A) \[\frac{1}{10\lambda }\]
B) \[\frac{1}{11\lambda }\]
C) \[\frac{11}{10\lambda }\]
D) \[\frac{1}{9\lambda }\]
Correct Answer: D
Solution :
[d] \[{{N}_{1}}={{N}_{0}}{{e}^{-10\lambda t}}\] and \[{{N}_{2}}={{N}_{0}}{{e}^{-\lambda t}}.\] \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{e}^{-10\lambda t}}}{{{e}^{-\lambda t}}}=\frac{1}{{{e}^{9\lambda t}}}\] Given \[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{e};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\,\,\,\frac{1}{{{e}^{9\lambda t}}}=\frac{1}{e}\] or, \[9\lambda t=1\] or \[t=\left( \frac{1}{9\lambda } \right)\]You need to login to perform this action.
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