Question

A vessel of $831\text{cc}$ contains ${}_1{\text{H}}^3$ at $0.6\text{atm}$ and $27{}^{\circ }\text{C}$. If the half life of ${}_1{\text{H}}^3$ is $12.3\text{years}$, then the activity of the gas is

• A. $3.04\times {10}^{11}\text{dps}$
• B. $582\text{Ci}$
• C. $2.15\times {10}^{13}\text{dps}$
• D. $823\text{Ci}$

Answer 1

Answer: (B), (C)

$2.15\times {10}^{13}\text{dps}$

Given that,

$P=0.6\text{atm}=0.6\times {10}^5\text{Pa}$
$V=831\text{cc}=831\times {10}^{-6}{\text{m}}^3$
$T=27{}^{\circ }\text{C}=300\text{K}$

The number of moles of gas is given by,

$n=\frac{PV}{RT}$

$\Rightarrow n=\frac{(0.6\times {10}^5)(831\times {10}^{-6})}{\left(8.31\right)\left(300\right)}$

$\Rightarrow n=0.02$

The number of nuclei $N=n{N}_A$

Activity,
$A=\lambda N=\frac{\left(0.693\right)n{N}_A}{T_{\frac{1}{2}}}$

Given that, $T_{\frac{1}{2}}=12.3\text{years}=12.3\times 3.15\times {10}^7\text{s}$

$\Rightarrow A=\frac{\left(0.693\right)\left(0.02\right)(6.02\times {10}^{23})}{12.3\times 3.15\times {10}^7}$

$\Rightarrow A=2.15\times {10}^{13}\text{dps}$

We know that, $1\text{Ci}=3.7\times {10}^{10}\text{dps}$

$\Rightarrow A=\frac{2.15\times {10}^{13}}{3.7\times {10}^{10}}\approx 582Ci$

Hence, option $\left(B\right)\text{and}\left(C\right)$ are correct.

Why this question? Activity of a sample (or decay rate) It is the rate of decay of a radioactive sample. $A=-\frac{dN}{dt}=N\lambda \text{or}A=A_0{e}^{-\lambda t}$