Question

${}_{84}{\text{Po}}^{210}$ decays with $\alpha$ to ${}_{82}^{206}\text{Pb}$ with a half life of 138.4 days. If 1.0 g of $P{o}^{210}$ is placed in a closed tube, how much helium accumulate in 69.2 days at STP?

• A. 32 mL
• B. 16 mL
• C. 64 mL
• D. 8 mL

Answer 1

Answer: (A)

32 mL

The decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{138.4}=0.005$
$t=\frac{2.303}{\lambda }log\frac{No}{N}$

$69.2=\frac{2.303}{0.005}log\frac{1.0}{N}$

$0.15=log\frac{1.0}{N}$

$\frac{1}{N}=1.414$

$N=0.707$

Mass of Po decayed $=No-N=1-0.707=0.293$ g

Moles of Po decayed $=\frac{0.293}{210}=0.00139$

Moles of He formed $=0.00139$

At STP $1$ mol $He$ occupies a volume of $22.4L$

So $0.00139$ moles $He$ will occupy a volume of $0.00139\times 22.4=0.032L$ or $32$ mL