Question

${}_{84}^{210}$Po decays with $\alpha$-particle to ${}_{82}^{206}Pb$ with a half-life of 138.4 days. If 1.0 g of ${}_{84}^{201}Po$ is placed in a sealed tube how much of helium will accumulate in 69.2 days? Express the answer in $cc$ at STP?

• A. 31.24 cc at STP
• B. 32.44 cc at STP
• C. 33.34 cc at STP
• D. None of these

Answer 1

The correct option is A 31.24 cc at STP

${}_{84}^{210}Po⟶{}_2^{4}He+{}_{82}^{2036}Pb$

$1gPo=\frac{1}{210}molPo$

$Po$ left in 69.2 days $=C=C_0{\left(\frac{1}{2}\right)}^y$

where, $y=\frac{totaltime}{t_{50}}=\frac{69.2}{138.4}=\left(\frac{1}{2}\right)$

$C=1{\left(\frac{1}{2}\right)}^{1/2}=0.7071g$

$Po$ changed to helium $=1-0.7071g$

$=0.2929g$

$=1.3947\times {10}^{-3}mol=n_{He}$

Helium formed $=31.24cc$ at STP

Note: Gas form is He only.