1 g of an α-emitting nuclide _zX^A (t_1/2 = 10 h) was placed in a sealed container. The time require — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - He Atoms Accumulated Equal Decayed Nuclei Each $\alpha$-decay ejects one $\alpha$-particle (helium nucleus), which captures electrons to form a He atom. So: $$\Delta N_{\text{decayed}} = N_{\text{He}} = 4.52 \times 10^{23}$$ Step 2 - Initial and Remaining Nuclei "1 g of nuclide" = 1 g-atom = 1 mole: $$N_0 = N_A = 6.02 \times 10^{23} \text{ nuclei}$$ Remaining active nuclei: $$N = N_0 - \Delta N = 6.02 \times 10^{23} - 4.52 \times 10^{23} = 1.50 \times 10^{23}$$ Step 3 - Number of Half-Lives $$\frac{N}{N_0} = \frac{1.50 \times 10^{23}}{6.02 \times 10^{23}} \approx 0.25 = \left(\frac{1}{2}\right)^2 \implies n = 2$$ Step 4 - Total Time $$t = n \times t_{1/2} = 2 \times 10 \text{ h} = \boxed{20.0 \text{ h}}$$ Step 5 - Evaluate Options - **(A) 4.52 h** — Incorrect distractor (copies the coefficient of He atoms). - **(B) 9.40 h** — Incorrect; does not correspond to an integer number of half-lives. - **(C) 10.0 h** — Only 1 half-life; $3.01 \times 10^{23}$ He atoms would accumulate, not $4.52 \times 10^{23}$. - **(D) 20.0 h** — Correct. Two half-lives elapsed; three-fourths of nuclei decayed.