Question

A certain radioactive isotope ${}_Z{X}^A(t_{1/2}=10day)$ decays to give ${}_{Z-4}{X}^{A-8}$. If $1g$ atom of ${}_Z{X}^A$ is kept in a sealed vessel, how much of $He$ will accumulate in $20$ days at $STP$?

Answer 1

The nuclear reaction is:
${}_Z^A\text{X}⟶{}_{Z-2}^{A-4}\text{Y}{+}_2^4\text{He}$

Given, the half life period $t_{1/2}=10day$, total time of decay $T=20days$, Initial amount of radioisotope $N_0=1g-atoms$

The number of half life periods in the total time of decay $n=\frac{T}{t_{1/2}}=\frac{20}{10}=2$

Amount left in 2 half lives $=\frac{1}{2^2}=\frac{1}{4}g-atom$
Amount of X decayed in 2 half lives $=(1-\frac{1}{4})=\frac{3}{4}g-atom$

Amount of He formed$=\frac{3}{4}g-atom$

(since 1 g-atom of X gives 1 g-atom of He)

Volume of He formed at STP $=\frac{3}{4}\times 22400=16800mL$