A) 50 years
B) 67 years
C) 500 years
D) 600 years
Correct Answer: B
Solution :
\[K=\frac{0.693}{10}yea{{r}^{-1}}\] \[K=\frac{2.303}{t}\log \frac{a}{a-0.99a}\] \[\therefore \] \[\frac{0.693}{10}=\frac{2.303}{t}\log {{10}^{2}}\] \[t=\frac{10\times 2.303\times 2}{0.693}\] \[=66.5\,years\,\approx 67\,years\]
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