9.3 Nuclear Kinetics
All radioactive decay processes follow first-order kinetics.
9.3.1 First-order Rate Law
\[\mathrm{rate} = \lambda N\]
- rate is the rate of reaction
- λ is the decay constant
- N is the number of nuclei (in g, mol, or number of atoms)
9.3.2 First-order Integrated Rate Law
\[\ln\dfrac{N_t}{N_0} = -\lambda t\]
- N0 is the number of initial nuclei (in g, mol, or number of atoms)
- Nt is the number of nuclei remaining after time t (in g, mol, or number of atoms)
- t is time
- λ is the decay constant
9.3.4 Radioactive Half-life
Cobalt-60 is a radioactive isotope that is used to treat cancer. It has a first-order half-life of 5.27 years. Every 5.27 years, half of a sample of cobalt-60 decays into nickel-60 via a β decay and emits strong gamma rays via the following process:
\[^{60}_{27}\mathrm{Co} \longrightarrow ^{60}_{28}\mathrm{Ac} + ^{\phantom{-}0}_{-1}e + \gamma\]
The intensity of the radiation decreases as the sample of cobalt-60 decays. Therefore, cobalt-60 sources must be replaced regularly.
)](files/ch21/cobalt-60-half-life.png)
Figure 9.1: Binding energy diagram for various isotopes (Image from openStax)
9.3.5 Example: First-order kinetics
Cobalt-60 (molar mass = 59.93 g mol–1) has a half-life of 5.27 years. What is the decay constant (in y–1) for this process?
Use the first-order half-life equation and solve for k.
\[\begin{align*} t_{1/2} &= \dfrac{0.693}{\lambda} \\[1.5ex] \lambda &= \dfrac{0.693}{t_{1/2}}\\[1.5ex] &= \dfrac{0.693}{5.27~\mathrm{y}} \\[1.5ex] &= 0.131~\mathrm{y^{-1}} \end{align*}\]
If you had 100.0 g of cobalt-60, how much (in g) of cobalt-60 would remain after 30 years?
Solve the first-order integrated rate law. Use the decay constant we previously solved for.
\[\begin{align*} \ln\dfrac{N_t}{N_0} &= -\lambda t \\ N_t &= N_0 e^{-\lambda t}\\ &= (100~\mathrm{g})e^{(-0.131~\mathrm{y^{-1}})(30~\mathrm{y})}\\ &= 1.96~\mathrm{g} \end{align*}\]