NCERT Exemplar Solutions Class 12 Physics Nuclei — Multiple Choice Questions Q1–10

NCERT Exemplar Problems, Class 12 Physics — Nuclei, MCQ I and MCQ II. All 10 questions solved.

The book numbers these questions 13.1 to 13.10; here they are Questions 1 to 10. Questions 1 to 7 are MCQ I (exactly one option is correct) and Questions 8 to 10 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Radioactive decay is random. Each nucleus has a fixed chance of decaying, so a half-life describes an average over many atoms, not a certainty for any one sample.
• Energy decides what can happen. A decay or reaction proceeds only if it lowers the energy, and the Q-value is a difference of masses times $c^2$.
• Two forces compete. The short-range nuclear attraction binds nucleons; the long-range Coulomb repulsion between protons pulls the nucleus apart, which is why heavy nuclei need extra neutrons and why fusion needs great energy.
• Rates, not amounts. Decay rate is $R = \lambda N$, so a larger decay constant means a faster fall, not necessarily a larger rate at every instant.

Key insight. Most of these are not calculations but a decision about which effect dominates, randomness or averages, attraction or repulsion, energy gained or lost. Questions 4, 9 and 10 need a little algebra.

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

Question 1

Suppose we consider a large number of containers each containing initially $10000$ atoms of a radioactive material with a half life of $1$ year. After $1$ year,

Solution. Each atom independently has a probability $\tfrac12$ of surviving one half-life. The number left in a container is therefore a random variable with mean $10000\times\tfrac12 = 5000$ and a spread (standard deviation) of $\sqrt{10000\times\tfrac12\times\tfrac12} = 50$. Different containers give different counts scattered about $5000$, and the average over many containers comes very close to $5000$. Options (a) and (b) claim every container is identical, which would make decay deterministic. Option (d) is false because a container can, by chance, retain slightly more than $5000$ atoms.

Question 2

The gravitational force between a H-atom and another particle of mass $m$ will be given by Newton’s law $F = G\dfrac{M\,m}{r^2}$, where $r$ is in km and

Solution. Gravitational mass equals inertial mass, and the inertial mass of a bound system is its total energy divided by $c^2$. A hydrogen atom has less energy than a free proton and electron by exactly the binding energy $B = 13.6$ eV, so its mass is $$M = m_{proton} + m_{electron} – \frac{B}{c^2}.$$ This rules out (a), which ignores the mass defect, and (c), since $M$ is just the atom’s mass. Option (d) uses the potential energy, $|V| = 27.2$ eV, which is twice the binding energy; the kinetic energy of the electron ($+13.6$ eV) partly offsets it, and only the net $13.6$ eV is the mass defect.

Question 3

When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom

Solution. The electronic energy levels depend on the nuclear charge $Ze$ that binds the electrons. In $\alpha$-decay $Z$ falls by $2$ and in $\beta$-decay it changes by $1$, so the new atom has different electron levels (the daughter is a different element). In $\gamma$-decay the nucleus only drops from an excited to a lower state with the same $Z$ and the same number of electrons, so the electron levels are untouched.

Question 4

$M_x$ and $M_y$ denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The $Q$-value for a $\beta^-$ decay is $Q_1$ and that for a $\beta^+$ decay is $Q_2$. If $m_e$ denotes the mass of an electron, then which of the following statements is correct?

Solution. Use atomic masses throughout, so electrons are counted correctly.

• $\beta^-$ decay, ${}^A_ZX \to {}^A_{Z+1}Y + e^- + \bar\nu$. The parent atom has $Z$ electrons; the daughter atom has $Z+1$. The extra electron in the daughter atom is exactly the one emitted, so the electron masses cancel and $Q_1 = (M_x – M_y)c^2$.
• $\beta^+$ decay, ${}^A_ZX \to {}^A_{Z-1}Y + e^+ + \nu$. The daughter atom has only $Z-1$ electrons, so one electron of the parent is surplus, and a positron is also created. The products’ masses, counted with the neutral daughter atom, include $M_y + m_e$ (surplus electron) $+ m_e$ (positron), so $Q_2 = (M_x – M_y – 2m_e)c^2$.

Question 5

Tritium is an isotope of hydrogen whose nucleus Triton contains $2$ neutrons and $1$ proton. Free neutrons decay into $p + \bar e + \bar\nu$. If one of the neutrons in Triton decays, it would transform into He$^3$ nucleus. This does not happen. This is because

Solution. A neutron inside a nucleus can decay only if the final nucleus has lower energy than the initial one. That is the book’s point: whether a bound neutron decays depends on the energies of the nuclei, not on the neutron alone. Options (b) and (d) are not the reason: the electron is simply emitted, as in any $\beta^-$ decay, and decay needs no external perturbation. Option (c) is wrong because one neutron changing into a proton gives $2$ protons, which is He$^3$ ($Z=2$), not $3$.

An honest note. Real tritium does in fact $\beta$-decay to He$^3$, with a half-life of about $12.3$ years and a $Q$-value of about $18.6$ keV, so the question’s premise is a simplification. The principle the book wants is still the right one: the energy of the final nucleus decides whether the decay is allowed.

Question 6

Heavy stable nuclei have more neutrons than protons. This is because of the fact that

Solution. The nuclear force between two nucleons is, to a good approximation, the same for $pp$, $nn$ and $np$ pairs, and it acts only between close neighbours. The Coulomb repulsion between protons acts over the whole nucleus, and its total grows roughly as $Z^2$. Adding neutrons increases the attractive nuclear binding without adding any repulsion, so a heavy nucleus needs extra neutrons to stay stable. Option (a) is irrelevant to stability, (c) describes a process, not a cause, and (d) is false because the nuclear force is charge-independent.

Question 7

In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because

Solution. The neutrons lose energy in elastic collisions. When a light neutron strikes a very heavy nucleus it bounces back almost like a ball off a wall: the heavy nucleus hardly recoils and the neutron keeps nearly all its speed. The fraction of energy transferred is largest when the two masses are equal, which is why hydrogen-rich materials (water) and carbon (graphite) work as moderators. The other options give reasons that have nothing to do with slowing the neutrons.

Question 8

Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact:

Solution. The attractive nuclear force acts only over about $10^{-15}$ m, so two deuterons must be brought that close before they bind: (a). But both nuclei are positive, and the Coulomb repulsion between them rises steeply as they approach; they must climb this Coulomb barrier, so they need very high kinetic energy, hence the very high temperatures: (b). Complete ionisation (c) is a consequence of high temperature, not a cause of the difficulty. Nuclei do not have to break up first (d); that is false.

Question 9

Samples of two radioactive nuclides A and B are taken. $\lambda_A$ and $\lambda_B$ are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?

Solution. The decay rate is $R = R_0e^{-\lambda t}$, so the ratio of the two rates is $$\frac{R_A}{R_B} = \frac{R_{A0}}{R_{B0}}\,e^{-(\lambda_A-\lambda_B)t}.$$ The rates are equal when this ratio equals $1$.

• (a) With $\lambda_A=\lambda_B$ the ratio stays at its initial value $2$ for ever. Never equal.
• (b) The ratio starts at $2$ and, since $\lambda_A > \lambda_B$, falls steadily towards $0$. It passes through $1$ when $e^{-(\lambda_A-\lambda_B)t} = \tfrac12$, i.e. at $t = \dfrac{\ln 2}{\lambda_A-\lambda_B}$. Possible.
• (c) The ratio starts at $\tfrac12$ and A decays faster, so the ratio only falls further below $1$. Never equal.
• (d) The condition $R_B(0) = R_A(2\ \text{h})$ means $R_B(0) < R_A(0)$, so the ratio $R_{A0}/R_{B0}$ is greater than $1$ and A falls faster than B ($\lambda_A>\lambda_B$). The ratio falls from a value above $1$ through $1$, at a time $t = \dfrac{2\lambda_A}{\lambda_A-\lambda_B}$ hours, which is positive. Possible.

Question 10

The variation of decay rate of two radioactive samples A and B with time is shown in Fig. 13.1.

The figure plots $dN/dt$ against $t$ for two falling curves. Curve A starts higher on the axis and falls more steeply; curve B starts lower and falls more gently. The two curves cross at one point, after which B lies above A and both continue to flatten towards the time axis.

Which of the following statements are true?

Solution. The decay rate is $R = R_0e^{-\lambda t}$, and on a graph of $R$ against $t$ the faster-falling curve belongs to the larger decay constant. Curve A falls more steeply, so $\lambda_A > \lambda_B$. A started with the higher rate, but because it decays faster the curves meet at a certain instant, and after that B’s rate exceeds A’s. So (c) is true: A has the larger decay constant but is not the faster decayer (higher rate) at all times. (d) is also true: B, with the smaller constant, catches up with A’s rate at the crossing point. Option (a) claims “always”, which the crossing disproves, and (b) reverses the decay constants.

Common mistakes

• Thinking every container has exactly $5000$ atoms in Question 1. Decay is random; only the average is close to $5000$.
• Subtracting $|V|$ instead of $B$ in Question 2. The mass defect is the binding energy, $13.6$ eV, not the $27.2$ eV potential energy.
• Forgetting the $2m_e$ in $\beta^+$ decay in Question 4. With atomic masses, one electron is surplus and a positron is created, so $2m_e$ comes off.
• Believing heavy nuclei slow neutrons best in Question 7. A light neutron bounces off a heavy nucleus almost unchanged in speed.
• Forgetting that a faster-decaying sample cannot catch up from below in Question 9. In (c) the ratio only moves further from $1$.
• Confusing decay constant with decay rate in Question 10. A larger $\lambda$ means a faster fall, not a larger rate at every time.

Practise next

• Exemplar Nuclei: Very Short, Short and Long Answer Questions — where the same ideas are applied to binding energy, decay chains and a $\beta$-decay of the neutron.
• NCERT textbook Chapter 13 — Nuclei for the theory of binding energy, radioactive decay law, $Q$-values, fission and fusion.
• Exemplar Atoms — the previous unit, whose energy-level and binding-energy ideas carry over.
• Exemplar Semiconductor Electronics — the next unit, on materials, junctions and simple circuits.