Question

${}_{84}^{210}\text{Po}$ decay with $\alpha$-particle to ${}_{82}^{206}\text{Pb}$ with a half-life of 138.4 day. If 1.0 g of ${}_{84}^{210}\text{Po}$ is placed in a sealed tube, the volume (in mL at STP) of helium will accumulate in 69.2 day is

(Write the answer to the nearest integer)

Answer 1

Given, the half life period $t_{1/2}=138.4$ day and the decay period $t=69.2$ day
$\because t=n\times t_{1/2}\therefore n=\frac{1}{2}$
Mass of Po left after $\frac{1}{2}$ half lives $=1/\left(\sqrt{2}\right)=0.707g$
Mass of Po used in $\frac{1}{2}$ half lives $=1-0.707=0.293g$
The nuclear reaction is ${}_{84}^{210}\text{Po}{⟶}_{82}^{206}\text{Pb}{+}_2^4\text{He}$
$210$ g Po on decay gives 4 g He
0.293 g Po on decay will give $\left[\right(4\times 0.293\left)/210\right]$
$=5.581\times {1.0}^{-3}gHe$
Now, the ideal gas equation is $PV=nRT$
$1\times V=\left[\right(5.581\times {10}^{-3}\left)/4\right]\times 0.0821\times 273$
The volume (in mL at STP) of helium will accumulate in 69.2 day is $V=31.2\times {10}^{-3}L=31.2mL$