Hello dear learner. Welcome to HSLC GURU. In this lesson we present the complete English-medium question-and-answer set, key concepts, derivations, formula sheet and additional important questions of the thirteenth chapter — Nuclei — of the ASSEB (Assam State School Education Board) Class 12 Physics syllabus. After learning that the atom is mostly empty space with a tiny positively charged nucleus at its centre (Chapter 12), we now zoom into that nucleus itself: what is it made of, how big is it, how is it held together, why do some nuclei spontaneously emit radiation, and how does fission of uranium or fusion of hydrogen liberate the colossal energies that power reactors and stars? These ideas underlie nuclear medicine, carbon-14 dating, the operation of every commercial reactor in India, and the energy production inside the Sun.
Summary
The nucleus of an atom is a dense central core containing protons and neutrons (collectively called nucleons). A nuclide is denoted $_Z^A X$ where $Z$ is the atomic (proton) number, $N=A-Z$ is the neutron number and $A=Z+N$ is the mass number. Nuclear masses are conveniently quoted in atomic mass units: $1\,\text{u} = 1.660539\times 10^{-27}\,\text{kg}$ which by Einstein’s relation corresponds to $931.5\,\text{MeV}/c^2$.
Experiments by Rutherford, Chadwick and others showed that the nuclear radius is well represented by $R = R_0 A^{1/3}$ with $R_0 \approx 1.2\,\text{fm}$. Because volume scales as $R^3 \propto A$, the nuclear density is essentially the same for every nucleus: roughly $2.3\times 10^{17}\,\text{kg/m}^3$ — about $10^{14}$ times the density of ordinary matter. The mass of any nucleus is always less than the sum of the free masses of its constituents; this missing mass — the mass defect $\Delta m$ — appears as the binding energy $B=\Delta m\,c^2$. The plot of $B/A$ versus $A$ rises steeply for light nuclei, peaks near $^{56}$Fe (≈8.75 MeV/nucleon) and falls slowly for heavy nuclei. Fusion of light nuclei and fission of heavy nuclei both move products toward the peak and therefore release energy.
The force binding nucleons together — the strong nuclear force — is short-ranged (≈1–2 fm), charge-independent, attractive in this range with a strongly repulsive core, and shows saturation. Henri Becquerel’s 1896 discovery of radioactivity revealed that some unstable nuclei spontaneously emit $\alpha$ (a $^{4}$He nucleus), $\beta^{\pm}$ (electron/positron with neutrino or antineutrino) or $\gamma$ (high-energy photon) radiation. Radioactive decay obeys the law $N(t) = N_0 e^{-\lambda t}$. Two convenient lifetimes are the half-life $t_{1/2} = (\ln 2)/\lambda$ and the mean life $\tau = 1/\lambda$. Activity $R=\lambda N$ is measured in becquerel (Bq) or curie (Ci, $1\,\text{Ci}=3.7\times 10^{10}\,\text{Bq}$). Nuclear fission of $^{235}$U absorbing a slow neutron releases about 200 MeV per event and a chain reaction sustains a reactor; fusion of light nuclei powers the Sun via the proton-proton chain.
Key Formulas
| Quantity | Formula | SI / Common Unit |
|---|---|---|
| Nuclide notation | $_Z^A X$, with $A=Z+N$ | — |
| Atomic mass unit | $1\,\text{u} = 1.660539\times 10^{-27}\,\text{kg}$ | kg |
| Energy equivalent of 1 u | $1\,\text{u}\cdot c^2 = 931.5\,\text{MeV}$ | MeV |
| Nuclear radius | $R = R_0 A^{1/3}$, $R_0 \approx 1.2\,\text{fm}$ | m |
| Nuclear volume | $V = \tfrac{4}{3}\pi R_0^3 A$ | m$^3$ |
| Nuclear density | $\rho \approx 2.3\times 10^{17}$ | kg/m$^3$ |
| Mass defect | $\Delta m = Z\,m_p + (A-Z)\,m_n – M$ | u |
| Binding energy | $B = \Delta m\,c^2$ | MeV |
| Binding energy per nucleon | $B/A$ | MeV/nucleon |
| Decay law | $N(t) = N_0 e^{-\lambda t}$ | — |
| Activity | $R = \lambda N = R_0 e^{-\lambda t}$ | becquerel (Bq) |
| Half-life | $t_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$ | s |
| Mean life | $\tau = \dfrac{1}{\lambda} = \dfrac{t_{1/2}}{\ln 2}$ | s |
| Curie | $1\,\text{Ci} = 3.7\times 10^{10}\,\text{Bq}$ | Bq |
| Q-value (reaction) | $Q = (m_i – m_f)c^2$ | MeV |
| Energy per fission of $^{235}$U | $\approx 200\,\text{MeV}$ | MeV |
13.1 Composition of the Nucleus
Every nucleus is made of two kinds of particles, jointly called nucleons.
- Proton (p) — charge $+e$, mass $m_p = 1.007276\,\text{u} \approx 1.6726\times 10^{-27}\,\text{kg}$.
- Neutron (n) — chargeless, mass $m_n = 1.008665\,\text{u} \approx 1.6749\times 10^{-27}\,\text{kg}$. Discovered by Chadwick in 1932.
The number of protons $Z$ fixes the chemical identity of the element; the total number of nucleons $A$ fixes the mass. Isotopes have the same $Z$ but different $A$ (e.g. $^1_1$H, $^2_1$H, $^3_1$H). Isobars have the same $A$ but different $Z$ (e.g. $^3_1$H and $^3_2$He). Isotones have the same $N$ (e.g. $^3_1$H and $^4_2$He both have $N=2$).
13.2 Nuclear Size and Density
Rutherford-style scattering and electron-scattering experiments establish that the nuclear radius depends on $A$ as
$$R = R_0 A^{1/3}, \quad R_0 \approx 1.2\,\text{fm} = 1.2\times 10^{-15}\,\text{m}.$$
Hence the volume $V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A$ is proportional to $A$. Taking the mass of a nucleus of mass number $A$ as $M \approx A m_u$ (with $m_u \approx 1.66\times 10^{-27}$ kg), the density
$$\rho = \dfrac{M}{V} = \dfrac{A m_u}{\tfrac{4}{3}\pi R_0^3 A} = \dfrac{m_u}{\tfrac{4}{3}\pi R_0^3} \approx 2.3\times 10^{17}\,\text{kg/m}^3$$
is independent of $A$. Every nucleus has essentially the same density — a consequence of the saturation property of the nuclear force, which lets a nucleon interact only with its immediate neighbours.
13.3 Mass-Energy and Binding Energy
Einstein’s mass-energy relation $E = mc^2$ tells us that mass and energy are interconvertible. A nucleus is bound: when $Z$ free protons and $A-Z$ free neutrons assemble into a nucleus of mass $M$, the resulting object is lighter than the sum of its parts. This mass defect
$$\Delta m = Z\,m_p + (A-Z)\,m_n – M$$
is liberated as the binding energy
$$B = \Delta m\,c^2.$$
If a problem provides atomic masses (which include $Z$ electrons), we must use atomic mass of $^1_1$H (which already includes one electron) in place of $m_p$ so that the electron masses cancel. Dividing $B$ by $A$ gives the binding energy per nucleon — a measure of nuclear stability. Tightly bound nuclei (large $B/A$) are difficult to break apart.
Salient features of the $B/A$ vs $A$ curve:
- $B/A$ rises rapidly for very light nuclei.
- It is roughly constant ($\approx 8.5$ MeV/nucleon) for $30 < A < 170$.
- Maximum is about $8.75$ MeV/nucleon near $^{56}$Fe.
- For $A>170$ the curve gradually falls — heavy nuclei can release energy by splitting.
- Light nuclei have small $B/A$ — combining them releases energy (fusion).
13.4 Nuclear Force
The strong nuclear force has these defining features:
- Short range: effective only up to $\sim 1$–$2$ fm; falls to zero rapidly beyond.
- Strong: about $10^2$ times stronger than the electrostatic force at nuclear distances and $\sim 10^{38}$ times stronger than gravity.
- Charge-independent: forces between p-p, n-n and p-n pairs are essentially equal.
- Saturation: each nucleon interacts only with its nearest neighbours, so $B \propto A$ approximately.
- Repulsive core: at separations below $\sim 0.7$ fm the force becomes strongly repulsive, preventing nuclear collapse.
- Spin-dependent: depends on the relative orientations of nucleon spins.
13.5 Radioactivity
Radioactivity was discovered by Henri Becquerel in 1896 while studying uranium salts. Marie and Pierre Curie isolated polonium and radium and named the phenomenon. Three types of natural radiation were identified by Rutherford:
| Type | Identity | Charge | Penetration | Effect on $Z$, $A$ |
|---|---|---|---|---|
| $\alpha$ | $^4_2$He nucleus | +2e | stopped by paper, range cm in air | $Z\to Z-2,\ A\to A-4$ |
| $\beta^-$ | electron (with $\bar{\nu}_e$) | $-e$ | stopped by thin Al; mm range | $Z\to Z+1,\ A$ unchanged |
| $\beta^+$ | positron (with $\nu_e$) | $+e$ | similar to $\beta^-$ | $Z\to Z-1,\ A$ unchanged |
| $\gamma$ | high-energy photon | 0 | stopped only by thick lead | both unchanged |
Examples:
$$_{92}^{238}\text{U} \;\to\; _{90}^{234}\text{Th} + _{2}^{4}\text{He}$$
$$_{6}^{14}\text{C} \;\to\; _{7}^{14}\text{N} + e^- + \bar{\nu}_e$$
$$_{11}^{22}\text{Na} \;\to\; _{10}^{22}\text{Ne} + e^+ + \nu_e$$
13.6 Law of Radioactive Decay
Experimentally, the number of nuclei disintegrating per unit time at any instant is proportional to the number $N$ of un-decayed nuclei present at that instant:
$$\dfrac{dN}{dt} = -\lambda N,$$
where $\lambda$ is the decay constant. Integrating with $N(0)=N_0$,
$$N(t) = N_0 e^{-\lambda t}.$$
The activity $R = -dN/dt = \lambda N = R_0 e^{-\lambda t}$ is measured in becquerel (Bq, 1 disintegration per second) or curie (1 Ci $= 3.7\times 10^{10}$ Bq).
Setting $N = N_0/2$ gives the half-life
$$t_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}.$$
The mean life is the average lifetime of a nucleus before decay:
$$\tau = \dfrac{1}{\lambda}, \qquad t_{1/2} = \tau \ln 2 = 0.693\,\tau.$$
13.7 Nuclear Fission and Fusion
Fission is the splitting of a heavy nucleus into two medium-mass fragments. When $^{235}_{92}$U absorbs a slow (thermal) neutron it forms $^{236}_{92}$U which fragments — for example as
$$_{92}^{235}\text{U} + n \;\to\; _{56}^{144}\text{Ba} + _{36}^{89}\text{Kr} + 3n + Q$$
with $Q \approx 200\,\text{MeV}$ released per event. The released neutrons can induce further fission events, sustaining a chain reaction. Controlled chain reactions in moderators (which slow neutrons) and control rods (which absorb excess neutrons) form the basis of nuclear reactors. Uncontrolled chain reactions in pure $^{235}$U or $^{239}$Pu form the basis of fission weapons.
Fusion is the joining of two light nuclei into a heavier one. Fusion requires extreme temperatures because the colliding nuclei must overcome their mutual Coulomb repulsion. The Sun’s energy comes mainly from the p-p chain:
$$_{1}^{1}\text{H} + _{1}^{1}\text{H} \to _{1}^{2}\text{H} + e^+ + \nu_e + 0.42\,\text{MeV}$$
$$_{1}^{2}\text{H} + _{1}^{1}\text{H} \to _{2}^{3}\text{He} + \gamma + 5.49\,\text{MeV}$$
$$_{2}^{3}\text{He} + _{2}^{3}\text{He} \to _{2}^{4}\text{He} + 2\,_{1}^{1}\text{H} + 12.86\,\text{MeV}$$
The net process $4\,_{1}^{1}\text{H} \to _{2}^{4}\text{He} + 2e^+ + 2\nu_e + 26.7\,\text{MeV}$ powers stars, including our Sun.
NCERT / ASSEB Exercise Solutions
Q1. (a) Two stable isotopes of lithium $_3^6$Li and $_3^7$Li have respective abundances of 7.5% and 92.5%. Their masses are 6.01512 u and 7.01600 u. Find the atomic mass of lithium.
Answer: Atomic mass = weighted mean of isotopic masses.
$$m = \dfrac{7.5 \times 6.01512 + 92.5 \times 7.01600}{100}$$
$$m = \dfrac{45.1134 + 648.98}{100} = \dfrac{694.0934}{100} = 6.9409\,\text{u}.$$
(b) Boron has two stable isotopes $_5^{10}$B and $_5^{11}$B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances.
Answer: Let abundance of $^{10}$B be $x$%, then abundance of $^{11}$B is $(100-x)$%. Then
$$10.811 = \dfrac{x\times 10.01294 + (100-x)\times 11.00931}{100}.$$
$$1081.1 = 10.01294 x + 1100.931 – 11.00931 x$$
$$0.99637 x = 19.831 \;\Rightarrow\; x \approx 19.91\%.$$
So $^{10}$B is about 19.9% and $^{11}$B is about 80.1%.
Q2. The three stable isotopes of neon $_{10}^{20}$Ne, $_{10}^{21}$Ne and $_{10}^{22}$Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u respectively. Obtain the average atomic mass of neon.
$$m = \dfrac{90.51\times 19.99 + 0.27\times 20.99 + 9.22\times 21.99}{100}$$
$$m = \dfrac{1809.694 + 5.667 + 202.747}{100} = 20.181\,\text{u}.$$
Answer: Average atomic mass of neon $\approx 20.18\,\text{u}$.
Q3. Obtain the binding energy (in MeV) of a nitrogen nucleus $_7^{14}$N, given $m(_7^{14}\text{N}) = 14.00307\,\text{u}$.
Answer: Using $m_p = 1.00783\,\text{u}$ (mass of $^1$H atom — electron masses cancel) and $m_n = 1.00867\,\text{u}$,
$$\Delta m = 7\,m_p + 7\,m_n – m(_7^{14}\text{N})$$
$$\Delta m = 7(1.00783) + 7(1.00867) – 14.00307$$
$$\Delta m = 7.05481 + 7.06069 – 14.00307 = 0.11243\,\text{u}.$$
$$B = 0.11243 \times 931.5 = 104.72\,\text{MeV}.$$
Binding energy per nucleon $= 104.72/14 \approx 7.48\,\text{MeV}$.
Q4. Obtain the binding energy of $_{26}^{56}$Fe and $_{83}^{209}$Bi. Given $m(_{26}^{56}\text{Fe}) = 55.93494\,\text{u}$ and $m(_{83}^{209}\text{Bi}) = 208.98037\,\text{u}$.
Answer: For $_{26}^{56}$Fe ($Z=26$, $N=30$):
$$\Delta m = 26(1.00783) + 30(1.00867) – 55.93494 = 0.52854\,\text{u}.$$
$$B = 0.52854 \times 931.5 \approx 492.3\,\text{MeV}.$$
So $B/A = 492.3/56 \approx 8.79\,\text{MeV/nucleon}$ — close to the curve’s peak.
For $_{83}^{209}$Bi ($Z=83$, $N=126$):
$$\Delta m = 83(1.00783) + 126(1.00867) – 208.98037 = 1.76157\,\text{u}.$$
$$B = 1.76157 \times 931.5 \approx 1640.3\,\text{MeV}.$$
So $B/A = 1640.3/209 \approx 7.85\,\text{MeV/nucleon}$ — smaller than for iron, showing why heavy nuclei tend to fission.
Q5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. The coin is entirely made of $_{29}^{63}$Cu atoms (mass 62.92960 u).
Answer: Number of atoms in 3.0 g of Cu
$$N = \dfrac{6.022\times 10^{23}\times 3.0}{62.92960} = 2.871\times 10^{22}.$$
For one $_{29}^{63}$Cu atom ($Z=29$, $N=34$):
$$\Delta m = 29(1.00783) + 34(1.00867) – 62.92960 = 0.59225\,\text{u}.$$
$$B_{\text{atom}} = 0.59225 \times 931.5 = 551.4\,\text{MeV}.$$
Total energy
$$E = 2.871\times 10^{22}\times 551.4\,\text{MeV} \approx 1.58\times 10^{25}\,\text{MeV} \approx 2.53\times 10^{12}\,\text{J}.$$
Q6. Write nuclear reaction equations for: (i) $\alpha$-decay of $_{88}^{226}$Ra, (ii) $\alpha$-decay of $_{94}^{242}$Pu, (iii) $\beta^-$-decay of $_{15}^{32}$P, (iv) $\beta^-$-decay of $_{83}^{210}$Bi, (v) $\beta^+$-decay of $_{6}^{11}$C, (vi) $\beta^+$-decay of $_{43}^{97}$Tc, (vii) electron capture of $_{54}^{120}$Xe.
Answer:
$$\text{(i)}\;_{88}^{226}\text{Ra} \to _{86}^{222}\text{Rn} + _2^4\text{He}$$
$$\text{(ii)}\;_{94}^{242}\text{Pu} \to _{92}^{238}\text{U} + _2^4\text{He}$$
$$\text{(iii)}\;_{15}^{32}\text{P} \to _{16}^{32}\text{S} + e^- + \bar{\nu}_e$$
$$\text{(iv)}\;_{83}^{210}\text{Bi} \to _{84}^{210}\text{Po} + e^- + \bar{\nu}_e$$
$$\text{(v)}\;_{6}^{11}\text{C} \to _{5}^{11}\text{B} + e^+ + \nu_e$$
$$\text{(vi)}\;_{43}^{97}\text{Tc} \to _{42}^{97}\text{Mo} + e^+ + \nu_e$$
$$\text{(vii)}\;_{54}^{120}\text{Xe} + e^- \to _{53}^{120}\text{I} + \nu_e$$
Q7. A radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value?
Answer: Activity follows $R = R_0 e^{-\lambda t}$. Take ratio $R/R_0$.
(a) $3.125\% = 1/32 = (1/2)^5$, so 5 half-lives are needed.
$$t = 5\,T \;\text{years}.$$
(b) $1\% = 0.01$. Use $0.01 = e^{-\lambda t}$ with $\lambda = (\ln 2)/T$.
$$t = \dfrac{\ln 100}{\ln 2}\,T = \dfrac{4.6052}{0.6931}T \approx 6.65\,T\;\text{years}.$$
Q8. The normal activity of living carbon-containing matter is found to be about 15 decays per minute per gram of carbon. This activity arises from the small proportion of $_6^{14}$C in the carbon, which is being maintained by cosmic-ray bombardment of nitrogen. When a plant or animal dies, the intake of $^{14}$C from the atmosphere ceases and the activity decays. From the known half-life (5730 y) of $^{14}$C, and the measured activity (9 decays/min/g) of an old sample from Mohenjo-daro, estimate its approximate age.
Answer: $R_0 = 15$, $R = 9$, so
$$\dfrac{R}{R_0} = e^{-\lambda t} \;\Rightarrow\; t = \dfrac{1}{\lambda}\ln\dfrac{R_0}{R} = \dfrac{T_{1/2}}{\ln 2}\ln\dfrac{15}{9}.$$
$$t = \dfrac{5730}{0.693}\times \ln(1.667) = 8268\times 0.5108 \approx 4223\;\text{years}.$$
Q9. Obtain the amount of $_{27}^{60}$Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of $^{60}$Co is 5.3 years.
Answer: Activity $R = \lambda N$, so $N = R/\lambda$.
$$\lambda = \dfrac{0.693}{5.3 \times 365 \times 86400} = 4.144\times 10^{-9}\;\text{s}^{-1}.$$
$$R = 8.0\times 10^{-3}\times 3.7\times 10^{10} = 2.96\times 10^{8}\;\text{Bq}.$$
$$N = \dfrac{2.96\times 10^{8}}{4.144\times 10^{-9}} = 7.14\times 10^{16}.$$
$$\text{Mass} = \dfrac{60 \times 7.14\times 10^{16}}{6.022\times 10^{23}} = 7.12\times 10^{-6}\;\text{g} \approx 7.12\,\mu\text{g}.$$
Q10. The half-life of $_{38}^{90}$Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:
$$N = \dfrac{6.022\times 10^{23}\times 15\times 10^{-3}}{90} = 1.004\times 10^{20}.$$
$$\lambda = \dfrac{0.693}{28\times 365\times 86400} = 7.85\times 10^{-10}\;\text{s}^{-1}.$$
$$R = \lambda N = 7.85\times 10^{-10}\times 1.004\times 10^{20} = 7.88\times 10^{10}\;\text{Bq}.$$
This equals about 2.13 Ci.
Q11. Obtain approximately the ratio of the nuclear radii of the gold isotope $_{79}^{197}$Au and the silver isotope $_{47}^{107}$Ag.
Answer: $R \propto A^{1/3}$, hence
$$\dfrac{R_{\text{Au}}}{R_{\text{Ag}}} = \left(\dfrac{197}{107}\right)^{1/3} = (1.841)^{1/3} \approx 1.226.$$
Q12. Find the Q-value and the kinetic energy of the emitted $\alpha$-particle in the $\alpha$-decay of (a) $_{88}^{226}$Ra and (b) $_{86}^{220}$Rn. Given $m(_{88}^{226}\text{Ra}) = 226.02540$ u, $m(_{86}^{222}\text{Rn}) = 222.01750$ u, $m(_{86}^{220}\text{Rn}) = 220.01137$ u, $m(_{84}^{216}\text{Po}) = 216.00189$ u and $m(_2^4\text{He}) = 4.00260$ u.
Answer: The Q-value is $Q = (m_{\text{parent}} – m_{\text{daughter}} – m_\alpha)c^2$. Since the parent is at rest, by momentum conservation the kinetic energy of the $\alpha$-particle is $K_\alpha = Q\cdot (A-4)/A$.
(a) For $_{88}^{226}$Ra:
$$\Delta m = 226.02540 – 222.01750 – 4.00260 = 0.00530\,\text{u}.$$
$$Q = 0.00530\times 931.5 = 4.94\,\text{MeV}.$$
$$K_\alpha = Q\times \dfrac{222}{226} \approx 4.85\,\text{MeV}.$$
(b) For $_{86}^{220}$Rn:
$$\Delta m = 220.01137 – 216.00189 – 4.00260 = 0.00688\,\text{u}.$$
$$Q = 0.00688\times 931.5 = 6.41\,\text{MeV}.$$
$$K_\alpha = Q\times \dfrac{216}{220} \approx 6.29\,\text{MeV}.$$
Q13. The radionuclide $^{11}$C decays according to $_6^{11}\text{C} \to _5^{11}\text{B} + e^+ + \nu_e$ with $T_{1/2} = 20.3$ min. The maximum energy of the emitted positron is 0.960 MeV. Given $m(_6^{11}\text{C}) = 11.011434$ u and $m(_5^{11}\text{B}) = 11.009305$ u, calculate $Q$ and compare it with the maximum positron energy.
Answer: When using atomic masses for $\beta^+$-decay, $Q = [m_a(_6^{11}\text{C}) – m_a(_5^{11}\text{B}) – 2m_e]c^2$ because the parent atom has 6 electrons but the daughter atom has only 5 (the extra electron mass plus the positron mass equals $2m_e$).
$$Q = (11.011434 – 11.009305)\times 931.5 – 2\times 0.511 = 1.983 – 1.022 = 0.961\,\text{MeV}.$$
This matches the observed maximum positron energy ($\approx 0.960$ MeV) since the positron and neutrino share the available kinetic energy and the recoil of $_5^{11}$B is negligible.
Q14. The nucleus $_{10}^{23}$Ne decays by $\beta^-$ emission. Write down the $\beta$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given: $m(_{10}^{23}\text{Ne}) = 22.994466$ u and $m(_{11}^{23}\text{Na}) = 22.989770$ u.
Answer:
$$_{10}^{23}\text{Ne} \to _{11}^{23}\text{Na} + e^- + \bar{\nu}_e$$
For $\beta^-$-decay using atomic masses, $Q = [m_a(_{10}^{23}\text{Ne}) – m_a(_{11}^{23}\text{Na})]c^2$ because the daughter atom carries one more electron, automatically accounting for the emitted electron.
$$Q = (22.994466 – 22.989770)\times 931.5 = 0.004696\times 931.5 \approx 4.374\,\text{MeV}.$$
The maximum KE of the emitted electron equals $Q$ when the antineutrino takes negligible energy: $K_{\max} \approx 4.37\,\text{MeV}$.
Q15. The Q-value of a nuclear reaction $A + b \to C + d$ is defined by $Q = [m_A + m_b – m_C – m_d]c^2$. Find whether the reactions are exothermic or endothermic. (i) $_1^1\text{H}+_1^3\text{H}\to _1^2\text{H}+_1^2\text{H}$, (ii) $_6^{12}\text{C}+_6^{12}\text{C}\to _{10}^{20}\text{Ne}+_2^4\text{He}$. Atomic masses: $m(_1^1\text{H})=1.007825$ u, $m(_1^2\text{H})=2.014102$ u, $m(_1^3\text{H})=3.016049$ u, $m(_6^{12}\text{C})=12.000000$ u, $m(_{10}^{20}\text{Ne})=19.992439$ u, $m(_2^4\text{He})=4.002603$ u.
Answer: If $Q>0$ the reaction is exothermic; if $Q<0$ it is endothermic.
(i)
$$Q = [(1.007825 + 3.016049) – 2(2.014102)]\times 931.5$$
$$Q = [4.023874 – 4.028204]\times 931.5 = -4.03\,\text{MeV}.$$
Hence (i) is endothermic.
(ii)
$$Q = [2(12.000000) – 19.992439 – 4.002603]\times 931.5$$
$$Q = [24 – 23.995042]\times 931.5 = 0.004958\times 931.5 \approx 4.62\,\text{MeV}.$$
Hence (ii) is exothermic.
Q16. Suppose, we think of fission of a $_{26}^{56}$Fe nucleus into two equal fragments $_{13}^{28}$Al. Is the fission energetically possible? Argue by working out $Q$ of the process. $m(_{26}^{56}\text{Fe})=55.93494$ u, $m(_{13}^{28}\text{Al})=27.98191$ u.
$$Q = [55.93494 – 2(27.98191)]\times 931.5 = (-0.02888)\times 931.5 \approx -26.9\,\text{MeV}.$$
Answer: Since $Q<0$, energy must be supplied — the reaction is energetically not possible spontaneously. This reflects the fact that $^{56}$Fe lies near the BE/A peak; splitting it would lower stability.
Q17. The fission properties of $_{94}^{239}$Pu are very similar to those of $_{92}^{235}$U. The average energy released per fission is 180 MeV. How much energy in MeV is released if all the atoms in 1 kg of pure $^{239}$Pu undergo fission?
$$N = \dfrac{6.022\times 10^{23}\times 10^3}{239} = 2.52\times 10^{24}.$$
$$E = N\times 180 = 4.54\times 10^{26}\,\text{MeV}.$$
Answer: $\approx 4.54\times 10^{26}\,\text{MeV}$ ($\approx 7.27\times 10^{13}\,\text{J}$ — roughly $2\times 10^{7}$ kWh).
Q18. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much $^{235}_{92}$U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from fission of $^{235}$U, and that this nuclide is consumed only by fission. Energy per fission = 200 MeV.
Answer: Energy delivered in 5 y at 80% load:
$$E = 10^9\times 0.8\times 5\times 365\times 86400 = 1.262\times 10^{17}\,\text{J}.$$
Energy per fission $= 200\times 1.6\times 10^{-13} = 3.2\times 10^{-11}$ J. Number of $^{235}$U atoms consumed:
$$N_{\text{used}} = \dfrac{1.262\times 10^{17}}{3.2\times 10^{-11}} = 3.94\times 10^{27}.$$
Mass consumed $= (235\times 3.94\times 10^{27})/(6.022\times 10^{23}) = 1538$ g $\approx 1.54$ kg. As only half the fuel is consumed, initial mass $= 2\times 1.54 \approx 3.08$ kg.
Q19. How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as $_1^2\text{H}+_1^2\text{H}\to _2^3\text{He}+n+3.27\,\text{MeV}$.
$$N = \dfrac{6.022\times 10^{23}\times 2000}{2} = 6.022\times 10^{26}\;\text{deuterons}.$$
Two deuterons fuse per reaction, so number of reactions $= N/2 = 3.011\times 10^{26}$.
$$E = 3.011\times 10^{26}\times 3.27\times 1.6\times 10^{-13} = 1.575\times 10^{14}\,\text{J}.$$
$$t = \dfrac{E}{P} = \dfrac{1.575\times 10^{14}}{100} = 1.575\times 10^{12}\,\text{s} \approx 4.99\times 10^{4}\;\text{years}.$$
Answer: About $5\times 10^{4}$ years — illustrating the enormous energy density of fusion fuel.
Q20. Calculate the height of the potential barrier for a head-on collision of two deuterons. Hint: take the height of the potential barrier as the Coulomb repulsion energy when the two deuterons just touch each other. Take the radius of a deuteron to be 2.0 fm.
Answer: When the deuterons just touch, separation $r = 2\times 2.0 = 4.0$ fm.
$$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r} = \dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{4.0\times 10^{-15}}$$
$$U = \dfrac{2.304\times 10^{-28}}{4.0\times 10^{-15}} = 5.76\times 10^{-14}\,\text{J} \approx 360\,\text{keV}.$$
Each deuteron must therefore approach with about 180 keV of kinetic energy.
Q21. From the relation $R = R_0 A^{1/3}$, 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 $A$).
Answer: Mass of nucleus $M \approx A\,m_u$. Volume:
$$V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A.$$
$$\rho = \dfrac{M}{V} = \dfrac{A m_u}{\tfrac{4}{3}\pi R_0^3 A} = \dfrac{3 m_u}{4\pi R_0^3},$$
independent of $A$. Numerically, with $R_0 = 1.2$ fm and $m_u = 1.66\times 10^{-27}$ kg, $\rho \approx 2.3\times 10^{17}$ kg/m$^3$.
Additional Important Questions
Multiple Choice Questions
1. The mass number of a nucleus is
- (a) always less than its atomic number
- (b) the sum of protons and neutrons
- (c) always greater than the binding energy
- (d) the number of protons only
Answer: (b) the sum of protons and neutrons.
2. The energy equivalent of 1 atomic mass unit is
- (a) 931.5 MeV
- (b) 1.6×10⁻¹⁹ J
- (c) 1.6×10⁻¹³ J
- (d) 13.6 eV
Answer: (a) 931.5 MeV.
3. Nuclear density is
- (a) directly proportional to $A$
- (b) inversely proportional to $A$
- (c) independent of $A$
- (d) proportional to $A^2$
Answer: (c) independent of $A$.
4. The peak of the binding energy per nucleon curve occurs at
- (a) $^4$He
- (b) $^{56}$Fe
- (c) $^{235}$U
- (d) $^{12}$C
Answer: (b) $^{56}$Fe.
5. Half-life and mean life are related by
- (a) $t_{1/2} = \tau$
- (b) $t_{1/2} = \tau/\ln 2$
- (c) $t_{1/2} = \tau\ln 2$
- (d) $t_{1/2} = 2\tau$
Answer: (c) $t_{1/2} = \tau\ln 2 = 0.693\,\tau$.
6. In $\beta^-$-decay the daughter nucleus has
- (a) $Z$ unchanged, $A$ unchanged
- (b) $Z$ increased by 1, $A$ unchanged
- (c) $Z$ decreased by 1, $A$ unchanged
- (d) $Z$ decreased by 2, $A$ decreased by 4
Answer: (b).
7. One curie equals
- (a) $3.7\times 10^{10}$ Bq
- (b) $1$ Bq
- (c) $10^6$ Bq
- (d) $1.6\times 10^{-19}$ Bq
Answer: (a).
8. The energy released per fission of $^{235}$U is approximately
- (a) 2 MeV
- (b) 20 MeV
- (c) 200 MeV
- (d) 2000 MeV
Answer: (c) 200 MeV.
9. The Sun derives its energy mainly from
- (a) chemical reactions
- (b) gravitational contraction
- (c) nuclear fission
- (d) nuclear fusion of hydrogen
Answer: (d).
10. Two nuclei have mass numbers in the ratio 1:8. Their nuclear radii are in the ratio
- (a) 1:8
- (b) 1:2
- (c) 1:4
- (d) 1:6
Answer: (b) 1:2 since $R\propto A^{1/3}$.
Short-Answer Questions
Q. Define mass defect and binding energy.
Answer: The mass defect is the difference between the sum of the masses of the free nucleons and the mass of the nucleus they form: $\Delta m = Z m_p + (A-Z)m_n – M$. The binding energy is the energy equivalent of this mass defect, $B = \Delta m\,c^2$. It equals the work needed to dissociate the nucleus into its constituent nucleons.
Q. Distinguish between isotopes, isobars and isotones.
Answer: Isotopes have the same $Z$ but different $A$ (e.g. $^1$H, $^2$H). Isobars have the same $A$ but different $Z$ (e.g. $^{40}$Ar and $^{40}$K). Isotones have the same neutron number $N$ (e.g. $^{36}$S and $^{37}$Cl, both with $N=20$).
Q. State the properties of nuclear forces.
Answer: (i) Short-ranged ($\sim$1–2 fm); (ii) very strong, much greater than electromagnetic at this scale; (iii) charge-independent; (iv) saturated, i.e. each nucleon interacts only with neighbours; (v) attractive at distances above $\sim 0.7$ fm but strongly repulsive at smaller distances; (vi) spin-dependent.
Q. Why is the binding energy per nucleon nearly constant for $A$ between 30 and 170?
Answer: Because the nuclear force saturates: a nucleon interacts only with its nearest neighbours. So the total binding $B$ is roughly proportional to the number of nucleons $A$, making $B/A$ approximately constant.
Q. Why does the binding energy per nucleon decrease for very heavy nuclei?
Answer: The Coulomb repulsion between the increasing number of protons is long-range and grows as $Z^2$, while the binding nuclear force is short-range. Beyond about $A=170$ the destabilising electrostatic energy reduces $B/A$, eventually making heavy nuclei unstable to $\alpha$-decay or fission.
Q. State the law of radioactive decay and define decay constant.
Answer: The number of nuclei disintegrating per unit time is proportional to the number of un-decayed nuclei present at that instant: $dN/dt = -\lambda N$, where $\lambda$ is the decay constant — the probability per unit time that a nucleus will decay. Integrating gives $N = N_0 e^{-\lambda t}$.
Q. Show that $t_{1/2} = 0.693\,\tau$.
Answer: The mean life is
$$\tau = \dfrac{\int_0^\infty t\,\lambda N_0 e^{-\lambda t}\,dt}{N_0} = \dfrac{1}{\lambda}.$$
The half-life is given by $N_0/2 = N_0 e^{-\lambda t_{1/2}} \Rightarrow t_{1/2} = (\ln 2)/\lambda$. Therefore $t_{1/2} = \tau\ln 2 = 0.693\,\tau$.
Q. Why are slow (thermal) neutrons used in fission reactors?
Answer: The fission cross-section of $^{235}$U is much larger for slow neutrons than for fast neutrons. Slow neutrons therefore have a high probability of inducing fission and sustaining the chain reaction. Moderators (graphite, heavy water, light water) reduce the fast neutrons’ kinetic energy without absorbing them.
Q. Why does fusion require very high temperature?
Answer: Two positively charged nuclei must approach within $\sim 1$ fm for the strong attractive force to take over. To overcome the Coulomb repulsion they need very large kinetic energies — equivalent to temperatures of order $10^7$–$10^8$ K. Only at such temperatures (found in stellar cores or H-bomb interiors) is fusion possible.
Q. What are the four basic forces of nature in increasing order of strength?
Answer: Gravitational (weakest), weak nuclear, electromagnetic, strong nuclear (strongest, about $10^2$ times stronger than electromagnetic at nuclear distances).
Q. Explain controlled and uncontrolled chain reaction.
Answer: In an uncontrolled chain reaction (atomic bomb) every fission produces several secondary neutrons, which are all available to cause more fissions. The reaction grows exponentially and releases enormous energy in a fraction of a second. In a controlled chain reaction (nuclear reactor) cadmium or boron control rods absorb the surplus neutrons so that on average exactly one neutron from each fission causes the next; the energy is then released steadily and can be used to drive turbines.
Q. Explain proton-proton (p-p) chain.
Answer: The p-p chain is the sequence of fusion reactions powering low- and medium-mass main-sequence stars, including the Sun. Net effect: $4\,_1^1\text{H} \to _2^4\text{He} + 2e^+ + 2\nu_e + 26.7\,\text{MeV}$. About 0.7% of the rest mass of hydrogen is converted into radiant energy.
Numerical Practice
P1. Calculate the radius of the $^{12}$C nucleus.
Answer: $R = R_0 A^{1/3} = 1.2\times (12)^{1/3}$ fm $= 1.2\times 2.289 = 2.75$ fm.
P2. The half-life of a radioactive sample is 5 hours. What fraction of the sample is left after 20 hours?
Answer: Number of half-lives $n = 20/5 = 4$. Fraction remaining $= (1/2)^4 = 1/16 = 6.25\%$.
P3. The activity of a radioactive sample falls from $10^4$ Bq to $1.25\times 10^3$ Bq in 30 minutes. Find the half-life.
Answer: $R/R_0 = 1.25\times 10^3/10^4 = 1/8 = (1/2)^3$. So 3 half-lives = 30 min, giving $t_{1/2} = 10$ min.
P4. Calculate the energy released when 1 gram of $^{235}$U undergoes complete fission, given energy per fission = 200 MeV.
Answer: $N = (6.022\times 10^{23})/235 = 2.563\times 10^{21}$ atoms. $E = 2.563\times 10^{21}\times 200 = 5.13\times 10^{23}$ MeV $= 8.21\times 10^{10}$ J.
P5. The mean life of a radioactive nucleus is 1000 s. Find its half-life and decay constant.
Answer: $\lambda = 1/\tau = 10^{-3}\,\text{s}^{-1}$. $t_{1/2} = 0.693\tau = 693$ s.
Glossary / Key Terms
| Term | Meaning |
|---|---|
| Nucleon | Either a proton or a neutron — the building block of a nucleus. |
| Atomic number $Z$ | Number of protons. |
| Mass number $A$ | Total number of nucleons; $A = Z + N$. |
| Atomic mass unit (u) | 1/12th of the mass of a $^{12}$C atom; $1\,\text{u} = 1.66\times 10^{-27}$ kg = 931.5 MeV/$c^2$. |
| Mass defect | Difference between sum of free nucleon masses and bound-nucleus mass. |
| Binding energy | Energy equivalent of mass defect; energy needed to dissociate the nucleus. |
| Isotopes | Same $Z$, different $A$. |
| Isobars | Same $A$, different $Z$. |
| Isotones | Same $N$, different $Z$. |
| Radioactivity | Spontaneous emission of $\alpha$, $\beta$ or $\gamma$ radiation by an unstable nucleus. |
| $\alpha$-particle | $_2^4$He nucleus. |
| $\beta^-$-particle | An electron emitted from the nucleus, accompanied by an antineutrino. |
| $\beta^+$-particle | A positron emitted from the nucleus, accompanied by a neutrino. |
| $\gamma$-ray | High-energy electromagnetic photon emitted from the nucleus. |
| Decay constant $\lambda$ | Probability per unit time that a given nucleus will decay. |
| Half-life $t_{1/2}$ | Time after which half of a radioactive sample remains; $t_{1/2}=0.693/\lambda$. |
| Mean life $\tau$ | Average lifetime of a nucleus before decay; $\tau = 1/\lambda$. |
| Activity | Decay rate; $R = \lambda N$, measured in Bq or Ci. |
| Becquerel (Bq) | SI unit of activity = 1 disintegration per second. |
| Curie (Ci) | $3.7\times 10^{10}$ Bq. |
| Q-value | Net energy released in a nuclear reaction, $Q = (m_i – m_f)c^2$. |
| Fission | Splitting of a heavy nucleus into two medium-mass fragments releasing $\sim 200$ MeV. |
| Fusion | Combination of light nuclei into a heavier one, releasing energy. |
| Chain reaction | Self-sustaining sequence of nuclear fissions where each event triggers further fissions. |
| Critical mass | Minimum mass of fissile material needed to sustain a chain reaction. |
| Moderator | Material (graphite, heavy water) that slows neutrons in a reactor. |
| Control rod | Cadmium or boron rod that absorbs surplus neutrons in a reactor. |
| Proton-proton chain | Sequence of fusion reactions powering the Sun. |
That completes the English-medium notes for ASSEB Class 12 Physics Chapter 13 — Nuclei. For more chapter-wise solutions, formula sheets and examination-style problems for every subject of Classes 6 to 12, please continue exploring HSLC GURU.