A) \[\frac{1}{e}\]
B) \[1-\frac{1}{e}\]
C) \[\frac{1}{{{e}^{2}}}\]
D) \[1-\frac{1}{{{e}^{2}}}\]
Correct Answer: B
Solution :
By using \[N={{N}_{0}}{{e}^{-\lambda t}}\] and average life time, \[t=\frac{1}{\lambda }\] So, \[N={{N}_{0}}\,{{e}^{-\lambda \times 1\lambda }}={{N}_{0}}{{e}^{-1}}\] \[\Rightarrow \] \[\frac{N}{{{N}_{0}}}={{e}^{-1}}=\frac{1}{e}\]
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