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Radioactivity Decay Laws

🎓 Class 12 Physics CBSE Theory Ch 13 – Nuclei ⏱ ~14 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Radioactivity Decay Laws

આ મૂલ્યાંકન આના પર આધારિત હશે: Radioactivity Decay Laws

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

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.

TypeParticle emittedChargePenetrationEffect on (Z, A)
α-decayHelium nucleus \(^{4}_{2}\mathrm{He}\)+2eStopped by paper / few cm of airZ → Z−2, A → A−4
β⁻-decayElectron + antineutrino \((\bar\nu)\)−eStopped by ~1 mm AlZ → Z+1, A unchanged
β⁺-decayPositron + neutrino \((\nu)\)+eStopped by ~1 mm AlZ → Z−1, A unchanged
γ-decayPhoton (≥ keV)0Several cm Pb(Z, A) unchanged; nucleus de-excites
Penetrating power: α, β, γ Source paper Al (~1 mm) Pb α β γ α: heavy & charged → stopped easily. β: lighter charged → stopped by metal. γ: photons → highly penetrating.
Fig 13.3: Penetrating power of α, β and γ radiations. α is least penetrating; γ is most penetrating.

α-Decay

A nucleus emits an α-particle (helium-4). The general scheme:

^{A}_{Z}X → ^{A−4}_{Z−2}Y + ^{4}_{2}He + Q

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:

n → p + e⁻ + ν̄ ^{A}_{Z}X → ^{A}_{Z+1}Y + e⁻ + ν̄

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:

p → n + e⁺ + ν ^{A}_{Z}X → ^{A}_{Z−1}Y + e⁺ + ν

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.

Part of the ²³⁸U decay series ²³⁸U Z=92 α ²³⁴Th Z=90 β⁻ ²³⁴Pa Z=91 β⁻ ²³⁴U Z=92 α ²³⁰Th… → ²⁰⁶Pb Each α step lowers A by 4 and Z by 2; each β⁻ step keeps A but increases Z by 1. The chain ends at the stable lead isotope ²⁰⁶Pb.
Fig 13.4: Section of the ²³⁸U → ²⁰⁶Pb radioactive decay chain.

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:

−dN/dt = λN

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:

R = −dN/dt = λN = λ N₀ e^(−λt) = R₀ e^(−λt)

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.

Three useful relationships:
  • \(N = N_0 e^{-\lambda t}\) — number remaining
  • \(T_{1/2} = 0.693/\lambda\) — half-life
  • \(\tau = 1/\lambda\) — mean life
Time t (in units of T₁/₂) N(t) / N₀ 1.0 0.5 0.25 0.125 0 1 2 3 4 5 ½ N₀ ¼ N₀ ⅛ N₀ N(t) = N₀ e^(−λt)
Fig 13.5: Exponential decay. After every T₁/₂, the population halves.

Worked Examples

Worked Example 13.4 — Half-life ↔ decay constant

A radioactive isotope has a half-life of 5.0 days. Find (a) its decay constant in s⁻¹ and (b) its mean life in days.

(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?

A sample initially contains 10²⁰ atoms of a radioisotope with half-life 30 minutes. How many atoms remain after 2 hours?

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

A 1.0 g sample of radium-226 (T₁/₂ = 1620 y, atomic mass ≈ 226) — find its activity in Bq.

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.)

Activity 13.3 — Coin-toss decay model

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.

What kind of curve do you expect? What is the "half-life" in terms of tosses?
Each coin has a 50% chance of "decaying" per round, independent of the others — exactly like radioactive decay. The plot is approximately exponential. The half-life is one toss (after one round, ~50 remain; after two, ~25; etc.). This shows that radioactive decay is a fundamentally statistical process.

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.

1.00 T½
Fraction left: 0.50 Activity ratio R/R₀: 0.50
t / T½ N/N₀

Competency-Based Questions

Q1 (MCQ). The relation between half-life and mean life is:

  • (a) T₁/₂ = τ
  • (b) T₁/₂ = ln 2 · τ ≈ 0.693 τ
  • (c) T₁/₂ = 2τ
  • (d) T₁/₂ = τ²
(b) T₁/₂ = (ln 2)/λ and τ = 1/λ, so T₁/₂ = 0.693 τ.

Q2 (MCQ). After 4 half-lives, the fraction of original nuclei remaining is:

  • (a) 1/2
  • (b) 1/8
  • (c) 1/16
  • (d) 1/4
(c) (1/2)⁴ = 1/16.

Q3 (Short Answer). Why are α-particles much less penetrating than β-particles, although they carry more energy?

α-particles are massive and doubly charged. They lose energy rapidly through dense ionisation as they collide with atoms in matter. β-particles are much lighter, less ionising per unit length, and so penetrate further before stopping.

Q4 (Numerical). The activity of a sample decreases from 8000 Bq to 1000 Bq in 9 hours. Find the half-life.

R/R₀ = 1000/8000 = 1/8 = (1/2)³ → 3 half-lives in 9 hours → T₁/₂ = 3 hours.

Q5 (HOTS). Why does β⁻-decay always come accompanied by an antineutrino?

Two clues forced the prediction of the neutrino (Pauli, 1930): (i) the energy spectrum of emitted electrons is continuous (not a single line), so the missing energy is carried away by an unseen particle. (ii) Conservation of angular momentum and lepton number requires an additional spin-1/2 antiparticle. The (anti)neutrino satisfies both — it is electrically neutral, nearly massless, and barely interacts with matter.

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.

(A) Both correct; the reason explains the assertion.

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.

(D) Assertion is false — half-life is intrinsic to the nuclide, independent of sample size. Reason is true. So (D).

Assertion: The mean life τ of a radioactive nuclide is greater than its half-life T₁/₂.

Reason: τ = T₁/₂ / ln 2.

(A) Both correct; the reason explains the assertion. Since ln 2 ≈ 0.693 < 1, τ > T₁/₂.

Frequently Asked Questions - Radioactivity Decay Laws

What is the main concept covered in Radioactivity Decay Laws?
In NCERT Class 12 Physics Chapter 13 (Nuclei), "Radioactivity Decay Laws" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Radioactivity Decay Laws useful in real-life applications?
Real-life applications of "Radioactivity Decay Laws" from NCERT Class 12 Physics Chapter 13 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Radioactivity Decay Laws?
Key formulas in "Radioactivity Decay Laws" (NCERT Class 12 Physics Chapter 13 Nuclei) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 13?
NCERT Class 12 Physics Chapter 13 (Nuclei) is structured so each part builds on the previous one. "Radioactivity Decay Laws" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Radioactivity Decay Laws?
CBSE board questions from "Radioactivity Decay Laws" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Radioactivity Decay Laws" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part II – NCERT (2025-26)
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