Question

${}_{84}P{o}^{210}$ decays with $\alpha$-particles to ${}_{82}P{b}^{206}$ with a half-life of $138.4$ days. If 1 g of ${}_{84}P{o}^{210}$ is placed in a sealed tube, the amount of helium(in $c{m}^3$) that will accumulate in $69.2$ days will be.

(Write the value to the nearest integer)

Answer 1

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

Number of half lives (n)$=\frac{t}{t_{1/2}}=\frac{69.2}{138.4}=\frac{1}{2}$

Amount of Po left after 1/2 halves $=\frac{1}{(2{)}^{1/2}}=0.707g$

Amount of Po used in 1/2 halves $=1-0.707$ g $=0.293$ g

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

$210$ g of $Po$ on decay will produce $=4$ g $He$

$0.293$ g $Po$ on decay will produce$=\frac{4\times 0.293}{210}$

$=5.581\times {10}^{-3}g$ of $He$

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

$=31.25mL=31.25c{m}^3$