Nuclear Physics & Radioactive Decay: Master Notes

Chapter Overview

Nuclear physics explores subatomic forces, mass-energy equivalence ($\mathrm{\Delta }E=\mathrm{\Delta }m{c}^2$), and radioactive kinetics. In IIT-JEE (Advanced), core topics include Binding Energy per Nucleon ($BE/A$) stability, $\alpha$-decay $Q$-value energy partitioning ($K_{\alpha }=Q\frac{A-4}{A}$), parallel and sequential radioactive decay series, and nuclear fission/fusion energetics.

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1. Nuclear Size, Density & Mass Defect

$$R=R_0{A}^{1/3}(R_0\approx 1.2\text{ fm}=1.2\times {10}^{-15}\text{ m})$$$$\text{Nuclear Volume: }V\propto R^3\propto A⟹\text{Nuclear Density }{\rho }_{\text{nuc}}\approx 2.3\times {10}^{17}{\text{ kg/m}}^3=\text{constant for all nuclei!}$$

1.1 Mass Defect ($\mathrm{\Delta }m$) & Binding Energy ($BE$)

$$\mathrm{\Delta }m=[Z{m}_p+(A-Z)m_n]-M_{\text{nucleus}}(A,Z)$$$$\boxed{{\displaystyle BE=\mathrm{\Delta }m\cdot c^2=(\mathrm{\Delta }m\text{ in amu})\times 931.5\text{ MeV}}}$$

Multi-Mode DiagramNuclear Physics: Binding Energy Curve & Radioactive Decay

Option 1: Publication-Grade Scientific Vector SVG

BE/A curve showing Fe-56 stability maximum (~8.75 MeV) and exponential decay law N(t) = N_0 * e^(-lambda*t).

Decay Law: $N(t) = N_0 e^{-\lambda t} = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$

Binding Energy: $BE = [Z m_p + (A-Z)m_n - M_{\text{nuc}}]c^2$

Option 2: Interactive 60 FPS Canvas Physics Engine

Radioactive decay population simulator.

Half-Life T₁/₂: 3 s

Decay Constant ($\lambda$):0.231 s⁻¹

Option 3: Interactive 3D Spatial Viewport

3D nuclear binding core model.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Decay Law:$N(t) = N_0 e^{-\lambda t}$

Option 5: Reactive Parametric Physics with Live Calculus

Half-life & remaining fraction calculator.

Half-Life T₁/₂: 4 yrs

Elapsed time t: 8 yrs

Remaining ($N/N_0$):$25.0$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

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2. Radioactive Decay Kinetics

$$-\frac{dN}{dt}=\lambda N(t)⟹\boxed{{\displaystyle N(t)=N_0{e}^{-\lambda t}}}$$$$\text{Activity: }A(t)=\lambda N(t)=A_0{e}^{-\lambda t}(1\text{ Curie (Ci)}=3.7\times {10}^{10}\text{ Bq/s})$$

2.1 Lifetime Relationships

• Half-Life ($T_{1/2}$): Time required for half of initial nuclei to decay:$$\boxed{{\displaystyle T_{1/2}=\frac{\ln 2}{\lambda }\approx \frac{0.693}{\lambda }}}$$
• Mean Life ($\tau$): Average lifetime of an active nucleus:$$\boxed{{\displaystyle \tau =\frac{1}{\lambda }=\frac{T_{1/2}}{\ln 2}\approx 1.443T_{1/2}}}$$

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2.2 Parallel & Simultaneous Decay Channels

<!-- Branch 1 (Daughter B) --> <line x1="110" y1="50" x2="220" y2="25" stroke="#10b981" stroke-width="2.5" marker-end="url(#arrow-green)"/> <rect x="220" y="10" width="160" height="32" rx="3" fill="var(--vp-c-bg)" stroke="#10b981" stroke-width="1.5"/> <text x="230" y="30" fill="#10b981" font-size="11" font-weight="700">Daughter B (&lambda;_1, T_1)</text> <!-- Branch 2 (Daughter C) --> <line x1="110" y1="70" x2="220" y2="95" stroke="#ef4444" stroke-width="2.5" marker-end="url(#arrow-red)"/> <rect x="220" y="80" width="160" height="32" rx="3" fill="var(--vp-c-bg)" stroke="#ef4444" stroke-width="1.5"/> <text x="230" y="100" fill="#ef4444" font-size="11" font-weight="700">Daughter C (&lambda;_2, T_2)</text> 

Figure Parallel Decay Channels: Total decay rate is the sum of independent branch rates (lambda_eff = lambda_1 + lambda_2).

$$\boxed{{\displaystyle {\lambda }_{\text{effective}}={\lambda }_1+{\lambda }_2⟹\frac{1}{T_{1/2,\text{eff}}}=\frac{1}{T_1}+\frac{1}{T_2}}}$$$$\text{Fraction decaying into Channel B: }{f}_B=\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2}$$

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3. Alpha Decay Energetics ($Q$-Value & Momentum Share)

$${}_Z^A\text{X}⟶{}_{Z-2}^{A-4}\text{Y}+{}_2^4\text{He}+Q$$$$Q=[M(X)-M(Y)-M(\alpha )]c^2$$

By conservation of linear momentum ($p_{\alpha }=p_Y$):

$$Q=K_{\alpha }+K_Y=\frac{p^2}{2m_{\alpha }}+\frac{p^2}{2m_Y}=K_{\alpha }(1+\frac{m_{\alpha }}{m_Y})=K_{\alpha }(1+\frac{4}{A-4})=K_{\alpha }\left(\frac{A}{A-4}\right)$$$$\boxed{{\displaystyle K_{\alpha }=Q\left(\frac{A-4}{A}\right),K_Y=Q\left(\frac{4}{A}\right)}}$$

(The lighter alpha particle carries away almost all ($\sim 98\mathrm{\%}$) of the decay energy!)

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 2) — Alpha Particle Kinetic Energy in Radium Decay

Question:
A stationary Radium nucleus ${}_{88}^{226}\text{Ra}$ undergoes $\alpha$-decay into Radon ${}_{86}^{222}\text{Rn}$. The total $Q$-value of the reaction is $Q=4.87\text{ MeV}$. Calculate the kinetic energy $K_{\alpha }$ of the emitted alpha particle.

Step-by-Step Solution:

Using the momentum conservation energy-split formula:

$$K_{\alpha }=Q\left(\frac{A-4}{A}\right)=4.87\text{ MeV}\times \left(\frac{226-4}{226}\right)=4.87\times \left(\frac{222}{226}\right)=\boxed{{\displaystyle 4.78\text{ MeV}}}$$$$K_{\text{recoil, Rn}}=Q-K_{\alpha }=4.87-4.78=\boxed{{\displaystyle 0.09\text{ MeV}=90\text{ keV}}}$$

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PYQ 2: JEE Advanced 2020 — Simultaneous Parallel Decay

Question:
A radioactive isotope decays simultaneously into two stable daughters with individual half-lives $T_1=12\text{ hours}$ and $T_2=6\text{ hours}$. Find:

1. The effective half-life $T_{\text{eff}}$ of the sample.
2. The time after which only $25\mathrm{\%}$ of the original parent isotope remains.

Step-by-Step Solution:

• 1. Effective Half-Life:

  $$\frac{1}{T_{\text{eff}}}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}=\frac{1}{4}⟹\boxed{{\displaystyle T_{\text{eff}}=4.0\text{ hours}}}$$

• 2. Time for $25\mathrm{\%}$ Remaining: $25\mathrm{\%}=\frac{1}{4}={\left(\frac{1}{2}\right)}^2⟹$ Exactly $2$ effective half-lives have elapsed!

  $$t=2\times T_{\text{eff}}=2\times 4.0=\boxed{{\displaystyle 8.0\text{ hours}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Nuclear Radius	$R=1.2A^{1/3}\text{ fm}$	Mass number $A$
Activity	$A=\lambda N=A_0{e}^{-\lambda t}$	$1\text{ Bq}=1\text{ decay/s}$
Parallel Decay Half-Life	$\frac{1}{T_{\text{eff}}}=\frac{1}{T_1}+\frac{1}{T_2}$	Combined decay
$\alpha$-Particle Kinetic Energy	$K_{\alpha }=Q\frac{A-4}{A}$	Momentum conservation
Nuclear Energy Unit	$1\text{ amu}\approx 931.5\text{ MeV}$	$E=\mathrm{\Delta }m{c}^2$