A radioactive sample has an initial activity of 28 dpm. Half hour later, the activity is 14 dpm. How — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Activity halved from 28 to 14 dpm in 0.5 hours, so $t_{1/2} = 30 \text{ min}$. Decay constant $\lambda = 0.7 / 30 \text{ min}^{-1}$. Initial activity $A_0 = \lambda N_0 \implies 28 = (0.7/30) N_0 \implies N_0 = 1200$. Therefore, correct answer is A.