NCERT Exemplar Solutions Class 12 Physics Atoms — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Atoms, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 12.1 to 12.13; here they are Questions 1 to 13. Questions 1 to 7 are MCQ I (exactly one option is correct) and Questions 8 to 13 are MCQ II (one or more options may be correct). Four ideas carry the page:

• Hydrogen-like scaling. For one electron around a nucleus of charge $Ze$, $r_n = a_0 n^2/Z$ and $E_n = -13.6\,Z^2/n^2$ eV.
• Where Bohr’s model stops. It needs a single electron moving under a central force in an inertial frame; many-electron atoms, molecules and the true ground-state angular momentum lie outside it.
• Energy bookkeeping. An atom takes only the energy of a transition ($10.2$ eV for $n=1\to2$), never part of it; a free electron and proton must radiate to bind.
• Spectral series. Lines of a series crowd together towards the series limit.

Key insight. Each question is a test of one condition behind the Bohr model or one rule about energy levels. Find the condition and the option follows.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

Taking the Bohr radius as $a_0 = 53$ pm, the radius of $\text{Li}^{++}$ ion in its ground state, on the basis of Bohr’s model, will be about

Solution. For a hydrogen-like ion the Bohr radius shrinks because the nucleus pulls harder: $r_n = \dfrac{n^2 a_0}{Z}$. $\text{Li}^{++}$ has $Z = 3$ and its electron is in $n = 1$, so

$$r = \frac{53}{3} \approx 17.7 \text{ pm} \approx 18 \text{ pm}.$$

Question 2

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is $B = -\dfrac{me^4}{8n^2\varepsilon_0^2h^2}$ ($m$ = electron mass).

If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be $B = -\dfrac{Me^4}{8n^2\varepsilon_0^2h^2}$ ($M$ = proton mass).

This last expression is not correct because

Solution. The electron is bound and so moves in a circle with a centripetal acceleration. A frame attached to an accelerating body is non-inertial, and Newton’s laws (and the Coulomb force balance used by Bohr) hold only in inertial frames. So replacing $m$ by $M$ by “swapping roles” is not legitimate. Option (a) is wrong because $n$ is still an integer in any case; (b) is wrong because quantisation of angular momentum can be applied to any bound orbiting particle; and (d) is wrong because the proton would still go in a circle, although this is irrelevant to why the formula fails.

Question 3

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because

Solution. In hydrogen the only force on the electron is the pull of the nucleus, which is central, so angular momentum is conserved and the orbits are simple circles. With several electrons, each one is also pushed by the others. The sum of the nuclear attraction and these repulsions does not point towards the nucleus, so the net force is not central and the Bohr orbits are no longer defined. Screening, in (c), is a consequence of this repulsion, not the primary reason the model breaks down. Electrons do not “collide” in the sense of (b), and the nucleus–electron force remains Coulombic, so (d) is false.

Question 4

For the ground state, the electron in the H-atom has an angular momentum $= \hbar$, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

Solution. In quantum mechanics the ground state of hydrogen has orbital angular momentum $L = 0$, not $\hbar$. The ground-state wavefunction is spherically symmetric, so there is no preferred direction and the paradox of “infinitely many orbits” never arises. The difficulty comes from the Bohr model giving the wrong value $\hbar$. The other options give invented reasons: all orientations of $\mathbf{L}$ have the same energy, so (b) is false; $\mathbf{L}$ need not be along the spin, so (c) is false; and there is no rule of horizontal orbits, so (d) is false.

Question 5

$\text{O}_2$ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

Solution. The nuclear force acts only over a range of about $1$ fm ($10^{-15}$ m). The two nuclei in $\text{O}_2$ sit about $1$ Å ($10^{-10}$ m) apart, a hundred thousand times further than the range of the force, so it is negligible between them. Binding in the molecule is purely electromagnetic, through the shared electrons. The balance of neutrons and protons has no bearing on this, which removes (d).

Question 6

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

Solution. In an inelastic collision, kinetic energy lost goes into internal energy of the atoms, and an atom can absorb only an amount equal to a difference between its energy levels. The smallest such amount for a ground-state hydrogen atom is the $n=1\to2$ excitation, $-3.4 – (-13.6) = 10.2$ eV. If the kinetic energy available is less than that, the collision must be elastic. The largest loss in the collision is therefore the excitation of one atom, $10.2$ eV. (Taking $13.6$ eV would ionise an atom, and $20.4$ eV would excite both atoms, but the maximum reduction that is guaranteed to be possible is one excitation. The question’s key takes $10.20$ eV.)

Question 7

A set of atoms in an excited state decays.

Solution. Spontaneous emission needs no outside agent, which rules out (b) and (d). Decay is a random, statistical process: each atom goes independently, at its own time, so they do not all go together, ruling out (c). Each excited atom may drop to any lower level allowed by the selection rules, with probabilities that depend on the transition. The set therefore decays in general into any lower-energy state.

Question 8

An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,

Solution.

• (a) The electron feels the pull of two centres, so the force on it is not central about any one of them, and a circular orbit is not possible. True.
• (b) The factor $2^4$ would come from $Z^2$ scaling with $Z = 4$ or a similar argument, which does not apply to two separated protons. The binding energy of $\text{H}_2^+$ is of the same order as hydrogen’s, not sixteen times larger. False.
• (c) The electron’s orbit encloses both protons, since the electron is attracted to each. True.
• (d) $\text{H}_2^+$ is a stable ion: the shared electron lowers the energy compared with a separated proton and H atom. False.

Question 9

Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

Solution. A free electron and proton have total energy at least zero (kinetic energy only), while a bound hydrogen atom has negative energy. To end in a bound state, the pair must give up energy, and nothing in a two-body encounter can take it away except radiation. So the capture cannot happen by energy conservation alone, and it can happen only if a photon is emitted at the same time: (a) and (b) are true. Momentum and angular momentum can be conserved perfectly well in the process (the photon carries them away and the atom as a whole can recoil), so (c) and (d) are not the obstacle.

Question 10

The Bohr model for the spectra of a H-atom

Solution. The model describes a single electron in the field of a single nucleus. In $\text{H}_2$ the electrons sit in the field of two nuclei and each other, so the force is not central and the model fails: (a) is true. A neutral He atom has two electrons which repel each other, again not covered, so (b) is true. Temperature is irrelevant to the energy levels, so (c) is false. The model predicts only discrete lines, from transitions between allowed orbits; the continuum is not predicted by it, so (d) is false.

Question 11

The Balmer series for the H-atom can be observed

Solution. The Balmer series consists of the transitions $n \to 2$ for $n = 3, 4, 5, \dots$, that is, from higher excited states down to the first excited state: (b) is true. Falls to the ground state belong to the Lyman series, so (a) is false, and not every transition is Balmer, so (c) is false. The frequencies are $\nu = cR\left(\dfrac14 – \dfrac1{n^2}\right)$. As $n$ grows the gap between successive lines shrinks and the lines pile up towards the series limit $cR/4$, so the higher frequencies are more closely packed: (d) is true.

Question 12

Let $E_n = \dfrac{-1}{8\varepsilon_0^2}\dfrac{me^4}{n^2h^2}$ be the energy of the $n^{\text{th}}$ level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $(E_2 – E_1)/h$ falls on it,

Solution. Each photon has exactly the energy $E_2 – E_1$ needed for the $n=1\to2$ jump, so the radiation is resonantly absorbed and some atoms are raised to the first excited state: (b) is true and (a) is false. Absorption is a probabilistic process and the radiation has a limited number of photons, so not all atoms are excited at once: (c) is false. A photon of this energy is too small for $n=1\to3$, which needs $E_3 – E_1 > E_2 – E_1$, and an atom accepts a photon only of the exact energy gap, so none reach $n = 3$: (d) is true.

Question 13

The simple Bohr model is not applicable to $\text{He}^4$ atom because

Solution. The neutral helium atom has two electrons. They repel each other, so the force on either electron is the nuclear pull plus a repulsion from the other, and this is not a central force. The Bohr model, built for a single electron, therefore does not apply directly: (c) and (d) together give the reason. Being an inert gas, (a), is a result of the closed shell, not the cause of the failure, and neutrons, in (b), only add to the nuclear mass without changing the electron’s force law.

Common mistakes

• Using $r_n \propto n^2$ without the $1/Z$ in Question 1. The $\text{Li}^{++}$ radius is $a_0/3$, not $a_0$ or $a_0/9$.
• Swapping electron and proton frames in Question 2. The electron is accelerating, so its rest frame is not inertial.
• Blaming screening or collisions in Question 3. The real obstacle is that the electron–electron repulsion destroys the central force.
• Choosing $13.6$ eV in Question 6. A collision can excite an atom by $10.2$ eV but cannot ionise it in this setting.
• Marking only one option in MCQ II. Questions 8 to 13 each need every true statement; for example, Question 12 has both (b) and (d).
• Confusing the series in Question 11. Balmer is $n \to 2$; $n \to 1$ is Lyman.

Practise next

• Exemplar Atoms: Very Short, Short and Long Answer Questions — the numerical and reasoning questions on Bohr’s model, reduced mass and positronium.
• NCERT textbook Chapter 12 — Atoms for the Bohr postulates, the energy-level diagram and the hydrogen spectral series.