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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 08 (Atoms & Nuclei)

Chapter 13: Nuclei

Master Revision Dossier • Nuclear Radius, Constant Density Proof, Mass Defect, Binding Energy Curve, Nuclear Forces, Fission & Fusion

1. Definitive Formula & Nuclear Parameters Matrix

Concept Mathematical Formulation Proportionality & Values Boundary Conditions & Notes
Nuclear Radius $$R = R_0 A^{1/3}$$ $$R_0 \approx 1.2\text{ fm} = 1.2 \times 10^{-15}\text{ m}$$ Radius scales strictly with the cube root of mass number $A$. Volume is directly proportional to $A$ ($V \propto A$).
Nuclear Density $$\rho = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$$ $$\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$$ Mass number $A$ cancels out completely. Nuclear density is a universal constant for all elements (Density ratio of any two nuclei is $1:1$).
Mass Defect ($\Delta m$) $$\Delta m = [Z m_p + (A - Z) m_n] - M_{\text{nucleus}}$$ $$1\text{ u} \approx 931.5\text{ MeV}/c^2$$ Total constituent mass is always strictly greater than the rest mass of the bound nucleus.
Binding Energy $$E_b = \Delta m \times c^2, \quad E_{bn} = \frac{E_b}{A}$$ $$E_{bn,\max} \approx 8.75\text{ MeV} \text{ (}^{56}\mathrm{Fe}\text{)}$$ Stability is governed entirely by $E_{bn}$ (per nucleon), not by total $E_b$. Flat plateau ($E_{bn} \approx 8.5\text{ MeV}$) for $30 < A < 170$.
Thermal Fission ($^{235}\mathrm{U}$) $${}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3({}^{1}_{0}\mathrm{n}) + Q$$ $$Q \approx 200.6\text{ MeV per event}$$ Triggered by slow thermal neutrons ($0.025\text{ eV}$). Sustained controlled power requires neutron multiplication factor $k = 1$.
Solar Fusion ($p-p$ Cycle) $$4({}^{1}_{1}\mathrm{H}) + 2\mathrm{e}^- \to {}^{4}_{2}\mathrm{He} + 2\nu_e + 6\gamma + Q$$ $$Q \approx 26.7\text{ MeV per } \alpha\text{ formed}$$ Requires ultra-high temperatures ($T \sim 10^7\text{ K}$) to overcome mutual proton Coulomb barrier ($V_C \approx 400\text{ keV}$).
Specific Energy Ratio $$\text{Fusion} \approx 6.7\text{ MeV/u} \quad \text{vs} \quad \text{Fission} \approx 0.85\text{ MeV/u}$$ $$\text{Ratio} \approx 4 \text{ to } 7 \times$$ Per unit mass of fuel, thermonuclear fusion is vastly more energy-dense than heavy-element fission.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 13.1: Mathematical Proof of Invariant Nuclear Density ($\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$)

1. The radius $R$ of a nucleus with mass number $A$ is given by the empirical relation: $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}\text{ m}$.

2. Taking the nucleus to be a sphere of radius $R$, its volume is:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$

3. If $m$ is the average mass of a single nucleon ($m \approx 1.66 \times 10^{-27}\text{ kg}$), the total mass of the nucleus containing $A$ nucleons is $M = m A$.

4. Therefore, the nuclear density $\rho$ is:

$$\rho = \frac{M}{V} = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3} = \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.1416 \times (1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17}\text{ kg/m}^3$$

5. Because the mass number $A$ cancels completely from the expression, nuclear density is strictly constant and independent of the size or identity of the nucleus.

Blueprint 13.2: Explanation of Energy Release in Fission and Fusion from the $E_{bn}$ Curve

1. The binding energy per nucleon curve shows a broad maximum of $\approx 8.5\text{ MeV/nucleon}$ in the range $30 < A < 170$, peaking at $^{56}\mathrm{Fe}$ ($8.75\text{ MeV/nucleon}$).

2. For Heavy Nuclei ($A > 200$): In a nucleus like $^{240}\mathrm{X}$, $E_{bn} \approx 7.6\text{ MeV/nucleon}$. When it splits into two middle-weight fragments of $A \approx 120$ where $E_{bn} \approx 8.5\text{ MeV/nucleon}$:

$$\Delta E_{bn} = 8.5 - 7.6 = 0.9\text{ MeV/nucleon} \implies Q = A \times \Delta E_{bn} = 240 \times 0.9\text{ MeV} \approx 216\text{ MeV}$$

3. For Light Nuclei ($A \le 10$): Light nuclei like $^{2}\mathrm{H}$ have low $E_{bn} \approx 1.1\text{ MeV/nucleon}$. When they fuse to form $^{4}\mathrm{He}$ ($E_{bn} \approx 7.1\text{ MeV/nucleon}$), the binding energy jump is enormous ($\Delta E_{bn} \approx 6.0\text{ MeV/nucleon}$), liberating massive kinetic energy.

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Critical Conceptual Distinctions in Nuclear Physics

Physical Scenario Common Student Misconception Correct Scientific Law Exam Consequence
Nuclear Density Ratio Density depends on mass number: $\rho \propto A$. $\rho_1 : \rho_2 = 1 : 1$ (Strictly Invariant) Density ratio of $^{12}\mathrm{C}$ to $^{238}\mathrm{U}$ is exactly $1$.
Nuclear Stability Metric Total binding energy $E_b$ decides stability ($^{238}\mathrm{U}$ has $1800\text{ MeV}$). $E_{bn} = E_b / A$ decides stability $^{56}\mathrm{Fe}$ ($8.75\text{ MeV/u}$) is far more stable than $^{238}\mathrm{U}$ ($7.6\text{ MeV/u}$).
Nuclear Force below $0.8\text{ fm}$ Force remains attractive at all small distances. Switches to STRONGLY REPULSIVE for $r < 0.8\text{ fm}$ Repulsive core prevents the collapse of atomic nucleus.
Energy Output Comparison Fission is always more energetic than fusion. Fission wins per event; Fusion wins per kg Per kg of fuel, fusion yields $\approx 4\times$ to $7\times$ more energy than fission.

⚠️ Trap 2: Charge Independence of Strong Force

Nuclear force is strictly non-electromagnetic. At a distance of $\approx 1\text{ fm}$:
$$\mathbf{F_{pp} \approx F_{nn} \approx F_{pn}}$$ Coulomb repulsion between protons is an additional, separate electromagnetic force.

⚠️ Trap 3: Thermal Neutrons and Moderator Role

Fission of $^{235}\mathrm{U}$ has an extremely small cross-section for fast neutrons ($\approx 2\text{ MeV}$).
Heavy water ($D_2O$) or graphite slows neutrons to thermal energies ($\approx 0.025\text{ eV}$) via elastic scattering to maintain $k = 1$.

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Nuclear Radius & Density Ratio Exemplar 01

Given the mass numbers of two nuclei are in the ratio $27 : 125$. Find: (i) the ratio of their nuclear radii, and (ii) the ratio of their nuclear densities.

Solution: Given $A_1 / A_2 = 27 / 125$.

(i) Ratio of nuclear radii ($R = R_0 A^{1/3}$):

$$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5} = \mathbf{3 : 5}$$

(ii) Ratio of nuclear densities ($\rho = \frac{3m}{4\pi R_0^3}$):

$$\frac{\rho_1}{\rho_2} = \mathbf{1 : 1} \quad (\text{Strictly independent of mass number } A)$$
Problem 2: Q-Value Calculation in Alpha Decay Exemplar 02

Find the $Q$-value of the alpha decay: ${}^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}$. Given: $m({}^{238}_{92}\mathrm{U}) = 238.05079\text{ u}$, $m({}^{234}_{90}\mathrm{Th}) = 234.04363\text{ u}$, and $m({}^{4}_{2}\mathrm{He}) = 4.00260\text{ u}$.

Solution: Mass defect $\Delta m = m_{\text{reactants}} - m_{\text{products}}$:

$$\Delta m = 238.05079\text{ u} - [234.04363\text{ u} + 4.00260\text{ u}] = 238.05079\text{ u} - 238.04623\text{ u} = 0.00456\text{ u}$$
$$Q = \Delta m \times 931.5\text{ MeV} = 0.00456 \times 931.5\text{ MeV} \approx \mathbf{4.25\text{ MeV}}$$

Since $Q > 0$, the alpha decay is energetically spontaneous.

5. Printable Pocket Flashcard Grid

Radius Law

$R = R_0 A^{1/3}$

Nuclear Density

$\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$

Iron Peak

$^{56}\mathrm{Fe} = 8.75\text{ MeV/u}$

Conversion Unit

$1\text{ u} = 931.5\text{ MeV}$