Question

The nuclide ratio of ${}_1^3\text{H}$ to ${}_1^1\text{H}$ in a sample of water is $8.0\times {10}^{-18}:1$. Tritium undergoes decay with a half-life period of 12.3 year. Number tritium atoms left if 10.0 g of such a sample is present, 40 year after the original sample is collected is :

• A. $5.625\times {10}^{5}atoms$
• B. $8.675\times {10}^{5}atoms$
• C. $4.725\times {10}^{5}atoms$
• D. None of these

Answer 1

The correct option is A $5.625\times {10}^{5}atoms$

The ratio of ${}^3\text{H}:{}^1\text{H}::8\times {10}^{-18}:1$
No. of H atoms in 18 g $H_{2}O$ is 2N
No. of ${}^3\text{H}$ atoms in 18 g $H_{2}O=2N\times 8\times {10}^{-18}$
$=2\times 6.023\times {10}^{23}\times 8\times {10}^{-18}$ atoms
No. of ${}^3\text{H}$ atoms in 10 g $H_{2}O$
$=\frac{2\times 6.023\times {10}^{23}\times 8\times {10}^{-18}\times 10}{18}atoms$
$=5.354\times {10}^6$ atoms
No. of atoms left after 40 year are derived as
Now, $t=\frac{2.303}{\lambda }\log \frac{N_0}{N}$
$40=\frac{2.303\times 12.3}{0.693}\log \frac{5.354\times {10}^6}{N}$
$\therefore N=5.625\times {10}^5$ atoms