NCERT Exemplar Solutions Class 12 Physics Nuclei — Very Short, Short and Long Answer Questions Q11–26

NCERT Exemplar Problems, Class 12 Physics — Nuclei, Very Short Answer Questions 11 to 15, Short Answer Questions 16 to 20 and Long Answer Questions 21 to 26. All 16 questions solved. The book numbers them 13.11 to 13.26.

Three ideas carry this page. Binding energy is a balance between the attractive short-range nuclear force and the Coulomb repulsion of the protons, which is why $N \ge Z$ for stable nuclei, why two mirror nuclei differ in binding energy and why magic numbers matter. Radioactive decay is first-order, so the rate is $R = \lambda N$ and every chain question reduces to exponentials in $\lambda$, not in the half-lives. Conservation of energy and momentum together decide what a decay or a photon absorption can do, and in Questions 15, 22 and 24 the momentum condition is the whole point.

Key insight. Write the decay constant $\lambda = \ln 2 / T_{1/2}$ before touching any chain or ratio, and in every particle process write momentum conservation next to energy conservation. Four printed answers on this page fail exactly there: Questions 20, 21, 23 and 24.

Very Short Answer Questions

Question 11

$\mathrm{He}^3_2$ and $\mathrm{He}^3_1$ nuclei have the same mass number. Do they have the same binding energy?

Solution. No. The second nucleus, which the book prints as $\mathrm{He}^3_1$, is the triton $^3_1\mathrm{H}$ (one proton, two neutrons). $^3_2\mathrm{He}$ has two protons and one neutron. Both have $A = 3$, but $^3_2\mathrm{He}$ has two protons repelling each other by the Coulomb force, which lowers its binding energy. The measured values are about $7.72$ MeV for $^3_2\mathrm{He}$ and $8.48$ MeV for $^3_1\mathrm{H}$ (total binding energies; per nucleon, about $2.57$ and $2.83$ MeV), so the triton is more tightly bound.

Question 12

Draw a graph showing the variation of decay rate with number of active nuclei.

Solution. The decay rate is $R = -\dfrac{dN}{dt} = \lambda N$. This is a straight line through the origin with slope $\lambda$: the rate is directly proportional to the number of undecayed nuclei, so a sample with twice as many nuclei decays twice as fast. If the signed rate $dN/dt = -\lambda N$ is plotted instead, as the key does, the line passes through the origin with negative slope $-\lambda$. The content is the same.

Question 13

Which sample, A or B shown in Fig. 13.2 has shorter mean-life?

Solution. Fig. 13.2 plots the decay rate $dN/dt$ against time for two samples. Both curves start from the same point on the vertical axis, and then curve A falls slowly while curve B falls faster. A faster fall at the same starting rate means a larger decay constant $\lambda$. The mean life is $\tau = 1/\lambda$, so the sample with the larger $\lambda$ has the shorter mean life.

Question 14

Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.

Solution. The question means $\gamma$-radiation. An excited nucleus emits a $\gamma$-ray as it drops to a lower level, with an energy of the order of MeV, because nuclear energy levels are MeV apart. An excited electron can certainly emit light, but the spacing of atomic levels is only of the order of eV, so an electron transition gives photons of eV energy (visible, ultraviolet or X-ray). It can never give MeV $\gamma$-rays. So the one that cannot emit $\gamma$-radiation is the excited electron.

Question 15

In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved?

Solution. In the frame in which the pair is at rest, the total momentum before annihilation is zero. A single photon of energy $E$ carries momentum $E/c \ne 0$, so one photon could not conserve momentum. Two photons are produced, moving in opposite directions with equal energy $m_ec^2$ each (at least), so their momenta are equal and opposite and add to zero. Energy is conserved too: $2m_ec^2 = 1.02$ MeV is shared between the two.

Short Answer Questions

Question 16

Why do stable nuclei never have more protons than neutrons?

Solution. Protons repel each other by the Coulomb force, and this repulsion acts between every pair of protons and so grows roughly as $Z^2$, while the nuclear attraction is short-ranged and acts equally between neutrons and protons. Extra neutrons add attraction without adding repulsion, and they also dilute the proton density. So to stay bound a nucleus needs $N \ge Z$, and the heavier it is, the more neutrons it needs, with $N > Z$.

The statement holds beyond the lightest nuclei. Two light exceptions exist, $^1_1\mathrm{H}$ and $^3_2\mathrm{He}$, which are stable with $Z > N$, because with so few protons there is almost no repulsion to worry about.

Question 17

Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence: A $\to$ B $\to$ C. Here B is an intermediate nuclei which is also radioactive. Considering that there are $N_0$ atoms of A initially, plot the graph showing the variation of number of atoms of A and B versus time.

Solution. A decays by a simple exponential, $N_A = N_0e^{-\lambda_At}$. B is made from A at the rate $\lambda_AN_A$ and lost at the rate $\lambda_BN_B$, so $$\frac{dN_B}{dt} = \lambda_AN_A – \lambda_BN_B.$$ Solving with $N_B(0) = 0$, $$N_B = \frac{N_0\lambda_A}{\lambda_B – \lambda_A}\left(e^{-\lambda_At} – e^{-\lambda_Bt}\right).$$ The graph therefore has these features.

• $N_A$ starts at $N_0$ and falls steadily and exponentially towards zero.
• $N_B$ starts at zero, rises, reaches a maximum and then decays to zero. At the maximum $dN_B/dt = 0$, which means $\lambda_AN_A = \lambda_BN_B$: B is being formed as fast as it decays. The $N_B$ curve crosses the $N_A$ curve, if it does, only after its own peak, never before.
• Finally both fall to zero, and all the atoms end up as stable C.

Question 18

A piece of wood from the ruins of an ancient building was found to have a $^{14}$C activity of 12 disintegrations per minute per gram of its carbon content. The $^{14}$C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of $^{14}$C is 5760 years.

Solution. The activity is proportional to the number of $^{14}$C nuclei, so $R = R_0e^{-\lambda t}$ with $\lambda = \dfrac{\ln 2}{5760}$ per year. Then $$\frac{R_0}{R} = \frac{16}{12} = e^{\lambda t} \;\Rightarrow\; t = \frac{1}{\lambda}\ln\frac{16}{12} = \frac{5760}{\ln 2}\ln\frac43.$$ With $\ln\frac43 = 0.28768$ and $\ln 2 = 0.69315$, $t = 8310 \times 0.28768 \approx 2391$ years.

Check: this is a little less than one half-life ($5760$ years), as it should be, since the activity has fallen to $75\%$ of its original value and not yet to $50\%$.

Question 19

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately $10^{-15}$ m.

Solution. To resolve a structure of size $d$, the probe’s de Broglie wavelength must be no larger than $d$: $\lambda \lesssim 10^{-15}$ m. The momentum is then $p = h/\lambda$. The electron’s rest energy is only $0.511$ MeV, but this momentum corresponds to $pc$ of order $10^{9}$ eV, so the electron is ultra-relativistic and $K \approx pc$: $$K \approx pc = \frac{hc}{\lambda} = \frac{1.24\times10^{-6}\ \text{eV m}}{10^{-15}\ \text{m}} = 1.24\times10^{9}\ \text{eV} \approx 1.2\ \text{GeV}.$$ So the electron needs a kinetic energy of about $1$ GeV. Using the non-relativistic $p^2/2m$ instead would give a badly wrong energy.

Printed key error. Minor: the key rounds $1.24$ GeV to $10^{9}$ eV and prints Planck’s constant as $6.63\times10^{34}$ (the sign of the exponent is dropped). The reasoning and the order of magnitude are right.

Question 20

A nuclide 1 is said to be the mirror isobar of nuclide 2 if $Z_1 = N_2$ and $Z_2 = N_1$. (a) What nuclide is a mirror isobar of $^{23}_{11}\mathrm{Na}$? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?

Solution. (a) For $^{23}_{11}\mathrm{Na}$ we have $Z_1 = 11$ and $N_1 = 23 – 11 = 12$. The mirror isobar has $Z_2 = N_1 = 12$ and $N_2 = Z_1 = 11$, so $A = 23$ again. It is magnesium: $^{23}_{12}\mathrm{Mg}$.

(b) The two nuclides have the same mass number and the same number of nucleon pairs, so the nuclear attraction is almost identical. The difference is the Coulomb repulsion. $^{23}_{12}\mathrm{Mg}$ has one more proton than $^{23}_{11}\mathrm{Na}$, hence more proton-proton repulsion, which lowers the binding energy. So $^{23}_{11}\mathrm{Na}$ is the more tightly bound. The measured values are about $186.6$ MeV for $^{23}_{11}\mathrm{Na}$ and $181.7$ MeV for $^{23}_{12}\mathrm{Mg}$.

Printed key error. The key says $^{23}_{12}\mathrm{Mg}$ has the greater binding energy “since $Z_2 > Z_1$”. It is the other way round: for the same $A$ a larger $Z$ means more Coulomb repulsion, so a lower binding energy.

Long Answer Questions

Question 21

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is: $^{38}$Sulphur $\xrightarrow{\text{half-life }=2.48\text{ h}}$ $^{38}$Cl $\xrightarrow{\text{half-life }=0.62\text{ h}}$ $^{38}$Ar (stable). Assume that we start with 1000 $^{38}$S nuclei at time $t = 0$. The number of $^{38}$Cl is of count zero at $t = 0$ and will again be zero at $t = \infty$. At what value of $t$, would the number of counts be a maximum?

Solution. Let S be nucleus 1 and Cl be nucleus 2, with decay constants $\lambda_1 = \ln2/2.48$ and $\lambda_2 = \ln2/0.62$ per hour. As in Question 17, $$\frac{dN_2}{dt} = \lambda_1N_1 – \lambda_2N_2, \qquad N_1 = N_0e^{-\lambda_1t},$$ and with $N_2(0) = 0$, $$N_2 = \frac{N_0\lambda_1}{\lambda_2-\lambda_1}\left(e^{-\lambda_1t} – e^{-\lambda_2t}\right).$$ The count is maximum when $dN_2/dt = 0$: $$-\lambda_1e^{-\lambda_1t} + \lambda_2e^{-\lambda_2t} = 0 \;\Rightarrow\; e^{(\lambda_2-\lambda_1)t} = \frac{\lambda_2}{\lambda_1} \;\Rightarrow\; t_{\max} = \frac{\ln(\lambda_2/\lambda_1)}{\lambda_2-\lambda_1}.$$ Now put $\lambda = \ln 2/T$. Then $\lambda_2/\lambda_1 = T_1/T_2$ and $\lambda_2 – \lambda_1 = \ln 2\,\dfrac{T_1-T_2}{T_1T_2}$, so $$t_{\max} = \frac{T_1T_2}{T_1-T_2}\cdot\frac{\ln(T_1/T_2)}{\ln 2} = \frac{2.48\times0.62}{2.48-0.62}\times\frac{\ln 4}{\ln 2} = \frac{1.5376}{1.86}\times 2 = 1.65\ \text{h}.$$ This is about $99$ minutes. The value is sensible: it is longer than the half-life of Cl, because the Cl must first accumulate, and it is shorter than the half-life of S.

Printed key error. The key puts the half-lives $2.48$ and $0.62$ in place of the decay constants, giving $\ln(2.48/0.62)/(2.48-0.62) = 0.745$, and labels it seconds. The formula needs $\lambda_1$ and $\lambda_2$, not $T_1$ and $T_2$, and the unit is hours; the correct answer is $1.65$ h.

Question 22

Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2$ MeV. A $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma$-ray. If $E = B$, show that this cannot happen. Hence calculate how much bigger than $B$ must $E$ be for such a process to happen.

Solution. Why $E = B$ fails. If $E = B$, all of the photon’s energy goes into breaking the bond and the $n$ and $p$ have no kinetic energy, so their total momentum is zero. But the photon brought momentum $E/c \ne 0$, and momentum must be conserved. This is a contradiction, so the process is impossible.

Threshold. The photon’s momentum $E/c$ must be carried by the $n$ and $p$ moving together (the minimum-energy case, in which they share the same velocity). Their total mass is $M = 2m$ with $mc^2 \approx 938$ MeV, so $Mc^2 \approx 1876$ MeV. Energy conservation gives $$E – B = K = \frac{p^2}{2M} = \frac{(E/c)^2}{2M} = \frac{E^2}{2Mc^2}.$$ Since $E – B$ is tiny compared with $B$, we put $E \approx B$ on the right-hand side: $$E – B \approx \frac{B^2}{2Mc^2} = \frac{(2.2)^2}{2\times1876}\ \text{MeV} \approx 1.3\times10^{-3}\ \text{MeV}.$$ So $E$ must exceed $B$ by only about $1.3$ keV. The key writes the same result as $B^2/4mc^2$ without a number, which is identical because $M = 2m$.

Question 23

The deuteron is bound by nuclear forces just as H-atom is made up of $p$ and $e$ bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge $e’$: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e’^2}{r}$, estimate the value of $(e’/e)$ given that the binding energy of a deuteron is $2.2$ MeV.

Solution. The hydrogen atom’s ground-state energy from Bohr’s theory is $$E = \frac{\mu e^4}{8\varepsilon_0^2h^2},$$ with $\mu$ the reduced mass. For hydrogen $\mu \approx m_e$ and $E = 13.6$ eV. For the deuteron the reduced mass is $\mu = \dfrac{m_nm_p}{m_n+m_p} \approx \dfrac{M}{2}$, where $M \approx 1836\,m_e$ is the nucleon mass, so $\mu \approx 918\,m_e$. Replacing $e$ by $e’$, $$\frac{E_d}{E_H} = 918\left(\frac{e’}{e}\right)^4 = \frac{2.2\times10^{6}}{13.6} = 1.62\times10^{5}$$ $$\Rightarrow\; \left(\frac{e’}{e}\right)^4 = \frac{1.62\times10^{5}}{918} \approx 176 \;\Rightarrow\; \frac{e’}{e} \approx 3.6.$$ (The force law as printed, with $1/r$ and not $1/r^2$, is a loose modelling choice by the book; we follow the question’s instruction, which is to use the Bohr formula with $e \to e’$.)

Printed key error. The key sets up the same equation but prints $e’/e \approx 11$. Check: $11^4 \times 918 \approx 1.3\times10^{7}$, not $1.6\times10^{5}$, so $11$ does not satisfy its own equation. The fourth root of $176$ is $3.6$.

Question 24

Before the neutrino hypothesis, the beta decay process was thought to be the transition $n \to p + \bar{e}$. If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.

Solution. Take the neutron at rest. Momentum conservation requires $\mathbf p_p = -\mathbf p_e$, so both particles have the same momentum magnitude $p$. Energy conservation gives $$m_nc^2 = \sqrt{p^2c^2 + m_p^2c^4} + \sqrt{p^2c^2 + m_e^2c^4}.$$ This is one equation in one unknown $p$, so $p$ has one fixed value, and hence so do both energies. Solving, $$E_e = \frac{m_n^2 + m_e^2 – m_p^2}{2m_n}\,c^2, \qquad E_p = \frac{m_n^2 + m_p^2 – m_e^2}{2m_n}\,c^2.$$ With $m_nc^2 = 939.57$ MeV, $m_pc^2 = 938.27$ MeV and $m_ec^2 = 0.511$ MeV, the energy released is $Q = (m_n – m_p – m_e)c^2 = 0.78$ MeV. Then

• electron total energy $E_e \approx 1.29$ MeV, so its kinetic energy is $\approx 0.78$ MeV, and its momentum is $pc = \sqrt{1.293^2 – 0.511^2} \approx 1.19$ MeV;
• proton total energy $\approx 938.27$ MeV, with kinetic energy $p^2/2m_p = (1.19)^2/(2\times938.27) \approx 0.75$ keV.

So the electron takes essentially all the released energy, and the proton, being so heavy, takes a very small fixed share. The experiment, however, found the electron energy spread continuously from zero up to about $0.78$ MeV. A two-body decay cannot give a continuous spectrum; a third particle must share the energy and momentum, which is the antineutrino.

Printed key error. The method in the key (fixed energies from two-body kinematics) is right, but its numbers are not. It uses $m_pc^2 \approx 936$ MeV and $m_nc^2 \approx 938$ MeV, a mass difference of $2$ MeV instead of the true $1.29$ MeV, and so gets $pc \approx 2$ MeV and $E_e \approx 2.06$ MeV. With the correct masses, $E_e \approx 1.29$ MeV and $pc \approx 1.19$ MeV.

Question 25

The activity $R$ of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

(i) Plot the graph of $R$ versus $t$ and calculate half-life from the graph. (ii) Plot the graph of $\ln(R/R_0)$ versus $t$ and obtain the value of half-life from the graph.

Solution. The successive ratios are $35.36/100 = 0.3536$, $12.51/35.36 = 0.3538$, $4.42/12.51 = 0.3533$ and $1.56/4.42 = 0.3529$, so the activity falls by the same factor each hour, which is the signature of exponential decay.

(i) The graph of $R$ against $t$ is a smooth exponential curve falling from $100$ MBq. Half of $100$ MBq is $50$ MBq, and reading across from $50$ MBq on the vertical axis, the curve reaches it at $t \approx 0.67$ h. Therefore $T_{1/2} \approx 0.67$ h $\approx 40$ minutes. (Check that $R$ halves again in the same interval: $R(1.33\ \text{h}) \approx 25$ MBq.)

(ii) Taking logarithms, $\ln(R/R_0) = -\lambda t$. The values are $$t = 1,2,3,4\ \text{h}: \quad \ln(R/R_0) = -1.04,\ -2.08,\ -3.11,\ -4.16.$$ These lie on a straight line through the origin whose slope is $-\lambda$, with $\lambda \approx 4.16/4 = 1.04\ \text{h}^{-1}$. Then $$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.04} \approx 0.67\ \text{h} \approx 40\ \text{min}.$$

Printed key error. Minor: the key writes $\lambda = 1.05$ with the unit “h”; the unit of a decay constant is $\text{h}^{-1}$. The half-life is unaffected.

Question 26

Nuclei with magic no. of proton $Z = 2, 8, 20, 28, 50, 52$ and magic no. of neutrons $N = 2, 8, 20, 28, 50, 82$ and $126$ are found to be very stable. (i) Verify this by calculating the proton separation energy $S_p$ for $^{120}$Sn ($Z = 50$) and $^{121}$Sb ($Z = 51$). The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by $S_p = (M_{Z-1,N} + M_H – M_{Z,N})c^2$. Given $^{119}$In $= 118.9058$ u, $^{120}$Sn $= 119.902199$ u, $^{121}$Sb $= 120.903824$ u, $^{1}$H $= 1.0078252$ u. (ii) What does the existence of magic number indicate?

Solution. (The question has two misprints: the magic proton number $52$ should be $82$, and “$^{121}$Sb $= (Z = 51)$” should read “$^{121}$Sb $(Z = 51)$”. Neither affects the working.)

(i) For $^{120}$Sn, removing a proton leaves $^{119}$In: $$S_p(^{120}\mathrm{Sn}) = (118.9058 + 1.0078252 – 119.902199)\,u\,c^2 = 0.0114262\ u\,c^2.$$ Using $1\ u\,c^2 = 931.5$ MeV, this is $\approx 10.64$ MeV.

For $^{121}$Sb, removing a proton leaves $^{120}$Sn: $$S_p(^{121}\mathrm{Sb}) = (119.902199 + 1.0078252 – 120.903824)\,u\,c^2 = 0.0062002\ u\,c^2 \approx 5.78\ \text{MeV}.$$ The tin nucleus, with the magic number $Z = 50$, needs nearly twice as much energy to lose a proton as antimony does, which has just one proton beyond the closed shell. So the closed-shell nucleus holds its last proton much more tightly, which is exactly what “very stable” means.

(ii) Magic numbers show that the nucleus has a shell structure, in which protons and neutrons occupy discrete energy levels grouped into shells, much as electrons do in an atom. A nucleus with a completely filled shell is unusually tightly bound, which is why binding energy per nucleon shows peaks at the magic numbers.

Printed key error. Minor: the key’s arithmetic prints $0.0114362$ u and $0.0059912$ u, which are slips in the last digits. The correct values are $0.0114262$ u and $0.0062002$ u. The conclusion is unchanged.

Common mistakes

• Q11 and Q20: using the nuclear force to explain the difference. The nuclear force is almost the same for both nuclei in each pair, so the difference in binding energy comes entirely from the Coulomb repulsion, which depends on $Z$, not on $A$. More protons means lower binding energy.
• Q18 and Q21: using half-lives where decay constants belong. Always convert $\lambda = \ln 2/T_{1/2}$ first. Putting $T_{1/2}$ straight into $e^{-\lambda t}$, or into the peak-time formula, is how the printed key of Question 21 went wrong.
• Q19: using $p^2/2m$ for an electron at GeV energies. The electron is ultra-relativistic, so use $K \approx pc = hc/\lambda$. The non-relativistic formula gives a value off by orders of magnitude.
• Q22 and Q24: conserving only energy. Always add momentum conservation. At $E = B$ there is energy to unbind the deuteron but no way to carry the photon’s momentum, and in beta decay momentum is what fixes the electron’s energy.
• Q23: raising $e’/e$ to the wrong power. The Bohr energy goes as $e^4$, so $e’/e$ is the fourth root of the energy ratio, not the square root.
• Q26: reading a bigger $S_p$ as a sign of weak binding. A large proton separation energy means the last proton is held tightly. Sn ($Z = 50$) at about $10.6$ MeV is the more stable, and Sb at about $5.8$ MeV is the less.

Practise next

• The other page of this unit: NCERT Exemplar Class 12 Physics Nuclei — Multiple Choice Questions Q1–10.
• The textbook chapter: NCERT Solutions Class 12 Physics Chapter 13 — Nuclei.
• The previous unit: NCERT Exemplar Class 12 Physics Atoms.
• The next unit: NCERT Exemplar Class 12 Physics Semiconductor Electronics: Materials, Devices and Simple Circuits.