Question

A vessel of volume $125{\text{cm}}^3$ contains tritium ${(}^3\text{H},t_{\frac{1}{2}}=12.3\text{years})$ at $500k\text{Pa}$ and $300\text{K}.$ Then,
$[\text{Take Bolzmann constant}{k}_B=1.38\times {10}^{-23}{\text{JK}}^{-1}]$

• A. The activity of the gas is $729\text{Ci}$.
• B. The activity of the gas is $540\text{Ci}$.
• C. The decay constant is $1.81\times {10}^{-9}{\text{s}}^{-1}$.
• D. The decay constant is $3.52\times {10}^{-9}{\text{s}}^{-1}$.

Answer 1

Answer: (A), (C)

The decay constant is $1.81\times {10}^{-9}{\text{s}}^{-1}$.

Given:
$V=125{\text{cm}}^{3}P=500k\text{Pa}T=300\text{K}$

$t_{\frac{1}{2}}=12.3\text{years}=3.82\times {10}^8\text{s}$

We know that,
$\text{Activity}A=\lambda \times N$

From, ideal gas equation,

$PV=\mu RT=N{k}_{B}T$

$\Rightarrow N=\frac{PV}{k_{B}T}$

$\Rightarrow N=\frac{500\times {10}^3\times 125\times {10}^{-6}}{1.38\times {10}^{-23}\times 300}$

$\Rightarrow N=1.5\times {10}^{22}\text{atoms}$

Also, we know that,

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$\Rightarrow \lambda =\frac{0.693}{3.82\times {10}^8}$

$\Rightarrow \lambda =1.81\times {10}^{-9}{\text{s}}^{-1}$

$\text{Activity}A=\lambda N$

$\Rightarrow A=1.81\times {10}^{-9}\times 1.5\times {10}^{22}$

$\Rightarrow A=2.7\times {10}^{13}\text{disintegration /sec}$

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

$\Rightarrow A=\frac{2.7\times {10}^{13}}{3.7\times {10}^{10}}\text{Ci}=729\text{Ci}$

Hence, option $\left(B\right)$ and $\left(D\right)$ are correct.