A) \[2T\]
B) \[\frac{1}{2}(ln2)T\]
C) \[4T\]
D) \[2(ln2)T\]
Correct Answer: D
Solution :
As \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] \[\frac{25}{100}={{e}^{-\lambda t}}=\frac{1}{4}\] or \[-\lambda t{{\log }_{e}}e=-{{\log }_{e}}4\] \[=-2{{\log }_{e}}2\] \[t=\frac{2{{\log }_{e}}2}{\lambda }\] Because mean life \[T=\frac{1}{\lambda }\] \[t=(2ln2)T\]
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