A) 69.3%
B) 63%
C) 50%
D) 36%
Correct Answer: B
Solution :
From the law of Rutherford and Soddy, we have \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] But mean life \[T=\frac{1}{\lambda }\] \[\therefore \] \[\frac{N}{{{N}_{0}}}={{e}^{-t/T}}\] At \[t=T\] \[\frac{N}{{{N}_{0}}}={{e}^{-1}}=\frac{1}{e}\] \[=0.37\] Therefore, percentage decay\[=1-0.37\] \[=0.63=63%\]
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