Question

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

• A. $28.21 \, cm^{3}$
• B. $31.25 \, cm^{3}$
• C. $36.85 \, cm^{3}$
• D. $38.47 \, cm^{3}$

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 $=\frac{1}{\left(2\right)^{1 / 2}}g=0.707 \, g$

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

Now $_{84}Po^{210} \rightarrow _{82}Pb^{206}+_{2}H e^{4}$

210 g Po on decay will produce = 4g He

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

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