One mole of a radioactive nuclide is present in a closed vessel and undergoes decay via the reaction — Nuclear Chemistry and Radioactivity Chemistry Question
Question
One mole of a radioactive nuclide $A$ is present in a closed vessel and undergoes decay via the reaction: ${}_{Z}^{M}A \rightarrow {}_{Z-4}^{M-8}B + 2 {}_{2}^{4}He$. If the half-life of $A$ is $10\text{ days}$, what is the total volume of Helium gas (in Litres) collected at NTP after exactly $20\text{ days}$? (Assume ideal gas behavior where $1\text{ mole}$ of gas occupies $22.4\text{ L}$ at NTP).
💡 Solution & Explanation
Time elapsed is $20\text{ days}$, which corresponds to $20 / 10 = 2$ half-lives. The fraction of $A$ remaining after $2$ half-lives is $(1/2)^2 = 1/4$. Thus, the fraction of $A$ that has decayed is $1 - 1/4 = 3/4$. Since we started with $1\text{ mole}$ of $A$, $0.75\text{ moles}$ of $A$ have decayed. The stoichiometry of the reaction shows that $1\text{ mole}$ of $A$ produces $2\text{ moles}$ of Helium gas. Therefore, $0.75\text{ moles}$ of $A$ will produce $0.75 \times 2 = 1.5\text{ moles}$ of $He$. The volume of $1.5\text{ moles}$ of ideal gas at NTP is $1.5 \times 22.4\text{ L} = 33.6\text{ Litres}$.