A) \[\frac{1}{6\lambda }\]
B) \[\frac{1}{3\lambda }\]
C) \[\frac{3}{6\lambda }\]
D) \[\frac{6}{9\lambda }\]
Correct Answer: B
Solution :
From \[{{N}_{1}}{{e}^{-{{\lambda }_{1}}\,t}}={{N}_{2}}{{e}^{-{{\lambda }_{2}}\,t}}\] \[\frac{{{N}_{1}}}{{{N}_{2}}}={{e}^{-({{\lambda }_{2}}-{{\lambda }_{1}})t}}\] \[\frac{1}{e}={{e}^{-({{\lambda }_{2}}-{{\lambda }_{1}})t}}\] \[\left( Given\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{e} \right)\] \[1=(6\lambda -3\lambda )\,t\] \[t=\frac{1}{3\lambda }\]You need to login to perform this action.
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