Question

A certain radioisotope ${}_Z^{A}X$ (Half life $=10$ days) decays to ${}_{Z-2}^{A-4}Y$. If $1g$ atom of ${}_Z^{A}X$ is kept in sealed vessel, how much helium will accumulated in $20$ days?

Answer 1

${}_Z^{A}X⟶$ ${}_{Z-2}^{A-4}Y{+}_2^{4}He$
In two half lives, $\frac{3}{4}$ of the isotope ${}_Z^{A}X$ has disintegrated, i.e., $\frac{3}{4}g$ atom of helium has been formed from $\frac{3}{4}g$ atom of ${}_Z^{A}X$.
Volume of $1g$ atom of helium $=22400mL$
So, Volume of $\frac{3}{4}g$ atom of helium $=\frac{3}{4}\times 22400mL$$=16800mL$