A-Level — physicsHard3 min read

Radioactivity and Nuclear Decay

α, β, γ radiation; half-life; N = N₀e^(-λt); activity A = λN

Tutor AI Team
Tutor AI Team

# Radioactivity and Nuclear Decay — A-Level Physics

Radioactivity is the spontaneous emission of radiation from unstable nuclei. At A-Level, we study the three types of radiation, the mathematics of exponential decay, and the concept of half-life.

1. Types of Radiation

Property Alpha (α) Beta-minus (β⁻) Gamma (γ)
Nature ²He nucleus Electron EM radiation
Charge +2 −1 0
Mass 4u ~0 0
Speed ~5% c Up to 99% c c
Ionising Strongly Moderately Weakly
Penetrating Paper stops Al (few mm) stops Pb (thick) reduces
Range in air ~5 cm ~1 m Infinite (inverse square)
Deflection Slight in B/E fields Large in B/E fields None

Decay Equations

Alpha: ZAX→Z−2A−4Y+24α^A_Z X \to ^{A-4}_{Z-2}Y + ^4_2\alpha

Beta-minus: ZAX→Z+1AY+−10β+νˉe^A_Z X \to ^A_{Z+1}Y + ^0_{-1}\beta + \bar{\nu}_e (A neutron → proton + electron + antineutrino)

Gamma: No change in A or Z — just energy release

2. Exponential Decay

Radioactive decay is random and spontaneous. For a large sample:

N=N0e−λt\boxed{N = N_0 e^{-\lambda t}} A=A0e−λt\boxed{A = A_0 e^{-\lambda t}}

Where:

  • NN = number of undecayed nuclei
  • A=λNA = \lambda N = activity (decays per second, Bq)
  • λ\lambda = decay constant (s⁻¹)

3. Half-Life

t1/2=ln⁡2λ=0.693λ\boxed{t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}

The time for half the nuclei to decay (or activity to halve).

4. Activity

A=λN=dNdtA = \lambda N = \frac{dN}{dt}

Units: becquerels (Bq) = decays per second.

Worked Example: Half-Life Calculation

Problem

A sample has a half-life of 8 days. After 24 days, what fraction remains?

Number of half-lives = 24/8 = 3 Fraction remaining = (1/2)3=1/8(1/2)^3 = 1/8 = 12.5%

Solution

Worked Example: Decay Constant

Problem

Carbon-14 has a half-life of 5730 years. Find the decay constant.

λ=ln⁡2/t1/2=0.693/(5730×365.25×24×3600)=0.693/1.807×1011=3.83×10−12\lambda = \ln 2/t_{1/2} = 0.693/(5730 \times 365.25 \times 24 \times 3600) = 0.693/1.807 \times 10^{11} = 3.83 \times 10^{-12} s⁻¹

Solution

Worked Example: Activity

Problem

A sample contains 4×10204 \times 10^{20} atoms of a radioisotope with λ=2.5×10−4\lambda = 2.5 \times 10^{-4} s⁻¹. Find the initial activity and the activity after 1 hour.

A0=λN0=2.5×10−4×4×1020=1017A_0 = \lambda N_0 = 2.5 \times 10^{-4} \times 4 \times 10^{20} = 10^{17} Bq A=A0e−λt=1017e−2.5×10−4×3600=1017e−0.9=1017×0.407=4.07×1016A = A_0 e^{-\lambda t} = 10^{17} e^{-2.5 \times 10^{-4} \times 3600} = 10^{17} e^{-0.9} = 10^{17} \times 0.407 = 4.07 \times 10^{16} Bq

Solution

Worked Example: Carbon Dating

Problem

A sample has 25% of the original C-14. How old is it?

N/N0=0.25=e−λtN/N_0 = 0.25 = e^{-\lambda t} λt=ln⁡4=1.386\lambda t = \ln 4 = 1.386 t=1.386/λ=1.386/(3.83×10−12)=3.62×1011t = 1.386/\lambda = 1.386/(3.83 \times 10^{-12}) = 3.62 \times 10^{11} s = 11{,}460 years

Or: 0.25=(1/2)n0.25 = (1/2)^n → n=2n = 2 → t=2×5730=11,460t = 2 \times 5730 = 11{,}460 years ✓

Solution

6. Practice Questions

    1. Write the decay equation for Uranium-238 undergoing alpha decay to Thorium. (2 marks)
    1. A radioisotope has a half-life of 6 hours. If the initial activity is 800 Bq, find the activity after 18 hours. (2 marks)
    1. Technetium-99m has t1/2=6t_{1/2} = 6 hours. A hospital receives a 500 MBq sample at 8 am. What is the activity at 8 pm? (3 marks)
    1. Explain why radioactive decay is described as random and spontaneous. (2 marks)

    Answers

    1. 92238U→90234Th+24α^{238}_{92}U \to ^{234}_{90}Th + ^4_2\alpha.

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Summary

  • Three types: α (He-4), β⁻ (electron + antineutrino), γ (EM photon)
  • N=N0e−λtN = N_0 e^{-\lambda t}; A=λNA = \lambda N; t1/2=ln⁡2/λt_{1/2} = \ln 2/\lambda
  • Decay is random and spontaneous
  • Carbon dating uses known half-life of C-14
Tags:
#a-level#physics#nuclear#radioactivity#half-life#decay#alpha#beta#gamma

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