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UNIT SYLLABUS

E.3 Radioactive decay

SL/HL 7 hours SL + 5 hours HL
Some nuclei are unstable, and their decay is nature's purest randomness: no trigger, no memory, just a fixed probability per unit time. Alpha, beta and gamma emissions each change the nucleus in characteristic ways, with penetrating powers spanning paper to lead. Binding energy — the mass defect via $E = mc^2$ — explains both why decay releases energy and why iron sits at the curve's peak, dividing fusion territory from fission territory. The half-life turns randomness into clockwork at scale: at HL the exponential decay law $N = N_0e^{-\lambda t}$ makes it quantitative, dating everything from archaeological remains to the Earth itself.

Guiding Questions

  • ? Why are some nuclei stable while others decay?
  • ? How can a process that is random for one nucleus be so predictable for many?
t N N₀ N₀/2 N₀/4 N₀/8 2T½ 3T½ N = N₀e^(−λt),  T½ = ln2 / λ
Exponential decay: each half-life halves the remaining nuclei. After n half-lives, a fraction 1/2ⁿ remains.

What the IB expects you to master

  • Define isotopes, nuclear binding energy and mass defect; use E=mc2E = mc^2 in nuclear reactions.
  • Interpret the binding-energy-per-nucleon curve, including its maximum near iron and the approximate constancy above A60A \approx 60 (HL detail).
  • Describe the strong nuclear force: short-range, attractive, acting between nucleons.
  • Explain the random, spontaneous nature of decay, and balance decay equations for α\alpha, β\beta^-, β+\beta^+ and γ\gamma emissions, including neutrinos and antineutrinos.
  • Compare the penetration and ionising power of alpha, beta and gamma radiation.
  • Use activity, count rate and half-life, with integer numbers of half-lives (SL); account for background radiation.
  • HL: explain nuclear stability via the neutron–proton ratio, and the evidence for nuclear energy levels (discrete alpha/gamma spectra) and for the neutrino (continuous beta spectrum).
  • HL: apply the decay law N=N0eλtN = N_0e^{-\lambda t}, activity A=λN=A0eλtA = \lambda N = A_0e^{-\lambda t}, and T1/2=ln2λT_{1/2} = \frac{\ln 2}{\lambda}.

1 Key Formulas

Mass–energy equivalence
E=mc2E = mc^{2}
Decay law (HL)
N=N0eλtN = N_{0}e^{-\lambda t}
Activity (HL)
A=λN=A0eλtA = \lambda N = A_{0}e^{-\lambda t}
Half-life and decay constant (HL)
T12=ln2λT_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

2 Exam Preparation & Topic Explanations

Half-life arithmetic and beyond

SL questions stay on integer half-lives: divide by two the right number of times, remembering to strip background first. HL unlocks arbitrary times via N=N0eλtN = N_0e^{-\lambda t} — get λ\lambda from ln2/T1/2\ln 2 / T_{1/2} and keep units consistent throughout.

Binding-energy calculations follow one recipe: mass defect in u → multiply by 931.5 for MeV. State whether energy is released (products more bound) or required.

Pro Exam Strategy
  • Alpha: helium nucleus, stopped by paper, heavily ionising. Beta: electron/positron, stopped by aluminium. Gamma: photon, attenuated by lead. Know all nine facts.

  • Decay is random per nucleus but statistically exponential for large N — be ready to articulate both halves.

  • HL neutron–proton ratio: light stable nuclei have NZN \approx Z; heavy ones need N>ZN > Z; the wrong side of the stability band decides β\beta^- vs β+\beta^+.

  • Discrete alpha/gamma spectra ↔ nuclear energy levels; continuous beta spectrum ↔ neutrino. Two pieces of evidence, two conclusions.

3 MCQ Practice

Q1. A sample contains 8.0×10208.0\times10^{20} radioactive nuclei with a half-life of 6.0 hours. After one day, the number remaining is:

  • 4.0×10204.0\times10^{20}
  • 2.0×10202.0\times10^{20}
  • 1.0×10201.0\times10^{20}
  • 0.5×10200.5\times10^{20}

Q2. In beta-minus decay, the emitted electrons show a continuous energy spectrum. This was evidence for:

  • Discrete nuclear energy levels
  • The existence of the antineutrino, sharing the decay energy
  • The quantisation of charge
  • The strong nuclear force

Q3. The binding energy per nucleon curve peaks near iron (A56A \approx 56). This explains why:

  • Iron is the most abundant element
  • Fusion releases energy below iron and fission above it
  • Iron cannot undergo any nuclear reactions
  • All nuclei decay to iron eventually

4 Short Answer Questions

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