A) \[e\]
B) \[\frac{1}{e}\]
C) \[\frac{e-1}{e}\]
D) \[\frac{e}{e-1}\]
Correct Answer: C
Solution :
\[N={{N}_{0}}{{e}^{-\lambda t}}\] and average year\[t=\frac{1}{\lambda }\] \[\therefore \] \[N={{N}_{0}}{{e}^{-\lambda \times 1/\lambda }}={{N}_{0}}{{e}^{-1}}\] \[\frac{N}{{{N}_{0}}}=\frac{1}{e}\] Reduced part \[=1-\frac{N}{{{N}_{0}}}=1-\frac{1}{e}=\frac{e-1}{e}\]
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