleus, $$r=0.$$

It should not be confused with the radial probability distribution, $$4\pi r^2|\psi|^2,$$ which is zero at the nucleus and has its maximum value at the Bohr radius, $$a_0.$$

(B) The electron can be found at a distance $$2a_0$$ from the nucleus.

This statement is correct. The wavefunction decreases asymptotically with increasing distance and becomes zero only at infinity. Hence, there is a finite probability of finding the electron at any finite distance from the nucleus, including $$2a_0.$$

(C) The $$1s$$ orbital is spherically symmetrical.

This statement is correct. Since the $$1s$$ orbital has $$l=0,$$ its wavefunction has no angular dependence, giving rise to a perfectly spherical electron density distribution around the nucleus.

(D) The total energy of the electron is maximum when it is at a distance $$a_0$$ from the nucleus.

This statement is incorrect. In a stationary state such as the $$1s$$ orbital, the total energy of the electron remains constant and is independent of its position.

For the ground state of the hydrogen atom, $$E=-13.6\ \text{eV},$$ regardless of the instantaneous distance of the electron from the nucleus. Although the kinetic and potential energies vary with distance, their sum remains constant.

Hence, statement (D) is incorrect and is the correct answer.

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.

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 energy of an electron in the $$n$$th Bohr orbit of hydrogen is given by:

$$E_n = \frac{E_1}{n^2}$$

Given $$E_1 = -2.18 \times 10^{-18}$$ J.

$$E_3 = \frac{E_1}{3^2} = \frac{E_1}{9} = \frac{-2.18 \times 10^{-18}}{9} = -0.2422 \times 10^{-18} \text{ J}$$

$$E_3 = \dfrac{1}{9}$$ of $$E_1$$. This means the energy in the third Bohr orbit is $$\dfrac{1}{9}$$th of the value of the energy in the first orbit.

The correct answer is Option A: $$\dfrac{1}{9}$$th of the value.

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 have the wave function for the $$2s$$ orbital:

A radial node occurs where $$\Psi = 0$$. Since the exponential factor $$e^{-r/2a_0}$$ is never zero for finite $$r$$, and the constant prefactor is non-zero, the node must come from

Solving for $$r$$,

Hence, $$r_0 = 2a_0$$. So, the answer is $$r_0 = 2a_0$$.

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

Using the Rydberg formula,

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

For the (He^+) transition with (Z=2), (n_2=4), and (n_1=2),

$$\frac{1}{\lambda}=R(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right).$$

Simplifying,

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

For hydrogen ((Z=1)),

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

Equating the two expressions,

$$\frac{3R}{4}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),$$

$$\frac{3}{4}=\frac{1}{n_1^2}-\frac{1}{n_2^2}.$$

Taking (n_1=1) and (n_2=2),

$$\frac{1}{1^2}-\frac{1}{2^2}=1-\frac{1}{4}=\frac{3}{4},$$

which satisfies the required condition.

Hence, the corresponding transition in the hydrogen spectrum is

$$\boxed{n=2\rightarrow n=1}.$$

Using the $$(n + l)$$ rule for ordering orbital energies (for multi-electron atoms):

A: $$n = 3, l = 0$$ → 3s orbital, $$(n+l) = 3$$

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

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

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

Higher $$(n+l)$$ means higher energy. For orbitals with same $$(n+l)$$, the one with higher $$n$$ has higher energy.

Ordering by decreasing energy:

D (n+l = 5) > B (n+l = 4, n = 4) > C (n+l = 4, n = 3) > A (n+l = 3)

The correct decreasing order is: D > B > C > A

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.

According to Moseley’s law:

$$\sqrt{\nu} \propto Z$$

or

$$\nu \propto Z^2$$

where

$$\nu = \text{frequency of characteristic X-rays}$$

and

$$Z = \text{atomic number}$$

Therefore:

- A plot of $$\sqrt{\nu}$$ versus $$Z$$ gives a straight line.
- A plot of $$\nu$$ versus $$Z$$ gives a parabola.

Hence the correct graph is:

Option C

i.e. the graph $$\sqrt{\nu}$$ versus $$Z$$ gives a straight line.

The maximum number of electrons that can be accommodated in a shell with principal quantum number $$n$$ is given by

For $$n = 4$$,

We can verify this by counting the subshells in the 4th shell: 4s has 2 electrons, 4p has 6, 4d has 10, and 4f has 14, giving a total of $$2 + 6 + 10 + 14 = 32$$ electrons.

So, the answer is $$32$$.

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

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

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.

Explanation

The ion contains 55 protons.

An element with atomic number

$$Z=55$$

is cesium, $$Cs$$.

The electronic configuration of neutral cesium is

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

or

$$[Xe],6s^1$$

The ion is in the unipositive state, which means it has a charge of $$+1$$.

To form $$Cs^+$$, the outermost electron is removed from the $$6s$$ orbital.

Therefore, the electronic configuration of $$Cs^+$$ becomes

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

Now count the electrons present in all the $$s$$ subshells.

From $$1s$$

$$2$$ electrons

From $$2s$$

$$2$$ electrons

From $$3s$$

$$2$$ electrons

From $$4s$$

$$2$$ electrons

From $$5s$$

$$2$$ electrons

The $$6s$$ orbital is empty in $$Cs^+$$.

Hence, the total number of $$s$$ electrons is

$$2+2+2+2+2=10$$

Therefore, the number of $$s$$ electrons present in the unipositive ion is

$$10$$

Hence, the correct answer is

$$10$$

For 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.

We begin by considering the Paschen series, which involves transitions to $$n = 3$$.

First, for the 1st line (4→3), we write $$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R \times \frac{7}{144}$$, and hence $$\lambda_1 = 720$$ nm.

Next, for the 2nd line (5→3), we have $$\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{25}\right) = R \times \frac{16}{225}$$.

Then, substituting these into the ratio, we find $$\frac{\lambda_2}{\lambda_1} = \frac{7/144}{16/225} = \frac{7 \times 225}{144 \times 16} = \frac{1575}{2304}$$,
and therefore $$\lambda_2 = 720 \times \frac{1575}{2304} = 492 \text{ nm}$$.

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.

1. d-Orbitals ($$d_{xy}, d_{yz}, d_{xz}, d_{z^2}, d_{x^2-y^2}$$):
   • Non-axial ($$d_{xy}, d_{yz}, d_{xz}$$): The electron density loops are oriented between the axes at a $$45^\circ$$ angle.
   • Axial ($$d_{z^2}, d_{x^2-y^2}$$): The electron density lies directly along the axes.
     • $$d_{x^2-y^2}$$ has lobes directly on the $$x$$ and $$y$$ axes.
     • $$d_{z^2}$$ has its main lobes directly along the $$z$$-axis.

Hence, Correct answer is 2

The orbital angular momentum of an electron is given by:

$$L = \sqrt{l(l+1)} \cdot \frac{h}{2\pi}$$

where $$l$$ is the azimuthal quantum number.

For the 3s orbital: $$n = 3$$, $$l = 0$$.

$$L = \sqrt{0(0+1)} \cdot \frac{h}{2\pi} = 0$$

Therefore $$x = 0$$.

This matches the answer key value of $$\mathbf{0}$$.

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 relationship between the radius of the 3rd and 4th Bohr orbits of a hydrogen atom.

The radius of the $$n^{th}$$ orbit in a hydrogen atom is given by $$r_n = a_0 \cdot n^2$$, where $$a_0$$ is the Bohr radius. Thus the radius is proportional to $$n^2$$.

Substituting $$n=3$$ yields $$r_3 = a_0 \cdot (3)^2 = 9a_0$$, and substituting $$n=4$$ yields $$r_4 = a_0 \cdot (4)^2 = 16a_0$$.

Taking the ratio, we find $$\frac{r_4}{r_3} = \frac{16a_0}{9a_0} = \frac{16}{9}$$, which implies $$r_4 = \frac{16}{9} r_3$$.

Therefore, the correct answer is Option B: $$r_4 = \frac{16}{9}r_3$$.

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$$^{-1}$$.

1. Identify the Number of Radial Nodes

The number of radial nodes (points where the probability of finding an electron drops to zero) is calculated using the formula:

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

For a 2s orbital:

• Principal quantum number ($$n$$) = $$2$$
• Azimuthal quantum number ($$l$$) = $$0$$ (for an s-orbital)

$$\text{Radial Nodes} = 2 - 0 - 1 = 1$$

This means the graph must touch the $$x$$-axis (where $$\psi^2 = 0$$) exactly once before eventually tapering off to zero at infinity.

2. Behavior at the Nucleus ($$r = 0$$)

For all s-orbitals, the electron wave function ($$\psi$$) has a finite, non-zero value at the nucleus because there is no angular node passing through the center. Therefore, the probability density $$\psi^2(r)$$ starts at a maximum value on the $$y$$-axis when $$r = 0$$.

Hence, Option B is correct because it matches all the points

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.

Let us evaluate each statement:

(A) The electronic configuration of Cr is [Ar] 3d$$^5$$4s$$^1$$.

This is correct. Chromium has an exceptional configuration due to the extra stability of a half-filled 3d subshell.

(B) The magnetic quantum number may have a negative value.

This is correct. The magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$, so it can take negative values.

(C) In the ground state of an atom, the orbitals are filled in order of their increasing energy order.

This is correct. This is the Aufbau principle, which states that electrons fill orbitals starting from the lowest energy level.

(D) The total number of nodes are given by n - 2.

This is incorrect. The total number of nodes (radial + angular) is given by $$n - 1$$, not $$n - 2$$.

Therefore, statements (A), (B), and (C) are correct.

Hence, the correct answer is Option D ((A), (B) & (C) only).

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

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