use λ = 0.693 / t

Radioactive Decay & Nuclear Physics (Cambridge AS & A‑Level 9702)

Learning Objectives (Syllabus 23 Nuclear physics)

• Explain nuclear structure and calculate mass‑defect, binding energy and binding‑energy‑per‑nucleon using E = mc².
• Derive the radioactive decay law, relate the decay constant (λ) to the half‑life (t_½) and calculate activity.
• Write balanced nuclear equations for the common decay modes (α, β⁻, β⁺, EC, γ) and comment on changes in mass number (A) and atomic number (Z).
• Connect β‑decay to the underlying quark‑flavour change and the weak interaction.
• Calculate the energy released in typical fission and fusion reactions.
• Apply logarithms, exponentials, unit conversion and error propagation to nuclear‑physics problems (AO3).
• Plan, carry out and analyse a simple half‑life experiment, including uncertainties and safety considerations.
• Identify real‑world applications (medical imaging, radiotherapy, carbon dating, nuclear power).

1. Nuclear Structure & Binding Energy

1.1 Mass‑defect (Δm)

The mass‑defect is the difference between the total mass of the separated nucleons and the actual mass of the nucleus.

\[

\Delta m = \bigl(Z\,mp + (A-Z)\,mn\bigr) - m_{\text{nuclide}}

\]

1.2 Binding energy (E_b)

Binding energy is the energy required to separate a nucleus into its constituent nucleons.

\[

E_b = \Delta m\,c^{2}

\]

1.3 Binding‑energy‑per‑nucleon

\[

\frac{E_b}{A}

\]

It allows direct comparison of nuclear stability. Plotting \(\frac{E_b}{A}\) against mass number A gives the familiar binding‑energy‑per‑nucleon curve:

• Peaks at A ≈ 56 (≈ 8.8 MeV nucleon⁻¹) – nuclei around iron‑56 are most stable.
• For A < 56, fusion releases energy (products have larger \(\frac{E_b}{A}\)).
• For A > 56, fission releases energy (products have larger \(\frac{E_b}{A}\)).

This curve underpins the syllabus statement that “the products of fission and fusion have a higher binding energy per nucleon than the reactants”.

1.4 Example – Binding energy of ¹²C

1. Mass of 6 p + 6 n = \(6(1.007276\;\text{u}) + 6(1.008665\;\text{u}) = 12.0959\;\text{u}\)
2. Atomic mass of ¹²C = 12.0000 u (by definition)
3. Δm = 0.0959 u; 1 u = 931.5 MeV c⁻²
4. \(E_b = 0.0959 \times 931.5 = 89.3\;\text{MeV}\)
5. \(E_b/A = 89.3/12 = 7.44\;\text{MeV nucleon}^{-1}\)

2. Radioactive Decay – The Decay Law

2.1 Decay equation

\[

N(t)=N_0 e^{-\lambda t}

\]

• N₀ – number of nuclei at \(t=0\)
• λ – decay constant (s⁻¹), the probability per unit time that a nucleus decays
• t – elapsed time (s)

2.2 Half‑life and the decay constant

At the half‑life \(t_{½}\) the number of undecayed nuclei is halved:

\[

\frac12 N0 = N0 e^{-\lambda t_{½}}

\;\Longrightarrow\;

e^{-\lambda t_{½}} = \frac12

\]

Taking natural logarithms gives the key relation required by the syllabus:

\[

\boxed{\lambda = \frac{\ln 2}{t{½}} = \frac{0.693}{t{½}}}

\]

Units: if \(t_{½}\) is expressed in years, first convert to seconds (1 yr = 3.156 × 10⁷ s) before using the formula.

2.3 Activity (A)

\[

A = \lambda N \qquad\text{(Bq = decays s⁻¹)}

\]

For a sample that started with activity \(A_0\):

\[

A(t)=A_0 e^{-\lambda t}

\]

2.4 Worked example – Determining λ and t_½

Problem: A sample has an activity of 2.5 × 10⁶ Bq and after 30 min the activity is 1.8 × 10⁶ Bq. Find λ and the half‑life.

1. Use \(A(t)=A0 e^{-\lambda t}\) → \(\frac{A}{A0}=e^{-\lambda t}\).
2. \(\frac{1.8\times10^{6}}{2.5\times10^{6}} = 0.72 = e^{-\lambda(1800\;\text{s})}\).
3. \(\lambda = -\frac{1}{1800}\ln(0.72)=2.01\times10^{-4}\;\text{s}^{-1}\).
4. Half‑life: \(t_{½}=0.693/\lambda = 0.693/(2.01\times10^{-4}) = 3.45\times10^{3}\;\text{s}=57.5\;\text{min}\).

3. Decay Modes, Nuclear Equations & Particle‑Physics Links

3.1 Particle‑physics perspective (β‑decay)

• β⁻ decay: a down quark (d) in a neutron converts to an up quark (u) via the weak interaction, emitting a virtual \(W^-\) boson that becomes an electron and an antineutrino.

  \[

  n(ddu) \rightarrow p(duu) + e^- + \bar\nu_e

  \]

• β⁺ decay / EC: an up quark converts to a down quark, emitting a virtual \(W^+\) boson (→ e⁺ + νₑ) or capturing an orbital electron.
• All decay modes conserve charge, baryon number, lepton number and total energy.

4. Energy from Fission & Fusion

4.1 Why energy is released

Because the products have a higher binding‑energy‑per‑nucleon than the reactants (see the curve in 1.3).

4.2 Fission example – ²³⁵U + n

\[

^{235}{92}\text{U} + ^10n \rightarrow ^{141}{56}\text{Ba} + ^{92}{36}\text{Kr} + 3\,^1_0n + Q

\]

Using atomic masses (u):

• m(⁽²³⁵⁾U) = 235.0439
• m(n) = 1.0087 (four neutrons involved)
• m(⁽¹⁴¹⁾Ba) = 140.9144
• m(⁽⁹²⁾Kr) = 91.9262

Mass defect:

\[

\Delta m = \bigl[235.0439 + 4(1.0087)\bigr] - \bigl[140.9144 + 91.9262 + 3(1.0087)\bigr] = 0.188\;\text{u}

\]

Energy released:

\[

Q = \Delta m\,c^{2}=0.188 \times 931.5 = 1.75\times10^{2}\;\text{MeV}\;\approx\;200\;\text{MeV}

\]

4.3 Fusion example – Deuterium–tritium reaction

\[

^{2}{1}\text{H} + ^{3}{1}\text{H} \rightarrow ^{4}{2}\text{He} + ^{1}{0}n + Q

\]

Mass defect ≈ 0.0189 u → \(Q \approx 17.6\;\text{MeV}\).

4.4 Worked example – Energy from 1 g of ²³⁵U

1. Number of nuclei: \(N = \dfrac{1.0\;\text{g}}{235\;\text{g mol}^{-1}} N_A = \dfrac{1}{235}\times6.022\times10^{23}=2.56\times10^{21}\).
2. Total energy: \(E = N \times 200\;\text{MeV}=2.56\times10^{21}\times200\times1.602\times10^{-13}\;\text{J}=8.2\times10^{10}\;\text{J}\).
3. ≈ 2 kg of gasoline would release the same energy.

5. Practical Skills (AO3) – Measuring a Half‑Life

1. Goal: Determine the half‑life of a short‑lived isotope (e.g., ⁶⁴Cu, \(t_{½}\approx12.7\) h) using a Geiger‑Müller (GM) counter.
2. Apparatus: sealed source, GM counter, timer, lead shielding, data‑log software, dose‑rate badge.
3. Method outline:

   • Place the source at a fixed geometry; record the initial count rate \(C_0\) after applying a dead‑time correction.
   • Take successive counts at regular intervals (e.g., every 30 min) for at least five half‑lives.
   • Convert counts to activity: \(A = C/\varepsilon\) (where \(\varepsilon\) is detector efficiency).
   • Plot \(\ln A\) versus \(t\); the slope equals \(-\lambda\).

4. Data analysis:

   • Uncertainty on each count: \(\sigma_C = \sqrt{C}\) (Poisson statistics).
   • Propagate to activity: \(\sigmaA = \sigmaC/\varepsilon\).
   • Linear regression gives \(\lambda\) and its standard error \(\sigma_\lambda\).
   • Half‑life: \(t{½}=0.693/\lambda\) with uncertainty \(\displaystyle\sigma{t{½}} = \frac{0.693}{\lambda^{2}}\sigma\lambda\).

5. Safety considerations:

   • Work in a designated radiation area; wear a personal dose‑rate badge.
   • Use lead shielding and keep a safe distance when possible.
   • Follow the institution’s radioactive‑material handling protocol (contamination checks, waste disposal).

6. Mathematical Tools (Key Concept: Mathematics as a Language)

• Natural logarithm: \(\ln x\) is the inverse of \(e^{x}\); used to linearise exponential decay.
• Exponential function: \(e^{x}\); fundamental to decay, growth and half‑life calculations.
• Unit conversion:

  • Time: years → seconds (1 yr = 3.156 × 10⁷ s).
  • Mass: atomic mass units → kilograms (1 u = 1.6605 × 10⁻²⁷ kg).
  • Energy: MeV → joules (1 MeV = 1.602 × 10⁻¹³ J).

• Error propagation (multiplication/division):

  \[

  \frac{\sigmaQ}{Q}= \sqrt{\left(\frac{\sigmaa}{a}\right)^2+\left(\frac{\sigmab}{b}\right)^2+\left(\frac{\sigmac}{c}\right)^2}

  \quad\text{for } Q=\frac{a\,b}{c}

  \]

7. Worked Example Problems

7.1 Decay constant & activity

Problem: A source of ⁹⁹Mo has an initial activity of \(5.0\times10^{18}\) Bq and a half‑life of 66 h. Find the activity after 200 h.

1. Convert half‑life: \(t_{½}=66\;\text{h}=2.376\times10^{5}\;\text{s}\).
2. Decay constant: \(\lambda = 0.693/t_{½}=2.92\times10^{-6}\;\text{s}^{-1}\).
3. Elapsed time: \(t=200\;\text{h}=7.20\times10^{5}\;\text{s}\).
4. Activity: \(A = A_0 e^{-\lambda t}=5.0\times10^{18}\,e^{-2.92\times10^{-6}\times7.20\times10^{5}} \approx 1.1\times10^{18}\;\text{Bq}\).

7.2 Energy from a fission reaction

Problem: Calculate the energy released when 1.0 g of ²³⁵U undergoes complete fission (average \(Q = 200\) MeV per fission).

1. Number of nuclei: \(N = \dfrac{1.0\;\text{g}}{235\;\text{g mol}^{-1}} N_A = 2.56\times10^{21}\).
2. Total energy: \(E = N \times 200\;\text{MeV}=2.56\times10^{21}\times200\times1.602\times10^{-13}\;\text{J}=8.2\times10^{10}\;\text{J}\).
3. Equivalent to burning ≈ 2 kg of gasoline.

8. Real‑World Applications

• Medical imaging: γ‑rays from ^99mTc (half‑life = 6 h) are used in single‑photon emission computed tomography (SPECT).
• Radiotherapy: High‑energy β⁻ or γ emitters (e.g., ⁶⁰Co, \(t_{½}=5.27\) y) deliver therapeutic doses to tumours.
• Carbon dating: β⁻ decay of ¹⁴C (\(t_{½}=5730\) y) provides ages up to ~50 000 y for archaeological samples.
• Nuclear power: Heat from the fission of ²³⁵U or ²³⁹Pu drives turbines in commercial reactors.

9. Summary

• Mass‑defect → binding energy → binding‑energy‑per‑nucleon curve explains why nuclei around A ≈ 56 are most stable.
• Radioactive decay follows \(N(t)=N0e^{-\lambda t}\); the decay constant is related to half‑life by \(\lambda = 0.693/t{½}\).
• Activity \(A=\lambda N\) decays with the same exponential law.
• Four principal decay modes (α, β⁻, β⁺/EC, γ) are described by balanced nuclear equations and quark‑level processes.
• Fission (A > 56) and fusion (A < 56) release energy because the products have a larger binding‑energy‑per‑nucleon.
• Mathematical tools (logarithms, exponentials, unit conversion, error propagation) are essential for quantitative analysis.
• Understanding half‑life measurement techniques and safety is a key AO3 requirement.
• Radioactive processes have vital applications in medicine, archaeology and energy production.