Atoms and Nuclei • Topic 3 of 3

Radioactivity & Nuclear Reactions

Radioactivity, discovered by Becquerel in 1896, is the spontaneous disintegration of an unstable nucleus with the emission of radiation. Three kinds of emission occur:

  • Alpha ($\alpha$) decay: the nucleus emits a helium nucleus ${}^{4}_{2}\text{He}$, so $Z$ falls by 2 and $A$ by 4: ${}^{A}_{Z}X \to {}^{A-4}_{Z-2}Y + {}^{4}_{2}\text{He}$. $\alpha$-particles are heavy, highly ionising but poorly penetrating (stopped by paper).
  • Beta ($\beta$) decay: a neutron converts to a proton, ejecting an electron and an antineutrino: ${}^{A}_{Z}X \to {}^{A}_{Z+1}Y + {}^{0}_{-1}e + \bar{\nu}$. Here $Z$ rises by 1, $A$ is unchanged. $\beta$-particles are lighter and more penetrating (stopped by aluminium).
  • Gamma ($\gamma$) decay: the nucleus, left excited after $\alpha$ or $\beta$ emission, releases a high-energy photon. There is no change in $Z$ or $A$. $\gamma$-rays are uncharged and the most penetrating (need thick lead or concrete).

Radioactive decay is random for a single nucleus but precise for large numbers. The number of nuclei decaying per unit time is proportional to the number present, giving the radioactive decay law $N = N_0 e^{-\lambda t}$, where $N_0$ is the initial number, $N$ the number left after time $t$, and $\lambda$ the decay constant (probability of decay per nucleus per second).

The activity $R$ is the rate of decay, $R = \lambda N = R_0 e^{-\lambda t}$, measured in becquerel (1 Bq = 1 decay/s) or curie (1 Ci $= 3.7 \times 10^{10}$ Bq).

The half-life $T_{1/2}$ is the time for half the nuclei to decay. Setting $N = \frac{N_0}{2}$ in the decay law gives $T_{1/2} = \frac{0.693}{\lambda}$. After $n$ half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^n$. The mean (average) life is $\tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693} = 1.44\, T_{1/2}$.

Large amounts of energy are also released in nuclear reactions. In nuclear fission, a heavy nucleus such as ${}^{235}_{92}\text{U}$ captures a slow neutron and splits into two medium nuclei plus a few neutrons, releasing about $200\ \text{MeV}$ per fission. The freed neutrons can trigger a self-sustaining chain reaction — controlled in a reactor, uncontrolled in a bomb. In nuclear fusion, light nuclei combine, as in the Sun where hydrogen fuses to helium: $4\,{}^{1}_{1}\text{H} \to {}^{4}_{2}\text{He} + 2e^{+} + \text{energy}$. Fusion releases even more energy per nucleon than fission but needs extremely high temperatures to overcome Coulomb repulsion. In every case the energy comes from the increase in binding energy per nucleon — mass is converted to energy through $E = \Delta m\, c^2$.

Radioactive decay: N vs t with half-lives T, 2T, 3T markedTime tNumber NN₀N₀/2N₀/4T2T3TN = N₀ e^(-λt)
Solved Examples
1
Worked Example
Write the nuclear reaction for the alpha decay of ${}^{238}_{92}\text{U}$ and identify the daughter nucleus.
Solution
  1. In $\alpha$ decay, $A$ decreases by 4 and $Z$ by 2.
  2. ${}^{238}_{92}\text{U} \to {}^{234}_{90}Y + {}^{4}_{2}\text{He}$.
  3. $Z = 90$ is thorium.

Answer: ${}^{238}_{92}\text{U} \to {}^{234}_{90}\text{Th} + {}^{4}_{2}\text{He}$; the daughter is thorium-234.

2
Worked Example
The half-life of a radioactive sample is 8 days. What fraction of the original nuclei remains after 24 days?
Solution
  1. Number of half-lives $n = \frac{24}{8} = 3$.
  2. Fraction remaining $= \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3$.
  3. $= \frac{1}{8}$.

Answer: One-eighth ($12.5\%$) of the original nuclei remain after 24 days.

3
Worked Example
A radioactive nuclide has a decay constant $\lambda = 0.0231\ \text{day}^{-1}$. Find its half-life and mean life.
Solution
  1. $T_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.0231}$.
  2. $T_{1/2} = 30\ \text{days}$.
  3. Mean life $\tau = \frac{1}{\lambda} = \frac{1}{0.0231} \approx 43.3\ \text{days}$ ($= 1.44\,T_{1/2}$).

Answer: Half-life $= 30$ days and mean life $\approx 43.3$ days.

4
Worked Example
A sample initially contains $4 \times 10^{20}$ nuclei with half-life 5 years. How many remain after 15 years, and what is the activity then if $\lambda = 0.1386\ \text{yr}^{-1}$?
Solution
  1. $n = \frac{15}{5} = 3$ half-lives, so $N = N_0 \left(\frac{1}{2}\right)^3 = \frac{4\times10^{20}}{8} = 5 \times 10^{19}$ nuclei.
  2. Activity $R = \lambda N$. Convert $\lambda = 0.1386\ \text{yr}^{-1} = \frac{0.1386}{3.156\times10^7}\ \text{s}^{-1} = 4.39 \times 10^{-9}\ \text{s}^{-1}$.
  3. $R = 4.39 \times 10^{-9} \times 5 \times 10^{19} = 2.2 \times 10^{11}\ \text{Bq}$.

Answer: $5 \times 10^{19}$ nuclei remain, with activity about $2.2 \times 10^{11}\ \text{Bq}$.

5
Worked Example
Complete the beta-decay reaction ${}^{14}_{6}\text{C} \to {}^{?}_{?}Y + {}^{0}_{-1}e + \bar{\nu}$ and name the daughter.
Solution
  1. In $\beta^-$ decay $A$ is unchanged and $Z$ increases by 1.
  2. $A = 14$ stays, $Z = 6 \to 7$.
  3. $Z = 7$ is nitrogen.

Answer: ${}^{14}_{6}\text{C} \to {}^{14}_{7}\text{N} + {}^{0}_{-1}e + \bar{\nu}$; the daughter is nitrogen-14 (the basis of carbon dating).

6
Worked Example
In the fission of ${}^{235}_{92}\text{U}$, about $200\ \text{MeV}$ is released per fission. How many fissions per second are needed to produce $1\ \text{MW}$ of power?
Solution
  1. Energy per fission $= 200\ \text{MeV} = 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\ \text{J}$.
  2. Power $P = 1\ \text{MW} = 10^{6}\ \text{J s}^{-1}$.
  3. Fissions per second $= \frac{P}{\text{energy per fission}} = \frac{10^{6}}{3.2 \times 10^{-11}} = 3.1 \times 10^{16}$.

Answer: About $3.1 \times 10^{16}$ fissions per second are required.

Key Points

  • $\alpha$ decay: $A \to A-4$, $Z \to Z-2$; $\beta^-$ decay: $A$ unchanged, $Z \to Z+1$; $\gamma$ decay: no change in $A$ or $Z$ (excited nucleus emits a photon).
  • Penetrating power increases $\alpha < \beta < \gamma$; ionising power decreases $\alpha > \beta > \gamma$.
  • Decay law $N = N_0 e^{-\lambda t}$; activity $R = \lambda N = R_0 e^{-\lambda t}$ (in becquerel).
  • Half-life $T_{1/2} = \frac{0.693}{\lambda}$; after $n$ half-lives a fraction $\left(\frac{1}{2}\right)^n$ remains; mean life $\tau = \frac{1}{\lambda} = 1.44\,T_{1/2}$.
  • Fission (heavy nuclei split, $\sim 200\ \text{MeV}$) and fusion (light nuclei combine, as in the Sun) both release energy by raising the binding energy per nucleon via $E = \Delta m\, c^2$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.In $\alpha$ decay, the mass number and atomic number of the parent change by:
Explanation: An $\alpha$ particle is ${}^{4}_{2}\text{He}$, so $A$ falls by 4 and $Z$ by 2.
Q2.The relation between half-life and decay constant is:
Explanation: Setting $N = N_0/2$ in the decay law gives $T_{1/2} = \frac{0.693}{\lambda}$.
Q3.After 3 half-lives, the fraction of a radioactive sample remaining is:
Explanation: Remaining fraction $= \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Q4.Energy released in the Sun is mainly due to:
Explanation: The Sun fuses hydrogen into helium, releasing energy via $E = \Delta m\, c^2$.
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