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

Step 1 - Calculate the Number of Elapsed Half-Lives $$n = \frac{t}{t_{1/2}} = \frac{12 \text{ days}}{3 \text{ days}} = 4$$ Four half-lives have elapsed during transit. Step 2 - Relate Remaining Mass to Initial Mass $$N = N_0 \left(\frac{1}{2}\right)^n \implies N_0 = N \cdot 2^n$$ Step 3 - Substitute and Compute Given $N = 3$ g and $n = 4$: $$N_0 = 3 \text{ g} \times 2^4 = 3 \times 16 = \boxed{48 \text{ g}}$$ Verification: $48 \xrightarrow{3\text{ d}} 24 \xrightarrow{3\text{ d}} 12 \xrightarrow{3\text{ d}} 6 \xrightarrow{3\text{ d}} 3$ g ✓ Step 4 - Evaluate Options - **(A) 12 g** — After 4 half-lives: $12 \times (1/2)^4 = 0.75$ g $\neq$ 3 g. Incorrect. - **(B) 24 g** — After 4 half-lives: $24 \times (1/2)^4 = 1.5$ g $\neq$ 3 g. Incorrect. - **(C) 36 g** — After 4 half-lives: $36 \times (1/2)^4 = 2.25$ g $\neq$ 3 g. Incorrect. - **(D) 48 g** — After 4 half-lives: $48 \times (1/2)^4 = 3$ g ✓. Correct.