# Kinetics of Nuclear Reactions

Nuclear decay processes follow first-order kinetics. Recall the first-order kinetics equations from Ch. 12 (see below). These equations modified to reflect number of nuclei (*N*) rather than concentrations since nuclear reactions occur outside of solution.

### 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)

### First-order Integrated Rate Law

\[\ln\dfrac{N_t}{N_0} = -\lambda t\]

*N*_{0}is the number of initial nuclei (in g, mol, or number of atoms)*N*_{t}is the number of nuclei remaining after time*t*(in g, mol, or number of atoms)*t*is time- λ is the decay constant

### First-order Half-Life

\[t_{1/2} = \dfrac{0.693}{\lambda}\]

- λ is the decay constant

## 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.

(Image from openStax)

### 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.

\[\begin{align*} \ln\dfrac{N_t}{N_0} &= -\lambda t \\ N_t &= N_0 e^{-\lambda t}\\ &= (100~\mathrm{g})e^{(-0.131~\mathrm{y^{-1}})(30~\mathrm{y})}\\ &= 1.96~\mathrm{g} \end{align*}\]