A) 24
B) 4
C) 2
D) 8
Correct Answer: B
Solution :
Since, all radioactive process follows first order kinetics, \[k\text{ }=\text{ }\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] \[=\frac{2.303}{12}\log \,8\] \[k=0.1733\] Half-life,\[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.1733}\] \[=3.99\] \[=4yr\]You need to login to perform this action.
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