A) \[{{f}_{1}}={{f}_{2}}\]
B) \[{{f}_{1}}=\frac{{{f}_{2}}}{2}\]
C) \[{{f}_{1}}<{{f}_{2}}\]
D) \[{{f}_{1}}>{{f}_{2}}\]
Correct Answer: C
Solution :
[c] \[N={{N}_{0}}\,{{e}^{-\lambda t}}\,\] \[t={{t}_{mean}}=\frac{1}{\lambda }\] \[\therefore \,{{f}_{1}}=\frac{N}{{{N}_{0}}}{{e}^{-1}}=\frac{1}{e}.\] \[t={{t}_{(half)}}=\frac{\ell n2}{\lambda }\] \[\therefore \,{{f}_{2}}=\frac{N}{{{N}_{0}}}{{e}^{-In2}}=\frac{1}{2}\] \[\therefore {{f}_{2}}>{{f}_{1}}\].
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