NCERT Class 12 Physics — Nuclei, Chapter 13 Exercises. All 10 questions solved.

The exercises at the end of Chapter 13 are about the link between mass and energy in the nucleus. Questions 13.1 to 13.3 find binding energies from mass defects; 13.5 and 13.6 find the energy released or absorbed in a reaction; 13.7 and 13.8 scale the energy of one fission or fusion up to a kilogram of fuel; 13.4 and 13.10 are about the size and density of nuclei; and 13.9 is about the Coulomb barrier that fusion has to overcome. Three results carry the arithmetic:

• Binding energy — for a nuclide of atomic number $Z$, mass number $A$ and atomic mass $M$, $E_b = \left[Zm_{\text{H}} + (A-Z)m_n – M\right]c^2$, with $1\ \text{u} = 931.5\ \text{MeV}/c^2$.
• The Q value of a reaction $A + b \to C + d$ — $Q = [m_A + m_b – m_C – m_d]\,c^2$. A positive $Q$ means energy is released (exothermic); a negative $Q$ means energy must be supplied (endothermic).
• Nuclear radius — $R = R_0A^{1/3}$, with $R_0 = 1.2\ \text{fm}$.

The book’s masses are atomic masses, which include the electrons. Using $m_{\text{H}}$, the mass of a whole hydrogen atom, in place of the proton mass takes care of this: the $Z$ electrons counted in $Zm_{\text{H}}$ cancel the $Z$ electrons included in $M$. In a reaction the electrons cancel in the same way, because the total atomic number is the same on both sides.

Key insight. Every energy in this exercise is a mass difference multiplied by $931.5\ \text{MeV}$ per u. A binding energy is the mass the separate nucleons have in excess of the nucleus (13.1 to 13.3); a $Q$ value is the mass the starting particles have in excess of the products (13.5, 13.6). Keep track of which way round the subtraction goes, and its sign tells you the physics — bound or not, exothermic or endothermic, possible or impossible.

The book gives the following data for use in these exercises: $e = 1.6\times10^{-19}\ \text{C}$; $N = 6.023\times10^{23}$ per mole; $1/(4\pi\varepsilon_0) = 9\times10^9\ \text{N m}^2/\text{C}^2$; $k = 1.381\times10^{-23}\ \text{J K}^{-1}$; $1\ \text{MeV} = 1.6\times10^{-13}\ \text{J}$; $1\ \text{u} = 931.5\ \text{MeV}/c^2$; $1\ \text{year} = 3.154\times10^7\ \text{s}$; $m_{\text{H}} = 1.007825\ \text{u}$; $m_n = 1.008665\ \text{u}$; $m({}^4_2\text{He}) = 4.002603\ \text{u}$; $m_e = 0.000548\ \text{u}$.

Question 13.1

Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({}^{14}_{7}\text{N}\right)$, given $m\left({}^{14}_{7}\text{N}\right) = 14.00307\ \text{u}$.

Solution. The binding energy is the energy needed to pull the nucleus apart into free nucleons, and it equals the mass the nucleus has lost compared with those nucleons. Nitrogen-14 has $7$ protons and $14 – 7 = 7$ neutrons. The separate particles (with the electrons included through $m_{\text{H}}$) have mass

$$7m_{\text{H}} + 7m_n = 7(1.007825) + 7(1.008665) = 7.054775 + 7.060655 = 14.115430\ \text{u}$$

The mass defect is $14.115430 – 14.00307 = 0.11236\ \text{u}$, so

$$E_b = 0.11236\times931.5 = 104.66\ \text{MeV}$$

Question 13.2

Obtain the binding energy of the nuclei ${}^{56}_{26}\text{Fe}$ and ${}^{209}_{83}\text{Bi}$ in units of MeV from the following data: $m\left({}^{56}_{26}\text{Fe}\right) = 55.934939\ \text{u}$, $m\left({}^{209}_{83}\text{Bi}\right) = 208.980388\ \text{u}$.

Solution. The method is the same as in Question 13.1. Iron-56 has $26$ protons and $30$ neutrons:

$$\Delta m = 26(1.007825) + 30(1.008665) – 55.934939 = 26.203450 + 30.259950 – 55.934939 = 0.528461\ \text{u}$$

$$E_b(\text{Fe}) = 0.528461\times931.5 = 492.26\ \text{MeV},\qquad \frac{E_b}{A} = \frac{492.26}{56} = 8.79\ \text{MeV per nucleon}$$

Bismuth-209 has $83$ protons and $126$ neutrons:

$$\Delta m = 83(1.007825) + 126(1.008665) – 208.980388 = 83.649475 + 127.091790 – 208.980388 = 1.760877\ \text{u}$$

$$E_b(\text{Bi}) = 1.760877\times931.5 = 1640.26\ \text{MeV},\qquad \frac{E_b}{A} = \frac{1640.26}{209} = 7.85\ \text{MeV per nucleon}$$

Bismuth has the larger total binding energy simply because it has more nucleons, but each nucleon in iron is bound more tightly. Iron sits near the peak of the binding-energy-per-nucleon curve, which is the fact Question 13.6 turns on. The answer key quotes the per-nucleon values, $8.79$ and $7.84\ \text{MeV}$; its $7.84$ is $7.848$ cut short rather than rounded.

Question 13.3

A given coin has a mass of $3.0\ \text{g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ${}^{63}_{29}\text{Cu}$ atoms (of mass $62.92960\ \text{u}$).

Solution. The energy needed is the binding energy of one copper nucleus multiplied by the number of nuclei in the coin.

One nucleus. Copper-63 has $29$ protons and $34$ neutrons:

$$\Delta m = 29(1.007825) + 34(1.008665) – 62.92960 = 29.226925 + 34.294610 – 62.92960 = 0.591935\ \text{u}$$

$$E_b = 0.591935\times931.5 = 551.39\ \text{MeV}$$

Number of nuclei. One mole of copper-63 has a mass of $62.92960\ \text{g}$ and contains $6.023\times10^{23}$ atoms, so $3.0\ \text{g}$ contains

$$\frac{3.0}{62.92960}\times6.023\times10^{23} = 2.871\times10^{22}\ \text{atoms}$$

Total.

$$E = (2.871\times10^{22})(551.39) = 1.583\times10^{25}\ \text{MeV} = 1.583\times10^{25}\times1.6\times10^{-13} = 2.53\times10^{12}\ \text{J}$$

The answer key’s $1.584\times10^{25}\ \text{MeV}$ and $2.535\times10^{12}\ \text{J}$ differ only in the rounding of the intermediate steps.

Question 13.4

Obtain approximately the ratio of the nuclear radii of the gold isotope ${}^{197}_{79}\text{Au}$ and the silver isotope ${}^{107}_{47}\text{Ag}$.

Solution. Nuclear radius depends only on the mass number, $R = R_0A^{1/3}$, and $R_0$ is the same for every nucleus, so it cancels in the ratio:

$$\frac{R_{\text{Au}}}{R_{\text{Ag}}} = \left(\frac{197}{107}\right)^{1/3} = (1.841)^{1/3} = 1.23$$

Gold has almost twice as many nucleons as silver, but its nucleus is only about a quarter larger in radius, because the volume, not the radius, is proportional to $A$.

Question 13.5

The $Q$ value of a nuclear reaction $A + b \to C + d$ is defined by $Q = [m_A + m_b – m_C – m_d]\,c^2$, where the masses refer to the respective nuclei. Determine from the given data the $Q$-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) ${}^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$

(ii) ${}^{12}_{6}\text{C} + {}^{12}_{6}\text{C} \to {}^{20}_{10}\text{Ne} + {}^{4}_{2}\text{He}$

Atomic masses are given to be $m\left({}^{2}_{1}\text{H}\right) = 2.014102\ \text{u}$, $m\left({}^{3}_{1}\text{H}\right) = 3.016049\ \text{u}$, $m\left({}^{12}_{6}\text{C}\right) = 12.000000\ \text{u}$, $m\left({}^{20}_{10}\text{Ne}\right) = 19.992439\ \text{u}$.

The definition uses nuclear masses, but atomic masses can be used directly: in each reaction the total atomic number is the same on both sides ($1 + 1 = 1 + 1$ and $6 + 6 = 10 + 2$), so the electrons included in the atomic masses cancel. The mass of ${}^1_1\text{H}$ is $m_{\text{H}}$ and that of ${}^4_2\text{He}$ is in the data list above.

Question 13.5 (i)

Solution.

$$m_A + m_b = 1.007825 + 3.016049 = 4.023874\ \text{u},\qquad m_C + m_d = 2(2.014102) = 4.028204\ \text{u}$$

$$Q = (4.023874 – 4.028204)\times931.5 = (-0.004330)(931.5) = -4.03\ \text{MeV}$$

The products are heavier than the reactants, so $Q$ is negative: the reaction can happen only if at least that much energy is supplied, for instance as kinetic energy of the incoming particle.

Question 13.5 (ii)

Solution.

$$m_A + m_b = 2(12.000000) = 24.000000\ \text{u},\qquad m_C + m_d = 19.992439 + 4.002603 = 23.995042\ \text{u}$$

$$Q = (24.000000 – 23.995042)\times931.5 = (0.004958)(931.5) = 4.62\ \text{MeV}$$

Here mass is lost, so energy is released.

Question 13.6

Suppose, we think of fission of a ${}^{56}_{26}\text{Fe}$ nucleus into two equal fragments, ${}^{28}_{13}\text{Al}$. Is the fission energetically possible? Argue by working out $Q$ of the process. Given $m\left({}^{56}_{26}\text{Fe}\right) = 55.93494\ \text{u}$ and $m\left({}^{28}_{13}\text{Al}\right) = 27.98191\ \text{u}$.

Solution. The process is ${}^{56}_{26}\text{Fe} \to {}^{28}_{13}\text{Al} + {}^{28}_{13}\text{Al}$, and the electrons again cancel ($26 = 13 + 13$). Its $Q$ value is

$$Q = \left[m\left({}^{56}_{26}\text{Fe}\right) – 2m\left({}^{28}_{13}\text{Al}\right)\right]c^2 = (55.93494 – 55.96382)\times931.5 = (-0.02888)(931.5) = -26.90\ \text{MeV}$$

$Q$ is negative: the two aluminium nuclei together weigh more than the iron nucleus, so the split would need $26.90\ \text{MeV}$ to be supplied. It cannot happen on its own.

The reason is the binding energy curve. Iron-56 has about $8.79\ \text{MeV}$ per nucleon (Question 13.2), while aluminium-28 works out, by the same method, to about $8.31\ \text{MeV}$ per nucleon. Splitting iron moves its nucleons to less tightly bound nuclei, and $56\times(8.79 – 8.31) \approx 26.9\ \text{MeV}$ — the same figure. Fission releases energy only for heavy nuclei such as uranium, whose fragments are more tightly bound than the parent.

Note on the book’s answer. The key prints “$Q = m({}^{56}_{26}\text{Fe}) – 2m({}^{28}_{13}\text{Al}) = 26.90\ \text{MeV}$; not possible”. With the given masses, $m(\text{Fe}) – 2m(\text{Al})$ is $-0.02888\ \text{u}$, so $Q = -26.90\ \text{MeV}$: the minus sign has been dropped. The conclusion “not possible” is right, and it follows precisely because $Q$ is negative.

Question 13.7

The fission properties of ${}^{239}_{94}\text{Pu}$ are very similar to those of ${}^{235}_{92}\text{U}$. The average energy released per fission is $180\ \text{MeV}$. How much energy, in MeV, is released if all the atoms in $1\ \text{kg}$ of pure ${}^{239}_{94}\text{Pu}$ undergo fission?

Solution. Each nucleus that splits releases $180\ \text{MeV}$, so we need the number of nuclei in $1\ \text{kg}$. A mole of plutonium-239 has a mass of about $239\ \text{g}$, so

$$N = \frac{1000}{239}\times6.023\times10^{23} = 2.520\times10^{24}\ \text{nuclei}$$

$$E = (2.520\times10^{24})(180) = 4.536\times10^{26}\ \text{MeV}$$

That is about $7.3\times10^{13}\ \text{J}$ from a single kilogram.

Question 13.8

How long can an electric lamp of $100\ \text{W}$ be kept glowing by fusion of $2.0\ \text{kg}$ of deuterium? Take the fusion reaction as

$${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{2}\text{He} + \text{n} + 3.27\ \text{MeV}$$

Solution. Number of deuterons. The atomic mass of deuterium is $2.014\ \text{u}$ (Question 13.5), so a mole has a mass of $2.014\ \text{g}$ and $2.0\ \text{kg}$ contains

$$N = \frac{2000}{2.014}\times6.023\times10^{23} = 5.98\times10^{26}\ \text{deuterons}$$

Energy. Each fusion uses up two deuterons, so the number of reactions is $N/2 = 2.99\times10^{26}$, and the energy released is

$$E = (2.99\times10^{26})(3.27) = 9.78\times10^{26}\ \text{MeV} = 9.78\times10^{26}\times1.6\times10^{-13} = 1.56\times10^{14}\ \text{J}$$

Time. A $100\ \text{W}$ lamp uses $100\ \text{J}$ every second:

$$t = \frac{1.56\times10^{14}}{100} = 1.56\times10^{12}\ \text{s} = \frac{1.56\times10^{12}}{3.154\times10^7} = 4.96\times10^4\ \text{years}$$

Taking the molar mass as exactly $2\ \text{g}$ gives $5.0\times10^4$ years instead. The answer key’s “about $4.9\times10^4$ y” is the same estimate, differing only in the rounding.

Question 13.9

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius $2.0\ \text{fm}$.)

Solution. Each deuteron carries charge $+e$. When the two hard spheres just touch, their centres are one diameter apart: $r = 2.0 + 2.0 = 4.0\ \text{fm} = 4.0\times10^{-15}\ \text{m}$. The Coulomb potential energy of two point charges at that separation is the height of the barrier:

$$U = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r} = \frac{(9\times10^9)(1.6\times10^{-19})^2}{4.0\times10^{-15}} = 5.76\times10^{-14}\ \text{J}$$

In electron volts, $U = \dfrac{5.76\times10^{-14}}{1.6\times10^{-19}} = 3.6\times10^5\ \text{eV} = 360\ \text{keV}$.

This is the total kinetic energy the pair needs to get into contact. If the two deuterons approach each other with equal speeds, each must bring $180\ \text{keV}$. Energies like this are why fusion happens only at the enormous temperatures found in the cores of stars.

Question 13.10

From the relation $R = R_0A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$).

Solution. A nucleus of mass number $A$ contains $A$ nucleons, each of mass about $m = 1.67\times10^{-27}\ \text{kg}$, so its mass is close to $Am$. Treating it as a sphere of radius $R = R_0A^{1/3}$, its volume is

$$V = \frac43\pi R^3 = \frac43\pi R_0^3A$$

The density is mass over volume:

$$\rho = \frac{Am}{\frac43\pi R_0^3A} = \frac{3m}{4\pi R_0^3}$$

The $A$ cancels, leaving only constants: every nucleus has the same density. The reason is that the volume, $R^3$, grows in proportion to $A$, just as the mass does. The density is only nearly constant because the nuclear mass is not exactly $A$ times the nucleon mass (binding energy and the neutron–proton mass difference change it slightly) and $R = R_0A^{1/3}$ is itself an approximate fit. With $R_0 = 1.2\ \text{fm}$,

$$\rho = \frac{3(1.67\times10^{-27})}{4\pi(1.2\times10^{-15})^3} = 2.3\times10^{17}\ \text{kg m}^{-3}$$

which is the value the chapter quotes — about $10^{14}$ times the density of water.

Common mistakes

• Question 13.1: using the proton mass with the atomic mass of nitrogen. The book’s $14.00307\ \text{u}$ includes seven electrons. Pairing it with $m_p = 1.007276\ \text{u}$ leaves those electrons uncancelled and gives $101.1\ \text{MeV}$, about $3.6\ \text{MeV}$ too low. Use $m_{\text{H}}$ with atomic masses.
• Question 13.2: confusing total and per-nucleon binding energy. Bismuth’s total, $1640\ \text{MeV}$, is larger than iron’s, $492\ \text{MeV}$, yet iron is the more tightly bound nucleus. The comparison that shows stability is the per-nucleon value — which is what the answer key quotes.
• Question 13.3: stopping at one nucleus. $551\ \text{MeV}$ is the binding energy of a single copper nucleus. The question is about the coin, so it has to be multiplied by the $2.87\times10^{22}$ nuclei it contains.
• Question 13.4: taking the ratio of the mass numbers. $197/107 = 1.84$ is the ratio of the volumes. The radius goes as $A^{1/3}$, so the ratio of radii is $1.23$.
• Question 13.5(i): dropping the sign of $Q$. The minus sign carries the answer: it is what makes the reaction endothermic. Writing $4.03\ \text{MeV}$ and calling it exothermic is the usual slip.
• Question 13.6: assuming that fission always releases energy. It does so only for heavy nuclei, whose fragments are more tightly bound than the parent. Iron-56 is already near the peak of the binding energy curve, so splitting it costs energy.
• Question 13.8: counting one reaction per deuteron. Each fusion consumes two deuterons, so the number of reactions is half the number of deuterons. Missing this doubles the answer to about $10^5$ years.
• Question 13.9: using the radius as the separation. Two touching spheres of radius $2.0\ \text{fm}$ have centres $4.0\ \text{fm}$ apart. Using $2.0\ \text{fm}$ doubles the barrier to $720\ \text{keV}$.

Practise next

• Chapter 14, Semiconductor Electronics: Materials, Devices and Simple Circuits — the last chapter of the book, from the energy levels of nuclei and atoms to the energy bands of solids.
• Chapter 12, Atoms — the alpha-particle scattering that revealed the nucleus, and the electron-volt energy scale that nuclear physics multiplies by a million.
• Chapter 2, Electrostatic Potential and Capacitance — the potential energy of two point charges, $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$, which is the whole of Question 13.9.