Question

Ig ${}_{84}^{210}Po(t_{1/2}=138.4days)$ is disintegrated in a sealed tube at NTP. How much helium will be accumulated after $69.2$ days?

• A. $32ml$
• B. $320ml$
• C. $68ml$
• D. $680ml$

Answer 1

The correct option is A $32ml$

$f_{1/2}=138.4$ days $,t=69.2$ day

$\therefore$ Number of half $=$ lifes $n=\frac{t}{t^{1/2}}=\frac{69.2}{138.4}=\frac{1}{2}$

$\therefore$ Amount of Po left after $\frac{1}{2}$ half life

$=\frac{1}{(2{)}^2}\cdot g=0.707g$

, Amount of Po used in $\frac{1}{2}$ half life $=1-0.707=0.293g$

Now ${{}_{84}PO}^{210}⟶{{}_{82}Pb}^{20C}+2{He}^4$

$210gPo$

in decay will produce $4gHe$

will produce $=\frac{4\times 0.293}{210}$

$={\mathrm{5.581.10}}^{-3}$

Volume of He at $STP=\frac{5.581\times {15}^{-3}\times 22400}{4}$

$\begin{array}{c}=31.25mL\\ =32.25{cm}^3\end{array}$

Correct Answer- A