A) \[2.01\times {{10}^{13}}s\]
B) \[201\times {{10}^{13}}s\]
C) \[201\times {{10}^{20}}s\]
D) \[0.201\times {{10}^{10}}s\]
Correct Answer: A
Solution :
Let, initial concentration \[=a\] Final concentration\[=a-\frac{2}{3}a=\frac{a}{3}\] \[{{t}_{2/3}}=\frac{2.303}{k}\log \frac{a}{a/3}\] \[=\frac{2.303}{5.48\times {{10}^{-14}}}\log 3\] \[=2.01\times {{10}^{13}}s\]
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