A radioactive isotope () decays to give stable atom. If 'm' g of 'A' is taken and kept in a sealed t — Nuclear Chemistry and Radioactivity Chemistry Question
Question
A radioactive isotope ${}_Z\text{A}^m$ ($t_{1/2} = 10 \text{ days}$) decays to give ${}_{Z-6}\text{B}^{m-12}$ stable atom. If 'm' g of 'A' is taken and kept in a sealed tube and $5.6 \times \text{'x'}$ litre of He gas at 1 atm and $0^\circ\text{C}$ is accumulated in 20 days, then the value of 'x' is
💡 Solution & Explanation
The change in mass number is $m - (m-12) = 12$, meaning $3$ alpha particles (${}_2\text{He}^4$) are emitted per decay. 'm' g of 'A' equals 1 mole of 'A'. Time elapsed is 20 days (2 half-lives). Moles of 'A' decayed $= 1 - (1/2)^2 = 0.75 \text{ moles}$. Moles of He produced $= 3 \times 0.75 = 2.25 \text{ moles}$. Volume of He at STP (1 atm, 0$^\circ$C) $= 2.25 \times 22.4 \text{ L} = 50.4 \text{ L}$. Setting this equal to $5.6x$, we get $5.6x = 50.4 \implies x = 9$. Therefore, correct answer is 9.