A) \[\frac{1}{\lambda }\]
B) \[\frac{1}{2\lambda }\]
C) \[\frac{1}{3\lambda }\]
D) \[\frac{1}{4\lambda }\]
Correct Answer: B
Solution :
We know that \[N={{N}_{0}}{{e}^{-\lambda t}}\] For A, \[{{N}_{A}}={{N}_{0}}{{e}^{-5\lambda t}}\] For B, \[{{N}_{B}}={{N}_{0}}{{e}^{-\lambda t}}\] Given, \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{1}{{{e}^{2}}}\] \[\frac{N}{{{N}_{B}}}=\frac{1}{{{e}^{4\lambda t}}}\] \[t=\frac{1}{2\lambda }\]You need to login to perform this action.
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