Class 12 Physics NCERT REPRINT 2026-27

Chapter 13: Nuclei

Complete concise study notes, atomic masses, discovery of neutron, nuclear radius and universal density, mass defect and binding energy curve, strong nuclear forces, nuclear fission, fusion in stars, solved examples 13.1–13.4, and exercises 13.1–13.10.

01 / Exam-Focused Notes

Chapter 13: Nuclei

Complete official NCERT textbook coverage with high-precision vector graphs: Unified Atomic Mass Unit ($1\text{ u} = 931.5\text{ MeV}/c^2$), Discovery of Neutron, Nuclear Radius ($R = R_0 A^{1/3}$) & Constant Density ($\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$), Mass Defect ($\Delta M$), Binding Energy Curve ($E_{bn}$), Characteristics of Nuclear Forces, Nuclear Fission ($Q \approx 200\text{ MeV}$), and Solar Thermonuclear Fusion ($p-p$ cycle).

100% SYLLABUS
Chapter Navigation • NCERT Sections 13.1 – 13.7
13.1-13.2Nucleus Composition 13.3Size & Density 13.4-13.5Binding Energy & Force 13.6-13.7Fission & Fusion
13.1 & 13.2
NCERT Sections

Atomic Masses, Composition of Nucleus & Discovery of Neutron

Nuclear Constituents & Units

13.2.1 Unified Atomic Mass Unit ($u$), Isotopes, Isobars, and Isotones

  • Unified Atomic Mass Unit ($u$): Defined as exactly $1/12^{\text{th}}$ of the mass of an unbound neutral Carbon-12 ($^{12}\text{C}$) atom at rest: $$1\text{ u} = \frac{1.992647 \times 10^{-26}\text{ kg}}{12} = 1.660539 \times 10^{-27}\text{ kg}$$ $$1\text{ u} \equiv 931.5\text{ MeV}/c^2 \quad (E = 931.5\text{ MeV})$$
  • Constituents of Nucleus (Nucleons):
    • Proton ($p$): Charge $= +e = +1.602 \times 10^{-19}\text{ C}$, Mass $m_p = 1.00727\text{ u} = 1.67262 \times 10^{-27}\text{ kg}$.
    • Neutron ($n$): Discovered by James Chadwick (1932) by bombarding Beryllium with $\alpha$-particles. Neutral ($q=0$), Mass $m_n = 1.00866\text{ u} = 1.6749 \times 10^{-27}\text{ kg}$. Free neutron is unstable ($\tau \approx 1000\text{ s}$, decays to $p + e^- + \bar{\nu}_e$), but stable inside bound nuclei.
  • Nuclide Classification ($^A_Z\text{X}$):
    • Isotopes: Same atomic number $Z$, different neutron number $N$ and mass number $A$ (e.g. $^1_1\text{H}, ^2_1\text{H}, ^3_1\text{H}$; $^{35}_{17}\text{Cl}, ^{37}_{17}\text{Cl}$). Same chemical properties.
    • Isobars: Same mass number $A$, different $Z$ and $N$ (e.g. $^3_1\text{H}$ and $^3_2\text{He}$; $^{40}_{18}\text{Ar}, ^{40}_{20}\text{Ca}$).
    • Isotones: Same neutron number $N = A - Z$, different $Z$ and $A$ (e.g. $^{198}_{\ 80}\text{Hg}$ and $^{197}_{\ 79}\text{Au}$ where $N = 118$).
13.3
NCERT Section

Size and Constant Density of the Nucleus

Nuclear Radius & Density

13.3.1 Radius Formula ($R = R_0 A^{1/3}$) & Universal Nuclear Density

  • Nuclear Radius Law: High-energy electron scattering measurements reveal: $$R = R_0 A^{1/3} \quad \text{where } R_0 \approx 1.2\text{ fm} = 1.2 \times 10^{-15}\text{ m}$$
  • Nuclear Volume & Mass Proportionality: $$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A \implies V \propto A$$
  • Universal Nuclear Density ($\rho$): $$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m}{4\pi R_0^3} \approx 2.3 \times 10^{17}\text{ kg m}^{-3}$$ (Remarkable Fact: Nuclear density is completely constant and independent of mass number $A$ for all elements; comparable to the density of astrophysical neutron stars).
Example 13.1

Nuclear Density Calculation of Iron ($^{56}\text{Fe}$)

Given the mass of iron nucleus $m_{\text{Fe}} = 55.85\text{ u}$ and $A = 56$, compute the nuclear density.

Mass of nucleus $M = 55.85 \times (1.66 \times 10^{-27}\text{ kg}) = 9.27 \times 10^{-26}\text{ kg}$.
Nuclear radius $R = R_0 A^{1/3} = (1.2 \times 10^{-15}\text{ m}) \times (56)^{1/3} = 4.59 \times 10^{-15}\text{ m}$.
$$\text{Density } \rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{9.27 \times 10^{-26}}{\frac{4}{3}\pi (1.2 \times 10^{-15})^3 \times 56} = 2.29 \times 10^{17}\text{ kg m}^{-3}$$
Nuclear density $\rho = 2.29 \times 10^{17}\text{ kg/m}^3$ ($2.3 \times 10^{14}$ times denser than water).
13.4 & 13.5
NCERT Sections

Mass-Energy, Binding Energy & Nuclear Forces

Einstein Relation & Mass Defect

13.4.1 Mass-Energy Equivalence ($E = \Delta M c^2$) & Binding Energy ($E_b$)

  • Einstein's Mass-Energy Relation: $E = m c^2$ ($1\text{ g}$ matter converts into $9 \times 10^{13}\text{ J}$).
  • Mass Defect ($\Delta M$): The difference between the sum of masses of free individual nucleons and the rest mass $M$ of the bound nucleus: $$\Delta M = [Z m_p + (A - Z) m_n] - M_N$$
  • Total Nuclear Binding Energy ($E_b$): $$E_b = \Delta M \times c^2 = (\Delta M\text{ in u}) \times 931.5\text{ MeV}$$
  • Binding Energy Per Nucleon ($E_{bn} = E_b / A$):
    Measures the stability of a nucleus against disassembly:
    • Practically constant ($\approx 8.0\text{ MeV/nucleon}$) across $30 < A < 170$ due to the saturation property of short-range nuclear forces.
    • Peak maximum at $^{56}\text{Fe}$ ($E_{bn} \approx 8.75\text{ MeV/nucleon}$, most stable nuclide).
    • Decreases for heavy nuclei ($E_{bn} \approx 7.6\text{ MeV}$ for $^{238}\text{U}$) due to Coulomb repulsion $\implies$ Fission is exothermic.
    • Low for very light nuclei ($A < 30$, $E_{bn} \approx 1.1\text{ MeV}$ for $^2\text{H}$) $\implies$ Fusion is exothermic.
Fig 13.1: Binding Energy per Nucleon (E_bn) vs Mass Number (A)
Mass Number A → E_bn (MeV / nucleon) → 0 2 4 6 8 Plateau Region: E_bn ≈ 8.0 MeV (30 < A < 170) ²H (1.11 MeV) ⁴He (7.07) ⁵⁶Fe Peak (8.75 MeV) ²³⁸U (7.6 MeV) FUSION ↗ ↖ FISSION
Fundamental Strong Interaction

13.5 Characteristics of the Nuclear Force

  1. Strongest Fundamental Force: $\sim 100$ times stronger than Coulomb electrostatic force, easily overcoming proton-proton repulsion inside the nucleus.
  2. Extremely Short-Range: Operates only over a few femtometres ($\sim 1\text{ to } 3\text{ fm}$), dropping rapidly to zero for $r > 3\text{ fm}$.
  3. Saturation Property: A nucleon interacts only with its immediate nearest neighbours.
  4. Repulsive Core at Ultra-Short Range: Attractive for $r > r_0 \approx 0.8\text{ fm}$, but becomes strongly repulsive for $r < 0.8\text{ fm}$, preventing nuclear collapse.
  5. Charge Independence: Force between $(p-p)$, $(p-n)$, and $(n-n)$ is identically strong. Non-central spin-dependent force.
Fig 13.2: Potential Energy (V) of Nucleon Pair vs Distance (r)
Separation r (fm) → Potential Energy V (MeV) V = 0 r₀ ≈ 0.8 fm (Minimum PE ≈ -100 MeV) REPULSIVE (r < r₀) ATTRACTIVE (r > r₀)
Example 13.2 & 13.3

Energy of 1 g Matter & Oxygen $^{16}\text{O}$ Binding Energy

(a) Calculate energy equivalent of $1\text{ g}$ substance.
(b) For $^{16}_{\ 8}\text{O}$ ($m = 15.99053\text{ u}$), find mass defect and total binding energy.

(a) $1\text{ g}$ mass energy: $E = mc^2 = (10^{-3}\text{ kg}) \times (3 \times 10^8\text{ m/s})^2 = 9.0 \times 10^{13}\text{ J}$.
(b) $^{16}\text{O}$ Mass Defect:
Constituents $= 8 m_p + 8 m_n = 8(1.00727) + 8(1.00866) = 16.12744\text{ u}$.
$\Delta M = 16.12744 - 15.99053 = 0.13691\text{ u}$.
$E_b = 0.13691 \times 931.5\text{ MeV} = 127.5\text{ MeV}$.
$E_{bn} = \frac{127.5\text{ MeV}}{16} = 7.97\text{ MeV/nucleon}$.
13.6 & 13.7
NCERT Sections

Radioactivity, Nuclear Fission & Thermonuclear Fusion

Nuclear Fission Mechanics

13.7.1 Nuclear Fission ($Q \approx 200\text{ MeV}$)

  • Fission: A heavy unstable nucleus ($A \sim 240$) splits into two medium-mass fragments ($A \sim 120$) upon absorbing a thermal neutron: $$^1_0\text{n} + {^{235}_{\ 92}\text{U}} \to {^{236}_{\ 92}\text{U}}^* \to {^{144}_{\ 56}\text{Ba}} + {^{89}_{36}\text{Kr}} + 3\ ^1_0\text{n} + Q$$
  • Energy Released: $\Delta E_{bn} = 8.5\text{ MeV} - 7.6\text{ MeV} = 0.9\text{ MeV/nucleon} \implies Q \approx 240 \times 0.9\text{ MeV} \approx 216\text{ MeV} \approx 200\text{ MeV/fission}$.
  • Energy Density: $1\text{ kg}$ $^{235}\text{U}$ yields $\sim 10^{14}\text{ J}$ ($\sim 2.5 \times 10^6$ times more than burning $1\text{ kg}$ coal).
Thermonuclear Fusion & Solar Cycles

13.7.2 Thermonuclear Fusion & Proton-Proton Cycle in Stars

  • Thermonuclear Fusion: Two light nuclei ($A \le 10$) combine at extremely high temperatures ($T \sim 10^7 - 10^8\text{ K}$) to overcome the Coulomb electrostatic barrier ($\sim 400\text{ keV}$).
  • Proton-Proton ($p-p$) Cycle in the Sun:
    $$4\ ^1_1\text{H} + 2\text{e}^- \to {^4_2\text{He}} + 2\nu_e + 6\gamma + 26.7\text{ MeV}$$
  • $4$ Hydrogen atoms fuse into one Helium nucleus with release of $26.7\text{ MeV}$ ($6.675\text{ MeV/nucleon}$).
Example 13.4

Conservation Laws in Nuclear vs Chemical Reactions

(a) How are nuclear reaction equations balanced compared to chemical equations?
(b) If proton and neutron numbers are conserved, where does mass-energy come from?
(c) Does mass-energy interconversion occur in chemical reactions?

(a) Chemical equations balance atoms of each element. Nuclear reactions transmute elements, but strictly conserve total charge ($Z$) and total nucleon/baryon number ($A$).
(b) The rest mass of free nucleons is identical on both sides, but the total nuclear binding energy of reactants differs from products. The difference in binding energies ($\Delta E_b = \Delta M c^2$) appears as kinetic energy/photons.
(c) Yes, chemical binding energy also creates a mass defect. However, chemical energy releases ($\sim \text{eV}$) involve mass changes a million times smaller than nuclear processes ($\sim \text{MeV}$), making chemical mass changes undetectable in ordinary measurements.
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering atomic mass units, nuclear density, mass defect, binding energy curve, nuclear force properties, fission, and fusion with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 13.1 – 13.10

Complete stepwise solutions for every exercise question in NCERT Class 12 Physics Chapter 13 Nuclei Reprint 2026-27.

10 QUESTIONS
Ex 13.1Binding Energy of Nitrogen-14

13.1 Obtain the binding energy (in MeV) of a nitrogen nucleus ($^{14}_{\ 7}\text{N}$), given $m(^{14}_{\ 7}\text{N}) = 14.00307\text{ u}$, $m_H = 1.007825\text{ u}$, $m_n = 1.008665\text{ u}$.

Binding Energy $E_b = 104.66\text{ MeV}$ ($E_{bn} = 7.48\text{ MeV/nucleon}$).
Constituent mass: $$7 m_H + 7 m_n = 7(1.007825) + 7(1.008665) = 7.054775 + 7.060655 = 14.115430\text{ u}$$ $$\Delta M = 14.115430 - 14.003070 = 0.112360\text{ u}$$ $$E_b = 0.112360 \times 931.5\text{ MeV} = 104.66\text{ MeV}$$
Ex 13.2Binding Energy of Iron-56 and Bismuth-209

13.2 Obtain the binding energy of $^{56}_{26}\text{Fe}$ ($m = 55.934939\text{ u}$) and $^{209}_{\ 83}\text{Bi}$ ($m = 208.980388\text{ u}$).

$^{56}\text{Fe}$: $E_b = 492.26\text{ MeV}$ ($E_{bn} = 8.79\text{ MeV/nucleon}$)
$^{209}\text{Bi}$: $E_b = 1640.3\text{ MeV}$ ($E_{bn} = 7.85\text{ MeV/nucleon}$).
For $^{56}_{26}\text{Fe}$:
$\Delta M = [26(1.007825) + 30(1.008665)] - 55.934939 = 56.463400 - 55.934939 = 0.528461\text{ u}$.
$E_b = 0.528461 \times 931.5\text{ MeV} = 492.26\text{ MeV} \implies E_{bn} = \frac{492.26}{56} = 8.79\text{ MeV/nucleon}$.
For $^{209}_{\ 83}\text{Bi}$:
$\Delta M = [83(1.007825) + 126(1.008665)] - 208.980388 = 210.741265 - 208.980388 = 1.760877\text{ u}$.
$E_b = 1.760877 \times 931.5\text{ MeV} = 1640.26\text{ MeV} \implies E_{bn} = \frac{1640.26}{209} = 7.85\text{ MeV/nucleon}$.
Ex 13.3Disassembly Energy of a 3.0 g Copper Coin

13.3 A coin has mass $3.0\text{ g}$. Calculate nuclear energy required to separate all nucleons from each other, assuming coin is pure $^{63}_{29}\text{Cu}$ ($m = 62.92960\text{ u}$).

Total Energy Required $E_{\text{total}} = 1.58 \times 10^{25}\text{ MeV} = 2.53 \times 10^{12}\text{ J}$ (equivalent to $\sim 700\text{ MWh}$).
Number of Cu atoms in $3.0\text{ g}$: $$N = \frac{3.0\text{ g}}{63\text{ g/mol}} \times (6.023 \times 10^{23}\text{ mol}^{-1}) = 2.868 \times 10^{22}\text{ atoms}$$
Mass defect per $^{63}\text{Cu}$ nucleus: $$\Delta M = [29(1.007825) + 34(1.008665)] - 62.92960 = 63.521535 - 62.92960 = 0.591935\text{ u}$$ $$E_b = 0.591935 \times 931.5\text{ MeV} = 551.39\text{ MeV}$$
$$E_{\text{total}} = N \times E_b = (2.868 \times 10^{22}) \times (551.39\text{ MeV}) = 1.58 \times 10^{25}\text{ MeV} = 2.53 \times 10^{12}\text{ J}$$
Ex 13.4Ratio of Nuclear Radii (Au / Ag)

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

Radius Ratio $\frac{R_{\text{Au}}}{R_{\text{Ag}}} = 1.225 \approx 1.23$.
$$R \propto A^{1/3} \implies \frac{R_{\text{Au}}}{R_{\text{Ag}}} = \left(\frac{197}{107}\right)^{1/3} = (1.8411)^{1/3} = 1.225$$
Ex 13.5Q-Values of Nuclear Reactions

13.5 Determine Q-values and state whether reactions are exothermic or endothermic:
(i) $^1_1\text{H} + {^3_1\text{H}} \to {^2_1\text{H}} + {^2_1\text{H}}$
(ii) $^{12}_{\ 6}\text{C} + {^{12}_{\ 6}\text{C}} \to {^{20}_{10}\text{Ne}} + {^4_2\text{He}}$
Given: $m(^2\text{H}) = 2.014102\text{ u}, m(^3\text{H}) = 3.016049\text{ u}, m(^{12}\text{C}) = 12.000000\text{ u}, m(^{20}\text{Ne}) = 19.992439\text{ u}, m(^4\text{He}) = 4.002603\text{ u}$.

(i) $Q = -4.03\text{ MeV}$ (Endothermic) • (ii) $Q = +4.61\text{ MeV}$ (Exothermic).
(i) $\Delta m = [m(^1\text{H}) + m(^3\text{H})] - 2m(^2\text{H}) = [1.007825 + 3.016049] - 2(2.014102) = 4.023874 - 4.028204 = -0.004330\text{ u}$.
$Q = -0.004330 \times 931.5\text{ MeV} = -4.033\text{ MeV} \implies$ Endothermic.
(ii) $\Delta m = 2m(^{12}\text{C}) - [m(^{20}\text{Ne}) + m(^4\text{He})] = 24.000000 - [19.992439 + 4.002603] = 24.000000 - 23.995042 = +0.004958\text{ u}$.
$Q = +0.004958 \times 931.5\text{ MeV} = +4.618\text{ MeV} \implies$ Exothermic.
Ex 13.6Fission of Iron-56 into Aluminum-28

13.6 Is the symmetric fission of $^{56}_{26}\text{Fe} \to 2\ {^{28}_{13}\text{Al}}$ energetically possible? Work out $Q$. ($m(^{56}\text{Fe}) = 55.93494\text{ u}, m(^{28}\text{Al}) = 27.98191\text{ u}$).

$Q = -26.90\text{ MeV} < 0 \implies$ Fission of Iron-56 is energetically impossible (highly endothermic).
$$\Delta m = m(^{56}\text{Fe}) - 2 m(^{28}\text{Al}) = 55.93494 - 2(27.98191) = 55.93494 - 55.96382 = -0.02888\text{ u}$$ $$Q = -0.02888 \times 931.5\text{ MeV} = -26.90\text{ MeV}$$ Since $Q < 0$, $^{56}\text{Fe}$ is at the peak of the binding energy curve and cannot spontaneously undergo fission.
Ex 13.7Energy Released in 1 kg Plutonium-239 Fission

13.7 Energy released per fission of $^{239}_{\ 94}\text{Pu}$ is $180\text{ MeV}$. How much energy (in MeV and Joules) is released if all atoms in $1\text{ kg}$ of pure $^{239}\text{Pu}$ undergo fission?

Energy Released $E = 4.53 \times 10^{26}\text{ MeV} = 7.26 \times 10^{13}\text{ J}$.
Number of $^{239}\text{Pu}$ nuclei in $1\text{ kg} = 1000\text{ g}$: $$N = \frac{1000\text{ g}}{239\text{ g/mol}} \times (6.023 \times 10^{23}\text{ mol}^{-1}) = 2.520 \times 10^{24}\text{ nuclei}$$
$$E = N \times (180\text{ MeV}) = (2.520 \times 10^{24}) \times 180 = 4.536 \times 10^{26}\text{ MeV}$$ $$E = (4.536 \times 10^{26}) \times (1.602 \times 10^{-13}\text{ J/MeV}) = 7.267 \times 10^{13}\text{ J}$$
Ex 13.8Glow Time of 100 W Lamp with 2.0 kg Deuterium

13.8 How long can a $100\text{ W}$ electric lamp be kept glowing by fusion of $2.0\text{ kg}$ deuterium? Reaction: $^2_1\text{H} + {^2_1\text{H}} \to {^3_2\text{He}} + \text{n} + 3.27\text{ MeV}$.

Time $t = 1.57 \times 10^{12}\text{ s} \approx 4.98 \times 10^4\text{ years} \approx 50,000\text{ years}$.
Number of deuterium nuclei in $2.0\text{ kg} = 2000\text{ g}$: $$N = \frac{2000}{2} \times 6.023 \times 10^{23} = 6.023 \times 10^{26}\text{ nuclei}$$ Two deuterons release $3.27\text{ MeV} \implies$ energy per deuteron $= \frac{3.27}{2} = 1.635\text{ MeV}$.
Total Energy: $$E = (6.023 \times 10^{26}) \times (1.635\text{ MeV}) = 9.848 \times 10^{26}\text{ MeV} = (9.848 \times 10^{26}) \times (1.602 \times 10^{-13}\text{ J}) = 1.578 \times 10^{14}\text{ J}$$
Time to consume at $P = 100\text{ W}$: $$t = \frac{E}{P} = \frac{1.578 \times 10^{14}\text{ J}}{100\text{ J/s}} = 1.578 \times 10^{12}\text{ s}$$ $$\text{In years: } \frac{1.578 \times 10^{12}\text{ s}}{3.154 \times 10^7\text{ s/yr}} \approx 50,030\text{ years} \approx 5.0 \times 10^4\text{ years}$$
Ex 13.9Coulomb Potential Barrier for Deuteron-Deuteron Fusion

13.9 Calculate the height of the Coulomb potential barrier for head-on collision of two deuterons (treated as hard spheres of radius $r = 2.0\text{ fm}$).

Potential Barrier Height $U = 360\text{ keV} = 5.76 \times 10^{-14}\text{ J}$.
At contact, separation distance $d = 2r = 2(2.0\text{ fm}) = 4.0\text{ fm} = 4.0 \times 10^{-15}\text{ m}$. $$U = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{d} = \frac{(9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (1.602 \times 10^{-19}\text{ C})^2}{4.0 \times 10^{-15}\text{ m}} = 5.77 \times 10^{-14}\text{ J}$$ $$U = \frac{5.77 \times 10^{-14}\text{ J}}{1.602 \times 10^{-16}\text{ J/keV}} = 360\text{ keV}$$
Ex 13.10Independence of Nuclear Matter Density from Mass Number A

13.10 From the relation $R = R_0 A^{1/3}$, show that nuclear matter density is nearly constant (independent of $A$).

Let $m$ be the average mass of a nucleon ($m \approx 1.66 \times 10^{-27}\text{ kg}$).
Total mass of a nucleus of mass number $A$ is $M = m A$.
Radius of the nucleus is $R = R_0 A^{1/3} \implies$ Volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$.
$$\text{Nuclear Density } \rho = \frac{M}{V} = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m}{4\pi R_0^3}$$ Since $m$, $\pi$, and $R_0$ are fundamental physical constants, the mass number $A$ cancels out completely. Hence, the nuclear matter density $\rho \approx 2.3 \times 10^{17}\text{ kg m}^{-3}$ is constant across all chemical elements.
04 / Rapid Reference

Chapter Summary & Points to Ponder

Core nuclear properties, radius and density relations, binding energy curve characteristics, fission/fusion equations, and official NCERT Points to Ponder for Chapter 13.

100% SYLLABUS

Universal Nuclear Density

R = R_0 A^{1/3} \implies \rho \approx 2.3 \times 10^{17}\text{ kg/m}^3

Nuclear volume is proportional to mass number $A$; density is constant for all nuclei regardless of size.

Mass Defect & 1 u Energy

1\text{ u} = 931.5\text{ MeV} • \Delta M c^2 = E_b

$\Delta M = [Z m_p + (A - Z) m_n] - M_N$. Released energy upon formation is binding energy $E_b$.

Binding Energy Curve

Peak at ^{56}\text{Fe} (8.75\text{ MeV/A})

Constancy at $\approx 8.0\text{ MeV}$ ($30 < A < 170$) due to force saturation. Drops for light ($A < 30$) and heavy ($A > 170$) nuclei.

Nuclear Force Properties

Short Range (\sim 1-3\text{ fm}) • Repulsive (<0.8\text{ fm})

Strongest fundamental force, charge-independent, saturating, spin-dependent, non-central.

Nuclear Fission

Q \approx 200\text{ MeV / fission}

$^{235}\text{U} + \text{n} \to {^{144}\text{Ba}} + {^{89}\text{Kr}} + 3\text{n} + 200\text{ MeV}$. $1\text{ kg}$ uranium $\sim 10^{14}\text{ J}$.

Thermonuclear Fusion

4\ ^1\text{H} \to {^4\text{He}} + 26.7\text{ MeV}

Solar energy source ($p-p$ cycle). Requires $T \sim 10^7 - 10^8\text{ K}$ to overcome $\approx 400\text{ keV}$ Coulomb barrier.

NCERT Official

Points to Ponder

  1. Universal Density: Nuclear density is completely constant and independent of the size/mass of the nucleus ($\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$). Atomic density does not obey this rule.
  2. Probe Dependency of Nuclear Size: Nuclear radius measured by high-energy electrons senses charge distribution ($Z$), whereas $\alpha$-particles sense the nuclear strong matter distribution.
  3. Unified Mass-Energy Conservation: Due to $E = mc^2$, mass and energy are unified. The $Q$-value is defined directly as $Q = (\sum M_{\text{reactants}} - \sum M_{\text{products}}) c^2$.
  4. Coulomb Barrier in Fusion: Light nuclei require extreme thermal kinetic energy ($T \sim 10^8\text{ K}$) to penetrate the mutual electrostatic repulsion barrier ($\sim 400\text{ keV}$) before the attractive strong force binds them.
  5. Neutron-to-Proton Ratio and Stability: Stable light nuclei have $N/Z \approx 1$. For heavy nuclei, repulsive Coulomb force necessitates $N/Z \approx 1.5$ (3:2) for stability. Only $\sim 10\%$ of all known nuclides are stable.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

24 QUESTIONS
Level 1 Active • 8 Questions