A radioactive isotope has an initial activity of . Its activity is reduced to exactly half an hour l — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

The activity halves from $28$ to $14\text{ dpm}$ in $30\text{ minutes}$. Therefore, the half-life $T_{1/2} = 30\text{ minutes}$. The decay constant $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{30} = 0.0231\text{ min}^{-1}$. The initial activity $A_0 = \lambda N_0$. Substituting the values: $28 = 0.0231 \times N_0$. Therefore, initial number of nuclei $N_0 = 28 / 0.0231 \approx 1212$.