Question

${}_{84}P{o}^{210}$ decays with a particle to ${}_{82}P{b}^{206}$ with a half-life of 138.4 days. If 1.0 g of ${}_{84}P{o}^{210}$ is placed in a sealed tube, how much helium will accumulate in 69.2 days. Express the answer in $c{m}^3$ at STP.

• A. $28.21c{m}^3$
• B. $31.25c{m}^3$
• C. $36.85c{m}^3$
• D. $38.47c{m}^3$

Answer 1

Answer: (B)

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

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

Amount of Po left $\frac{1}{2}$ after half-life $=\frac{1}{{\left(2\right)}^{1/2}}g=0.707g$

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

Now ${}_{84}P{o}^{210}{\to }_{82}P{b}^{206}{+}_{2}H{e}^4$

210 g Po on decay will produce = 4g He

0.293 g Po on decay will produce $=\frac{4\times 0.293}{210}=5.581\times 10^{-3}gHe$

Volume of He at STP $=\frac{5.581\times 10^{-3}\times 22400}{4}=31.25mL=31.25c{m}^3$