A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that the — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Number of half-lives $n = \frac{12 \text{ days}}{3 \text{ days}} = 4$. Remaining mass $N = N_o (1/2)^4 = N_o / 16$. Given $N = 3 \text{ g}$, the initial mass $N_o = 3 \times 16 = 48 \text{ g}$. Therefore, correct answer is D.