Question

${}_{84}P{o}^{210}$ decays with $\alpha$ - particle to ${}_{82}P{b}^{206}$ with a half life of 138.4 day. If 1.0 g of ${}_{84}P{o}^{210}$ is placed in a scaled tube.
(a) How much helium will accumulate in 69.2 day ? Express the answer in $c{m}^3$ at latm & 273 K.
(b) The volume of He formed if 1 g of $P{o}^{210}{O}_2$ is used.

• A. (a)$31.25c{m}^3\left(b\right)27.104c{m}^3$
• B. (a)$3.125c{m}^3\left(b\right)271.04c{m}^3$
• C. (a)$312.5c{m}^3\left(b\right)2.7104c{m}^3$
• D. None of these

Answer 1

The correct option is A (a)$31.25c{m}^3\left(b\right)27.104c{m}^3$

The half life period $t_{1/2}=138.4$ day and time $t=69.2$ day

Number of half lives $n=\frac{t}{t_{1/2}}=\frac{69.2}{138.4}=0.5$

The amount of $Po$ left after $69.2$ days $=1g\times \frac{1}{2^{0.5}}=0.707g$

The amount of $Po$ used in $69.2$ days $=1-0.707=0.293g$

${}_{84}^{210}Po{\to }_{82}^{206}{+}_2^{4}He$

$210$ g of $Po$ on decaying produces $4$ g of $He$
$0.293$ g of $Po$ decay will produce

$4\times \frac{0.293}{210}=5.58\times {10}^{-3}g$ of He

$1$ mole or $4$ g of $He$ contains $22400c{m}^3$ of $He$ at STP.

$5.58\times {10}^{-3}g$ of $He$ will contain $\frac{22400}{4}\times 5.58\times {10}^{-3}g=31.25c{m}^3$

$1$ g of ${}^{210}Po{O}_2$ is equal to $1\times \frac{210}{210+32}=0.8677g$ of Po-210

Volume of He obtained $=\frac{31.25\times 0.8677}{1}=27.104c{m}^3$