Question

In chernobyl disaster $6.0\times {10}^6$ Ci of ${}^{137}{C}_5$ is released to atmosphere. Mass of ${}^{137}{C}_5$ released if half life of ${}^{137}{C}_5$ is 30.2 years is approximately

• A. 70 kg
• B. 650 kg
• C. 5.8 kg
• D. 200 kg

Answer 1

The correct option is A 70 kg

Activity, $A_0=\lambda N_0$ $[\lambda =\frac{ln2}{T_{\frac{1}{2}}}]$

$6\times {10}^6\times 3.7\times {10}^{10}=\frac{0.693N_0}{30.2\times 365\times 24\times 3600}$

$N_0\approx 3\times {10}^{26}$

$6\times {10}^{23}$ atoms = 137 g = 0.137 kg

$\therefore 3\times {10}^{26}$ atoms $=\frac{0.137}{6\times {10}^{23}}\times 3\times {10}^{26}$ kg

= 68.5 kg