A) 37%
B) 87%
C) 63%
D) 69.3%
Correct Answer: A
Solution :
According to Rutherford-soddy law \[N={{N}_{0}}{{e}^{-\lambda \,r}}\] Where No is initial number of atoms present \[\lambda \] the decay constant and t the mean life-time. Applying the formula \[\tau =\frac{1}{\lambda },\]we have, \[N={{N}_{0}}{{e}^{-\lambda \tau }}={{N}_{0}}{{e}^{-1}}=\frac{{{N}_{0}}}{e}\] Percentage decayed \[=\frac{{{N}_{0}}-N}{{{N}_{0}}}\times 100\] \[=\frac{\frac{{{N}_{0}}-{{N}_{0}}}{e}}{{{N}_{0}}}\times 100=63%\]
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