This MCQ module is based on: Radioactivity Decay Laws
Radioactivity Decay Laws
This assessment will be based on: Radioactivity Decay Laws
Upload images, PDFs, or Word documents to include their content in assessment generation.
Radioactivity Decay Laws
13.6 Radioactivity — Discovery and Three Decay Modes
In 1896, the French physicist Henri Becquerel made one of physics' great accidental discoveries. He had wrapped uranium-potassium sulphate in black paper, placed a photographic plate underneath separated by a thin silver sheet, and stored everything in a dark drawer for several days. When he finally developed the plate, it was darkened — something invisible was passing through paper and metal. The compound was emitting nothing it was being given; the energy was coming from inside the atomic nucleus itself.
Marie and Pierre Curie soon isolated polonium and radium, and Rutherford named the three components of the mysterious radiation by how easily they were absorbed: α, β, γ. We now know that radioactivity is a nuclear phenomenon — an unstable nucleus spontaneously transforms by emitting a particle (or photon) and rearranging itself into a more stable configuration.
| Type | Particle emitted | Charge | Penetration | Effect on (Z, A) |
|---|---|---|---|---|
| α-decay | Helium nucleus \(^{4}_{2}\mathrm{He}\) | +2e | Stopped by paper / few cm of air | Z → Z−2, A → A−4 |
| β⁻-decay | Electron + antineutrino \((\bar\nu)\) | −e | Stopped by ~1 mm Al | Z → Z+1, A unchanged |
| β⁺-decay | Positron + neutrino \((\nu)\) | +e | Stopped by ~1 mm Al | Z → Z−1, A unchanged |
| γ-decay | Photon (≥ keV) | 0 | Several cm Pb | (Z, A) unchanged; nucleus de-excites |
α-Decay
A nucleus emits an α-particle (helium-4). The general scheme:
Example: \(^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}\) (Q ≈ 4.27 MeV). α-decay is energetically possible only if the parent's mass exceeds the combined mass of daughter + α.
β-Decay
In β⁻-decay a neutron in the nucleus converts to a proton, ejecting an electron and an antineutrino:
Example: \(^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + e^{-} + \bar\nu\). Useful in carbon dating.
In β⁺-decay a proton converts to a neutron, ejecting a positron and a neutrino:
γ-Decay
After α or β emission, the daughter nucleus is often left in an excited state. It drops to its ground state by emitting one or more γ-ray photons (energies ranging from a few keV to several MeV). Z and A do not change. γ-rays are simply nuclear EM radiation with very short wavelength.
Law of Radioactive Decay
Suppose we have a sample containing N(t) radioactive nuclei at time t. The radioactive decay of an individual nucleus is a statistical, spontaneous event — we cannot predict which nucleus will decay next or when. But for a large number of nuclei, the rate of decay is found experimentally to be proportional to N:
Here \(\lambda\) is the decay (or disintegration) constant — a characteristic property of each radioactive species. Solving the differential equation with the initial condition \(N(0) = N_0\):
\[ \boxed{\,N(t) = N_0 \, e^{-\lambda t}\,} \]This is the law of radioactive decay. The number of radioactive nuclei falls exponentially with time.
Activity (R)
The activity R = |dN/dt| is the rate at which decays occur in the sample:
The SI unit of activity is the becquerel (Bq): 1 Bq = 1 decay per second. An older unit still common in medicine is the curie (Ci): 1 Ci = 3.7 × 10¹⁰ Bq.
Half-life T₁/₂
The half-life is the time after which exactly half of the nuclei present at any instant have decayed.
Setting \(N = N_0/2\) at \(t = T_{1/2}\):
\[ \tfrac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \;\Rightarrow\; T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \]Mean life τ
The mean (average) life of the nuclei in the sample is
\[ \tau = \frac{1}{\lambda} \;\;\;\Rightarrow\;\;\; T_{1/2} = \tau \ln 2 = 0.693\,\tau \]Mean life is always slightly longer than half-life.
- \(N = N_0 e^{-\lambda t}\) — number remaining
- \(T_{1/2} = 0.693/\lambda\) — half-life
- \(\tau = 1/\lambda\) — mean life
Worked Examples
Worked Example 13.4 — Half-life ↔ decay constant
(a) T₁/₂ = 5.0 d = 5.0 × 86400 s = 4.32 × 10⁵ s.
\[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{4.32 \times 10^{5}} = 1.604 \times 10^{-6}\ \text{s}^{-1} \](b) τ = 1/λ = T₁/₂ / 0.693 = 5.0 / 0.693 ≈ 7.21 days.
Worked Example 13.5 — How many remain after time t?
Number of half-lives elapsed: n = 2 h / 30 min = 4.
\[ N = N_0 \left(\tfrac{1}{2}\right)^n = 10^{20} \times \left(\tfrac{1}{2}\right)^4 = 10^{20}/16 \approx 6.25 \times 10^{18}\ \text{atoms} \]Worked Example 13.6 — Activity
Number of nuclei: N = (1.0 g)(N_A) / (226 g/mol) = 6.022 × 10²³ / 226 ≈ 2.665 × 10²¹.
T₁/₂ = 1620 × 3.154 × 10⁷ s ≈ 5.11 × 10¹⁰ s, so λ = 0.693 / 5.11 × 10¹⁰ ≈ 1.356 × 10⁻¹¹ s⁻¹.
\[ R = \lambda N = (1.356 \times 10^{-11})(2.665 \times 10^{21}) \approx 3.61 \times 10^{10}\ \text{Bq} \approx 1\ \text{Ci} \](Indeed, 1 g of Ra-226 was the historical definition of 1 curie.)
Take 100 coins, all heads up. Toss them all; remove every coin that comes up tails. Repeat. Plot the number of coins remaining versus toss number.
Interactive — Half-Life Simulator
Decay-curve explorer
Choose an isotope and watch how N(t) and the activity R(t) evolve. The slider lets you scrub through time.
Competency-Based Questions
Q1 (MCQ). The relation between half-life and mean life is:
Q2 (MCQ). After 4 half-lives, the fraction of original nuclei remaining is:
Q3 (Short Answer). Why are α-particles much less penetrating than β-particles, although they carry more energy?
Q4 (Numerical). The activity of a sample decreases from 8000 Bq to 1000 Bq in 9 hours. Find the half-life.
Q5 (HOTS). Why does β⁻-decay always come accompanied by an antineutrino?
Assertion–Reason Questions
Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: γ-rays have no charge and no rest mass.
Reason: γ-rays are high-energy electromagnetic photons.
Assertion: Half-life of a radioactive substance depends on its initial mass.
Reason: The decay rate λN is proportional to the number of nuclei present.
Assertion: The mean life τ of a radioactive nuclide is greater than its half-life T₁/₂.
Reason: τ = T₁/₂ / ln 2.
Frequently Asked Questions - Radioactivity Decay Laws
What is the main concept covered in Radioactivity Decay Laws?
How is Radioactivity Decay Laws useful in real-life applications?
What are the key formulas in Radioactivity Decay Laws?
How does this part connect to other parts of Chapter 13?
What types of CBSE board questions come from Radioactivity Decay Laws?
How can students use the interactive simulation effectively?
🎯 Practise Physics
Sit a full paper on what you have been studying, marked question by question.
Board exam sample papers
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E