A) 37%
B) 50%
C) 63%
D) 69.3%
Correct Answer: C
Solution :
From Rutherford-Soddy law \[N={{N}_{0}}{{e}^{-\lambda \tau }}\] \[\therefore \]Using,\[\tau =\frac{1}{\lambda }\] we have \[N={{N}_{0}}{{e}^{-1}}=\frac{{{N}_{0}}}{e}\] Percentage decayed\[=\frac{{{N}_{0}}-N}{{{N}_{0}}}\times 100\] \[=\frac{({{N}_{0}}-{{N}_{0}}/e)}{{{N}_{0}}}\times 100\] 63%
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