NCERT Solutions for Class 12 Physics Chapter 13: Nuclei
All 10 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — mass defect and binding energy, nuclear radius and density, Q-values, fission and fusion energy and the Coulomb barrier — with binding-energy graphs linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
Chapter at a glance — complete revision notes
13.1 Introduction
- In every atom the positive charge and mass are concentrated in a nucleus whose radius is smaller than that of the atom by a factor of about 10⁴ — its volume is about 10⁻¹² of the atom’s volume; an atom is almost empty (if an atom were enlarged to the size of a classroom, the nucleus would be the size of a pinhead), yet the nucleus contains most of the atom’s mass. Questions: does the nucleus have a structure, what are its constituents, how are they held together? The chapter studies the size, mass and stability of nuclei and the phenomena of radioactivity, fission and fusion.
13.2 Atomic masses and composition of nucleus
- Atomic mass unit (u) = 1/12 of the mass of a ¹²C atom (the mass of a carbon atom is 1.992647 × 10⁻²⁶ kg): 1 u = 1.660539 × 10⁻²⁷ kg (= 931.5 MeV/c²). Atomic masses are close to integral multiples of the hydrogen atom mass, with exceptions (chlorine 35.46 u). Mass spectrometers measure atomic masses accurately and reveal isotopes — atoms of the same element (same chemical properties) with different masses (Greek: “same place”); e.g. chlorine has isotopes of 34.98 u and 36.98 u in the abundance 75.4 % and 24.6 %, giving an average 35.46 u; hydrogen has three isotopes (1.0078 u, 2.0141 u, 3.0160 u): protium (99.985 %), deuterium and tritium (unstable, artificially produced). The mass of the hydrogen nucleus (proton) = mass of the H atom − m_e (0.00055 u).
- Neutron: the nuclei of deuterium and tritium contain a proton and “neutral matter” in 1 and 2 units; in 1932 James Chadwick bombarded beryllium with α-particles and found a neutral radiation that could knock out protons from He, C and N — conservation of energy and momentum showed it was not photons but a new neutral particle of mass very nearly the proton’s: the neutron (Nobel Prize 1935). Electrons cannot be in the nucleus (quantum-theoretical argument). A free neutron is unstable (decays into a proton, an electron and an antineutrino, mean life ≈ 1000 s) but stable inside the nucleus.
- Composition: atomic number Z = number of protons (also of electrons); neutron number N; mass number A = Z + N (total number of nucleons). Notation ᴬ_Z X (e.g. ¹⁹⁷₇₉Au: 79 protons, 118 neutrons). Isotopes have the same Z but different N (gold has 32 isotopes from A = 173 to 204); isobars have the same A (e.g. ³H and ³He); isotones have the same N (e.g. ¹⁹⁸Hg and ¹⁹⁷Au).
13.3 Size of the nucleus
- From Rutherford scattering: the distance of closest approach of a 5.5 MeV α-particle to a gold nucleus is about 4.0 × 10⁻¹⁴ m, so the nucleus is smaller than this; at higher energies the scattering is affected by the short-range nuclear force (deviation from the Coulomb calculation) — which gives a measure of the size. Scattering of fast electrons gives accurate sizes: R = R₀A^(1/3), R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm (1 fm = 10⁻¹⁵ m). Volume ∝ R³ ∝ A, so the density of nuclear matter is the same for all nuclei (independent of A): ≈ 2.3 × 10¹⁷ kg m⁻³ (compared with 10³ kg m⁻³ for water) — nuclei behave like drops of a liquid of constant density; the matter in neutron stars has a comparable density. Example 13.1: iron nucleus (55.85 u, A = 56): R = 4.6 fm, density 2.3 × 10¹⁷ kg m⁻³.
13.4 Mass-energy and nuclear binding energy
13.4.1 Mass–energy
Einstein’s special relativity: mass is another form of energy: E = mc² (c ≈ 3 × 10⁸ m s⁻¹). Example 13.2: the energy equivalent of 1 g of matter is 9 × 10¹³ J. Mass and energy are conserved together (the energy associated with mass must be included); verified in nuclear reactions. Energy equivalent of 1 u: 1.4924 × 10⁻¹⁰ J = 931.5 MeV (Example 13.3).13.4.2 Nuclear binding energy
The mass M of a nucleus is always less than the sum of the masses of its nucleons: the mass defect ΔM = [Zm_p + (A − Z)m_n] − M. Example: ¹⁶₈O — mass defect 0.13691 u (nuclear mass 15.99493 u − 8 electrons). The equivalent energy ΔMc² is the binding energy E_b (energy to be supplied to break the nucleus into free nucleons; released when the nucleus forms). Binding energy per nucleon E_bn = E_b/A (Fig. 13.1): rises to ≈ 8 MeV for light nuclei, is nearly constant (≈ 8.0 MeV) between A = 30 and 170 (maximum ≈ 8.75 MeV near A = 56, iron), then falls slowly (≈ 7.6 MeV for U-238). Conclusions: (i) the nuclear force is attractive and strong; (ii) the near constancy for 30 < A < 170 results from the short range of nuclear forces (a nucleon interacts only with its few neighbours): saturation; (iii) a very heavy nucleus (A = 240, E_bn ≈ 7.6 MeV) splitting into two A ≈ 120 nuclei (E_bn ≈ 8.5 MeV) releases energy ≈ 0.9 MeV × 240 ≈ 216 MeV — fission; (iv) two very light nuclei (A ≤ 10) fusing into a heavier nucleus have a higher E_bn and release energy — fusion, the source of the Sun’s energy.
13.5 Nuclear force
- The binding energy per nucleon (≈ 8 MeV) is far larger than that of atoms, so a strong attractive force of a totally different kind (not Coulomb, not gravity) is needed. Features (obtained from experiments during 1930–1950s): (i) it is much stronger than the Coulomb force between protons and the gravitational force (in a nucleus it dominates the Coulomb repulsion); (ii) it is short-ranged — falls rapidly to zero beyond a few femtometres, leading to saturation and the constancy of E_bn; the potential energy between two nucleons has a minimum at r₀ ≈ 0.8 fm (Fig. 13.2): attractive for r > 0.8 fm and repulsive for r < 0.8 fm; (iii) it is charge-independent: approximately the same between n–n, p–n and p–p; (iv) unlike Coulomb’s or Newton’s law there is no simple mathematical expression for the nuclear force.
13.6 Radioactivity
- Discovered accidentally by A. H. Becquerel in 1896 (uranium–potassium sulphate wrapped in black paper blackened a photographic plate separated by silver). Radioactivity is a nuclear phenomenon: an unstable nucleus undergoes decay. Three types: α-decay (emission of a helium nucleus), β-decay (emission of electrons or positrons — particles with the electron’s mass but opposite charge) and γ-decay (emission of high-energy photons, hundreds of keV or more).
13.7 Nuclear energy
- The curve of E_bn shows that nuclei with less total binding energy transform to nuclei with greater binding energy with a net energy release: heavy → medium (fission), light → heavier (fusion). Energies are of the order of MeV compared with eV for chemical reactions — a million times more energy per unit mass: fission of 1 kg of uranium gives about 10¹⁴ J vs 10⁷ J for burning 1 kg of coal.
13.7.1 Fission
A neutron-induced reaction: ²³⁵₉₂U + n → ²³⁶₉₂U* → two intermediate-mass fragments (e.g. ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3n) with ≈ 200 MeV per fission (Q value); the energy appears first as the kinetic energy of the fragments and neutrons and then as heat in the surrounding matter. The fission fragments are radioactive (β emitters) and the extra neutrons can trigger further fissions — a chain reaction. The source of energy in nuclear reactors (electricity) and of the atom bomb (uncontrolled chain reaction).13.7.2 Nuclear fusion — energy generation in stars
Examples: p + p → ²H + e⁺ + ν (0.42 MeV); ²H + ²H → ³He + n (3.27 MeV); ²H + ²H → ³H + p. Light nuclei must overcome the Coulomb barrier (≈ 400 keV for two protons); the temperature for (3/2)kT ≈ 400 keV is T ≈ 3 × 10⁹ K — thermonuclear fusion. The interior of the Sun is at 1.5 × 10⁷ K (fusion involves protons far above average energy). The Sun’s proton–proton cycle burns hydrogen into helium: 4¹H → ⁴He + 2e⁺ + 2ν + 26.7 MeV. When the hydrogen core is depleted, the core contracts and heats to ≈ 10⁸ K, fusing helium into carbon, and so on up to nuclei near the peak of the binding energy curve (iron). The Sun’s age is ≈ 5 × 10⁹ y with about as long to go.13.7.3 Controlled thermonuclear fusion
Aim: steady power by heating the fuel to ≈ 10⁸ K — a plasma of ions and electrons; the challenge is to confine it since no container withstands such temperatures; several countries including India are developing techniques. Example 13.4: (a) nuclear reactions are not “balanced” like chemical equations in numbers of atoms (elements may be transmuted) but the number of protons and of neutrons are separately conserved (strictly total charge and baryon number); (b) mass is converted into energy through the difference in the total binding energies on the two sides (binding energy contributes a negative amount to mass); (c) chemical reactions also interconvert mass and energy (chemical binding-energy mass defects), but these are about a million times smaller than in nuclear reactions.
Summary & key points (NCERT)
- Nuclear size R = R₀A^{1/3}; constant nuclear density; mass defect and E_b = ΔMc²; 1 u = 931.5 MeV/c²; E_bn ≈ 8 MeV; nuclear force properties; fission and fusion energy sources; Q-value of a reaction = (initial masses − final masses)c² (positive: exothermic; negative: endothermic).
Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given m(¹⁴₇N) = 14.00307 u.
Concept and context. The mass of a nucleus is always less than the sum of the masses of its separate nucleons; the difference is the mass defect ΔM = [Zm_p + (A − Z)m_n] − M. By Einstein’s mass–energy relation E = mc² the equivalent energy E_b = ΔM c² is the binding energy: the energy that must be supplied to break the nucleus into its free protons and neutrons (or released when the nucleus is formed). With masses in atomic mass units the conversion is 1 u = 931.5 MeV/c². When atomic masses are given (as here, m(¹⁴₇N) = 14.00307 u is the mass of the neutral atom), the electron masses cancel if the proton mass is replaced by the mass of the hydrogen atom m_H = 1.007825 u (it includes one electron), because the atom has Z electrons.
Given: ¹⁴₇N: Z = 7, N = 14 − 7 = 7; m(¹⁴₇N) = 14.00307 u; m_H = 1.007825 u; m_n = 1.008665 u; 1 u = 931.5 MeV/c².
- Total mass of the constituents (7 hydrogen atoms + 7 neutrons): 7 × 1.007825 + 7 × 1.008665 = 7.054775 + 7.060655 = 14.115430 u.
- Mass defect: ΔM = 14.115430 − 14.00307 = 0.11236 u.
- Binding energy: E_b = ΔM × 931.5 MeV = 0.11236 × 931.5 = 104.66 MeV (≈ 104.7 MeV).
- Binding energy per nucleon: E_bn = 104.66/14 = 7.48 MeV.
E_b = [7m_H + 7m_n − m(¹⁴N)]c² = 0.11236 u × 931.5 MeV/u ≈ 104.7 MeV
| Quantity | Value |
|---|---|
| Z, N, A | 7, 7, 14 |
| 7 m_H + 7 m_n | 14.115430 u |
| Atomic mass of ¹⁴N | 14.00307 u |
| Mass defect | 0.11236 u |
| Binding energy | 104.66 MeV |
| Binding energy per nucleon | 7.48 MeV |
- Why the hydrogen atom mass? The atomic mass of ¹⁴N includes 7 electrons; 7 hydrogen atoms include 7 electrons as well (the tiny electron binding energies are neglected), so the electron masses cancel and the mass defect is that of the nucleus.
- Interpretation: 7.48 MeV per nucleon is typical of light nuclei (compare ⁴He: 7.07 MeV, ⁵⁶Fe: 8.79 MeV); it is roughly a million times the chemical binding energy (a few eV per atom).
Conclusion: the binding energy of the nitrogen nucleus is 104.7 MeV (7.48 MeV per nucleon).
Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi in units of MeV from the following data: m(⁵⁶₂₆Fe) = 55.934939 u, m(²⁰⁹₈₃Bi) = 208.980388 u.
Concept and context. The binding energy of a nucleus ᴬ_Z X with atomic mass M is E_b = {Z m_H + (A − Z) m_n − M} c², with m_H = 1.007825 u (hydrogen atom, to account for the Z electrons of the atom) and m_n = 1.008665 u; 1 u = 931.5 MeV/c². The binding energy per nucleon E_b/A measures how tightly the nucleons are bound: ⁵⁶Fe is near the maximum of the binding energy curve (most stable), whereas ²⁰⁹Bi, a heavy nucleus, has a lower value (NCERT Fig. 13.1).
(a) ⁵⁶₂₆Fe (Z = 26, N = 30, A = 56)
- Total mass of constituents: 26 × 1.007825 + 30 × 1.008665 = 26.20345 + 30.25995 = 56.46340 u.
- Mass defect: ΔM = 56.46340 − 55.934939 = 0.528461 u.
- E_b = 0.528461 × 931.5 = 492.26 MeV; per nucleon: 492.26/56 = 8.79 MeV.
(b) ²⁰⁹₈₃Bi (Z = 83, N = 126, A = 209)
- Total mass of constituents: 83 × 1.007825 + 126 × 1.008665 = 83.649475 + 127.091790 = 210.741265 u.
- Mass defect: ΔM = 210.741265 − 208.980388 = 1.760877 u.
- E_b = 1.760877 × 931.5 = 1640.26 MeV; per nucleon: 1640.26/209 = 7.85 MeV.
| Nucleus | Z / N / A | Mass defect (u) | E_b (MeV) | E_b/A (MeV) |
|---|---|---|---|---|
| ⁵⁶₂₆Fe | 26 / 30 / 56 | 0.528461 | 492.26 | 8.79 |
| ²⁰⁹₈₃Bi | 83 / 126 / 209 | 1.760877 | 1640.26 | 7.85 |
- Comparison: Bi has a larger total binding energy (more nucleons) but a smaller binding energy per nucleon; ⁵⁶Fe, with 8.79 MeV/nucleon, is among the most tightly bound nuclei. A heavy nucleus like Bi (and U) can release energy by splitting into medium-mass nuclei (fission), whereas Fe cannot (Exercise 13.6).
- Why E_b/A is smaller for heavy nuclei: the repulsive Coulomb force between the many protons grows as Z² (long range) while the attractive nuclear force is short-ranged and saturates, reducing the average binding per nucleon.
Conclusion: the binding energies are 492.26 MeV for ⁵⁶Fe (8.79 MeV/nucleon) and 1640.26 MeV for ²⁰⁹Bi (7.85 MeV/nucleon).
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. For simplicity assume that the coin is entirely made of ⁶³₂₉Cu atoms (of mass 62.92960 u).
Concept and context. The nuclear energy needed to separate all the protons and neutrons of a nucleus from each other is its binding energy E_b = ΔMc². For a macroscopic coin we find the binding energy of one ⁶³Cu nucleus and multiply by the number of nuclei (atoms) in the coin: N = (mass/molar mass) × Avogadro’s number (N_A = 6.023 × 10²³ per mole). The result illustrates the enormous scale of nuclear energies compared with everyday energies.
Given: mass of coin = 3.0 g; m(⁶³₂₉Cu) = 62.92960 u; Z = 29, N = 34; m_H = 1.007825 u; m_n = 1.008665 u; N_A = 6.023 × 10²³ mol⁻¹.
- Mass defect of one ⁶³Cu nucleus: ΔM = 29 × 1.007825 + 34 × 1.008665 − 62.92960 = 29.226925 + 34.294610 − 62.92960 = 0.591935 u.
- Binding energy per nucleus: E_b = 0.591935 × 931.5 = 551.4 MeV.
- Number of atoms in the coin: n = (3.0 g/62.9296 g mol⁻¹) × 6.023 × 10²³ = 0.047672 × 6.023 × 10²³ = 2.871 × 10²² atoms.
- Total nuclear energy required: E = n E_b = 2.871 × 10²² × 551.4 MeV = 1.58 × 10²⁵ MeV (= 1.58 × 10²⁵ × 1.6 × 10⁻¹³ J ≈ 2.5 × 10¹² J).
E ≈ 1.58 × 10²⁵ MeV ≈ 2.5 × 10¹² J
| Quantity | Value |
|---|---|
| Mass defect per nucleus | 0.591935 u |
| E_b per nucleus | 551.4 MeV (8.75 MeV/nucleon) |
| Atoms in 3.0 g | 2.871 × 10²² |
| Total energy | 1.58 × 10²⁵ MeV = 2.5 × 10¹² J |
- Perspective: 2.5 × 10¹² J is the energy that a 1 GW (10⁹ W) power station delivers in about 40 minutes — or roughly the energy of 600 tonnes of TNT (1 ton TNT = 4.2 × 10⁹ J). It is this enormous energy that makes nuclear processes such powerful energy sources.
- Note: the energy is the nuclear binding energy; the electrons’ binding (eV scale) is negligible and has been ignored.
Conclusion: the energy required to separate all the neutrons and protons in the coin is about 1.58 × 10²⁵ MeV (≈ 2.5 × 10¹² J).
Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and the silver isotope ¹⁰⁷₄₇Ag.
Concept and context. Experiments on the scattering of fast electrons by nuclei show that the radius of a nucleus of mass number A is R = R₀ A^(1/3), with R₀ = 1.2 × 10⁻¹⁵ m (1.2 fm), the same for all nuclei (NCERT Section 13.3). Consequently the volume of a nucleus (∝ R³) is proportional to A: nuclear matter has the same density in all nuclei, as if the nucleus were a drop of liquid of constant density. The ratio of the radii of two nuclei therefore depends only on the cube root of the ratio of their mass numbers.
- R_Au = R₀ (197)^(1/3), R_Ag = R₀ (107)^(1/3).
- Ratio: R_Au/R_Ag = (197/107)^(1/3) = (1.841)^(1/3).
- Cube root: 1.841^(1/3) = 1.23.
R_Au/R_Ag = (197/107)^(1/3) ≈ 1.23
| Nucleus | A | A^(1/3) | R = 1.2 A^(1/3) fm |
|---|---|---|---|
| ¹⁹⁷Au | 197 | 5.82 | 7.0 fm |
| ¹⁰⁷Ag | 107 | 4.75 | 5.7 fm |
| Ratio | — | 1.23 | 1.23 |
- Interpretation: although gold has about 1.84 times as many nucleons as silver, its radius is only 1.23 times larger — volume ∝ A (constant density): the ratio of volumes is 197/107 = 1.84 = (1.23)³ ✓.
- Check of consequence: the density of nuclear matter ρ = 3m_u/(4πR₀³) ≈ 2.3 × 10¹⁷ kg m⁻³ (see Exercise 13.10) is the same for both nuclei.
Conclusion: the ratio of the nuclear radii of gold and silver is approximately 1.23.
The Q value of a nuclear reaction A + b → C + d is defined by Q = [m_A + m_b − m_C − m_d]c², 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) ¹₁H + ³₁H → ²₁H + ²₁H (ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He. Atomic masses are given to be m(²₁H) = 2.014102 u, m(³₁H) = 3.016049 u, m(¹²₆C) = 12.000000 u, m(²⁰₁₀Ne) = 19.992439 u.
Concept and context. The Q-value of a nuclear reaction is the energy released (Q > 0, exothermic) or absorbed (Q < 0, endothermic) in the reaction: Q = (total initial mass − total final mass)c². It arises because the total binding energies on the two sides differ: the mass difference is converted into kinetic energy of the products. Since the numbers of protons and neutrons are conserved on both sides, the electron masses also cancel when atomic masses are used (the number of electrons is the same on both sides for a balanced reaction). The mass of the ordinary hydrogen atom is m(¹₁H) = 1.007825 u and of ⁴₂He is 4.002603 u (from the data list of the exercises).
(i) ¹₁H + ³₁H → ²₁H + ²₁H
- Initial mass = m(¹H) + m(³H) = 1.007825 + 3.016049 = 4.023874 u.
- Final mass = 2 × m(²H) = 2 × 2.014102 = 4.028204 u.
- Q = (4.023874 − 4.028204) u × 931.5 = −0.004330 u × 931.5 = −4.03 MeV — negative: the reaction is endothermic (energy of 4.03 MeV must be supplied).
(ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He
- Initial mass = 2 × 12.000000 = 24.000000 u.
- Final mass = 19.992439 + 4.002603 = 23.995042 u.
- Q = (24.000000 − 23.995042) u × 931.5 = 0.004958 × 931.5 = +4.62 MeV — positive: the reaction is exothermic (releases 4.62 MeV).
(i) Q = −4.03 MeV (endothermic) ; (ii) Q = +4.62 MeV (exothermic)
| Reaction | Initial mass (u) | Final mass (u) | Δm (u) | Q (MeV) | Type |
|---|---|---|---|---|---|
| (i) ¹H + ³H → ²H + ²H | 4.023874 | 4.028204 | −0.004330 | −4.03 | endothermic |
| (ii) ¹²C + ¹²C → ²⁰Ne + ⁴He | 24.000000 | 23.995042 | +0.004958 | +4.62 | exothermic |
- Interpretation of (i): the products ²H + ²H have more mass than the reactants, so the incident particle must supply at least 4.03 MeV (plus recoil energy) — the reaction occurs only if the projectile has more than this threshold energy.
- Interpretation of (ii): carbon burning in massive stars — fusion of ¹²C nuclei releases energy (it must still overcome a large Coulomb barrier, hence requires temperatures of ~10⁹ K).
Conclusion: (i) Q = −4.03 MeV — endothermic; (ii) Q = +4.62 MeV — exothermic.
Suppose, we think of fission of a ⁵⁶₂₆Fe nucleus into two equal fragments, ²⁸₁₃Al. Is the fission energetically possible? Argue by working out Q of the process. Given m(⁵⁶₂₆Fe) = 55.93494 u and m(²⁸₁₃Al) = 27.98191 u.
Concept and context. A nuclear process is energetically possible (without supplying external energy) only if its Q-value is positive — i.e. the total mass of the products is less than that of the original nucleus, so that the mass difference appears as kinetic energy. This is related to the binding energy curve: fission releases energy only if the fragments are more tightly bound per nucleon than the parent. Iron-56 lies at the very top of the binding energy curve (E_bn ≈ 8.8 MeV): splitting it into two ²⁸Al nuclei (E_bn lower, ≈ 8.3 MeV) would require energy.
Given: m(⁵⁶₂₆Fe) = 55.93494 u; m(²⁸₁₃Al) = 27.98191 u.
- Reaction: ⁵⁶₂₆Fe → ²⁸₁₃Al + ²⁸₁₃Al (the numbers of protons 26 = 13 + 13 and nucleons 56 = 28 + 28 balance).
- Mass of the products: 2 × 27.98191 = 55.96382 u.
- Mass difference: Δm = 55.93494 − 55.96382 = −0.02888 u.
- Q = Δm × 931.5 = −0.02888 × 931.5 = −26.9 MeV (≈ −26.9 MeV).
Q = [m(Fe) − 2m(Al)]c² = −26.9 MeV < 0
- Conclusion: Q is negative — the products are heavier than the parent nucleus, so the process is endothermic: 26.9 MeV must be supplied to split a ⁵⁶Fe nucleus into two ²⁸Al nuclei. The fission is not energetically possible (it will not happen spontaneously).
- Reason from the binding energy curve: fission releases energy only for heavy nuclei (A ≳ 200), whose E_bn (≈ 7.6 MeV) is lower than that of the medium-mass fragments (≈ 8.5 MeV). For A = 56 (the peak) any splitting leads to lower E_bn; similarly fusion of iron into heavier nuclei costs energy — which is why stellar fusion stops at iron.
| Quantity | Value |
|---|---|
| Parent mass | 55.93494 u |
| Mass of two ²⁸Al | 55.96382 u |
| Mass difference | −0.02888 u |
| Q-value | −26.9 MeV (endothermic) |
Conclusion: Q = −26.9 MeV is negative, so the fission of ⁵⁶Fe into two ²⁸Al nuclei is not energetically possible.
The fission properties of ²³⁹₉₄Pu are very similar to those of ²³⁵₉₂U. 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 ²³⁹₉₄Pu undergo fission?
Concept and context. Each fission of a plutonium-239 nucleus releases about 180 MeV (compare ≈ 200 MeV for ²³⁵U). The total energy from 1 kg is the energy per fission multiplied by the number of nuclei (atoms) in 1 kg, where the number of atoms follows from Avogadro’s number: N = (m/M) N_A, with the molar mass of ²³⁹Pu ≈ 239 g mol⁻¹. This shows the great energy density of nuclear fuel: about a million times that of chemical fuels.
Given: m = 1 kg = 1000 g; molar mass = 239 g mol⁻¹; N_A = 6.023 × 10²³ mol⁻¹; energy per fission = 180 MeV.
- Number of atoms: N = (1000/239) × 6.023 × 10²³ = 4.184 × 6.023 × 10²³ = 2.52 × 10²⁴ atoms.
- Total energy: E = N × 180 MeV = 2.52 × 10²⁴ × 180 = 4.5 × 10²⁶ MeV.
- In joules: 4.5 × 10²⁶ × 1.6 × 10⁻¹³ = 7.3 × 10¹³ J.
E = (1000/239) N_A × 180 MeV = 4.5 × 10²⁶ MeV ≈ 7.3 × 10¹³ J
| Quantity | Value |
|---|---|
| Atoms in 1 kg ²³⁹Pu | 2.52 × 10²⁴ |
| Energy per fission | 180 MeV |
| Total energy | 4.5 × 10²⁶ MeV |
| In joules | 7.3 × 10¹³ J |
| Equivalent mass of coal (3 × 10⁷ J/kg) | ≈ 2.4 × 10⁶ kg (2400 tonnes) |
- Comparison (NCERT): the fission of 1 kg of uranium generates about 10¹⁴ J, while burning 1 kg of coal gives about 10⁷ J — a nuclear fuel yields a million times more energy per unit mass.
- Fraction of mass converted: 7.3 × 10¹³ J/c² = 0.81 g — about 0.08 % of the mass of 1 kg is converted into energy.
Conclusion: about 4.5 × 10²⁶ MeV (≈ 7.3 × 10¹³ J) of energy is released when 1 kg of ²³⁹Pu undergoes fission.
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 ²₁H + ²₁H → ³₂He + n + 3.27 MeV.
Concept and context. In the fusion reaction ²H + ²H → ³He + n + 3.27 MeV, two deuterium nuclei react to release 3.27 MeV. The number of reactions that 2.0 kg of deuterium can undergo is therefore half the number of deuterium atoms. The total energy released divided by the power consumed by the lamp (100 W = 100 J s⁻¹) gives the time the lamp can be kept glowing. (1 year = 3.154 × 10⁷ s; 1 MeV = 1.6 × 10⁻¹³ J; N_A = 6.023 × 10²³ per mole.)
- Number of deuterium atoms: molar mass of ²H ≈ 2 g: 2.0 kg = 2000 g = 1000 mol ⇒ N = 1000 × 6.023 × 10²³ = 6.023 × 10²⁶ atoms.
- Number of fusion reactions (two nuclei per reaction): n = N/2 = 3.01 × 10²⁶.
- Energy released: E = n × 3.27 MeV = 3.01 × 10²⁶ × 3.27 = 9.85 × 10²⁶ MeV = 9.85 × 10²⁶ × 1.6 × 10⁻¹³ J = 1.58 × 10¹⁴ J.
- Time: t = E/P = (1.58 × 10¹⁴ J)/(100 W) = 1.58 × 10¹² s = 1.58 × 10¹²/3.154 × 10⁷ y = 5.0 × 10⁴ years.
t = E/P ≈ 1.6 × 10¹² s ≈ 5 × 10⁴ years
| Quantity | Value |
|---|---|
| Deuterium atoms in 2.0 kg | 6.0 × 10²⁶ |
| Number of reactions | 3.0 × 10²⁶ |
| Energy released | 9.85 × 10²⁶ MeV = 1.58 × 10¹⁴ J |
| Lamp power | 100 W |
| Duration | 1.58 × 10¹² s ≈ 5.0 × 10⁴ years |
- Perspective: a 100 W lamp could burn for fifty thousand years on 2 kg of deuterium — which shows the potential of controlled fusion (deuterium is abundant in sea water: about 1 in 6000 hydrogen atoms).
- Practical difficulties: fusion requires temperatures of ~10⁸ K to overcome the Coulomb barrier and confinement of the plasma (NCERT Section 13.7.3).
- (Using 3.27 MeV per pair of nuclei gives 4.9–5.0 × 10⁴ years depending on rounding.)
Conclusion: the lamp can be kept glowing for about 5.0 × 10⁴ years (≈ 1.6 × 10¹² s).
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 fm.)
Concept and context. For two nuclei to fuse they must come close enough (a few femtometres) for the short-range attractive nuclear force to dominate; but both are positively charged and repel each other through the long-range Coulomb force, so their electrostatic potential energy increases as they approach. This Coulomb barrier must be overcome (NCERT Section 13.7.2: ≈ 400 keV for two protons); its height is the Coulomb potential energy at the separation where the two nuclei just touch: U = (1/4πε₀)(q₁q₂/r), with r = sum of the radii. This barrier height determines the temperature required for thermonuclear fusion.
Given: each deuteron has charge +e = 1.6 × 10⁻¹⁹ C and radius 2.0 fm; at contact the centre-to-centre separation r = 2 × 2.0 fm = 4.0 fm = 4.0 × 10⁻¹⁵ m; 1/(4πε₀) = 9 × 10⁹ N m² C⁻².
- Barrier height: U = (1/4πε₀)(e × e)/r = (9 × 10⁹)(1.6 × 10⁻¹⁹)²/(4.0 × 10⁻¹⁵).
- (1.6 × 10⁻¹⁹)² = 2.56 × 10⁻³⁸; numerator = 9 × 10⁹ × 2.56 × 10⁻³⁸ = 2.304 × 10⁻²⁸ J m; divided by 4.0 × 10⁻¹⁵ m ⇒ U = 5.76 × 10⁻¹⁴ J.
- In MeV: U = 5.76 × 10⁻¹⁴/1.6 × 10⁻¹³ = 0.36 MeV = 360 keV.
U = e²/(4πε₀ × 2R) = 5.8 × 10⁻¹⁴ J = 360 keV
| Quantity | Value |
|---|---|
| Separation at contact | 4.0 fm |
| Barrier height | 5.76 × 10⁻¹⁴ J |
| In keV | 360 keV |
| Equivalent temperature (3/2)kT = U/2 per particle… see remark | ≈ 10⁹ K |
- Temperature estimate: if each deuteron must have the kinetic energy of half the barrier (so that their relative kinetic energy equals the barrier in a head-on collision) then (3/2)kT ≈ 180 keV ⇒ T = (2/3)(1.8 × 10⁵ × 1.6 × 10⁻¹⁹)/(1.38 × 10⁻²³) ≈ 1.4 × 10⁹ K (NCERT estimates ~3 × 10⁹ K for protons using 400 keV) — much higher than the Sun’s core 1.5 × 10⁷ K; fusion in the Sun proceeds via the high-energy tail of the distribution and via quantum-mechanical tunnelling through the barrier.
- Fusion reactors aim at ≈ 10⁸ K with plasma confinement (tokamak).
Conclusion: the height of the potential barrier for a head-on collision of two deuterons is about 0.36 MeV (360 keV).
From the relation R = R₀A^(1/3), where R₀ 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).
Concept and context. The empirical formula for the radius of a nucleus is R = R₀A^(1/3) with R₀ = 1.2 fm, obtained from electron-scattering experiments. The density of nuclear matter is the mass of the nucleus divided by its volume. The mass of a nucleus of mass number A is approximately A atomic mass units (A u, with 1 u = 1.66 × 10⁻²⁷ kg; the mass of a proton or neutron ≈ 1 u), and its volume (taking the nucleus as a sphere) is (4/3)πR³. Substituting the empirical radius shows that A cancels — the density is independent of A.
Proof
- Volume: V = (4/3)πR³ = (4/3)π(R₀A^(1/3))³ = (4/3)πR₀³ A.
- Mass: M = A u (mass number × atomic mass unit).
- Density: ρ = M/V = A u/[(4/3)πR₀³ A] = 3u/(4πR₀³) — the factor A cancels. Hence ρ is a constant, the same for all nuclei.
Numerical value (R₀ = 1.2 × 10⁻¹⁵ m; u = 1.66 × 10⁻²⁷ kg):
ρ = 3 × 1.66 × 10⁻²⁷/[4π × (1.2 × 10⁻¹⁵)³] = 4.98 × 10⁻²⁷/(2.17 × 10⁻⁴⁴) ≈ 2.3 × 10¹⁷ kg m⁻³
| Nucleus | A | R = 1.2A^(1/3) (fm) | Density (kg m⁻³) |
|---|---|---|---|
| ¹⁶O | 16 | 3.0 | 2.3 × 10¹⁷ |
| ⁵⁶Fe | 56 | 4.6 | 2.3 × 10¹⁷ |
| ¹⁹⁷Au | 197 | 7.0 | 2.3 × 10¹⁷ |
| ²³⁸U | 238 | 7.4 | 2.3 × 10¹⁷ |
- Meaning: nuclei behave like drops of an incompressible liquid of constant density 2.3 × 10¹⁷ kg m⁻³, about 2.3 × 10¹⁴ times the density of water (10³ kg m⁻³); a matchbox of nuclear matter would have a mass of ~ 10¹¹ kg. The matter in neutron stars has a comparable density.
- Link with the nuclear force: a constant density means each nucleon occupies the same volume (≈ (4/3)πR₀³), which reflects the short range and saturation of the nuclear force: nucleons cannot be squeezed beyond a certain proximity (repulsive core below 0.8 fm).
Conclusion: since ρ = 3u/(4πR₀³) is independent of A, the nuclear matter density is the same (≈ 2.3 × 10¹⁷ kg m⁻³) for all nuclei.