Radioactivity and Nuclear Reactions

Alpha, beta, gamma; half-life; decay equations; activity; N = N₀e^(-λt)

# Radioactivity and Nuclear Reactions — IB Physics

1. Types of Radiation

Alpha (α\alpha) Beta (β\beta^-) Gamma (γ\gamma)
Nature He-4 nucleus Electron EM photon
Charge +2 −1 0
Ionising Strongly Moderately Weakly
Penetrating Paper stops Al (mm) Pb (cm)
Deflection Slight (E/B) Large (E/B) None

2. Decay Equations

Alpha: ZAXZ2A4Y+24α^A_Z X \to ^{A-4}_{Z-2}Y + ^4_2\alpha

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

Gamma: No change in A or Z; nucleus loses energy.

3. Half-Life and Decay Law

N=N0eλtN = N_0 e^{-\lambda t} A=A0eλtA = A_0 e^{-\lambda t} t1/2=ln2λt_{1/2} = \frac{\ln 2}{\lambda}

A=λNA = \lambda N (activity = decay constant × number of nuclei).

Decay is random (cannot predict individual events) and spontaneous (not triggered externally).

Worked Example: Example 1

Problem

Ra-226 undergoes alpha decay. Write the equation. 88226Ra86222Rn+24α^{226}_{88}Ra \to ^{222}_{86}Rn + ^4_2\alpha

Solution

Worked Example: Half-Life

Problem

t1/2=5t_{1/2} = 5 days. Starting activity = 1600 Bq. After 20 days? 20/5 = 4 half-lives. A=1600/24=100A = 1600/2^4 = 100 Bq.

Solution

Worked Example: Decay Constant

Problem

C-14: t1/2=5730t_{1/2} = 5730 years. A sample has 1/8 of original C-14. Number of half-lives: 1/8=(1/2)31/8 = (1/2)^3 → 3 half-lives → age = 3×5730=17,1903 \times 5730 = 17{,}190 years.

Solution

5. Background Radiation

Sources: cosmic rays, radon gas, rocks/soil, medical (X-rays), food, nuclear testing.

Must subtract background count from measured count rate.

6. Practice Questions

    1. Write the equation for beta-minus decay of Carbon-14. (2 marks)
    1. A sample has half-life 8 hours. Activity drops from 480 Bq to 60 Bq. How long did this take? (2 marks)
    1. Explain why radioactive decay is described as random and spontaneous. (2 marks)

    Answers

    1. 614C714N+10β+νˉe^{14}_6 C \to ^{14}_7 N + ^0_{-1}\beta + \bar{\nu}_e.
    1. 60=480×(1/2)n60 = 480 \times (1/2)^n(1/2)n=1/8(1/2)^n = 1/8n=3n = 3. Time = 3×8=243 \times 8 = 24 hours.

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Summary

  • α (He-4), β⁻ (electron), γ (photon)
  • N=N0eλtN = N_0 e^{-\lambda t}; t1/2=ln2/λt_{1/2} = \ln 2/\lambda; A=λNA = \lambda N
  • Random and spontaneous; subtract background

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