A) \[\frac{1}{(10\lambda )}\]
B) \[\frac{1}{(11\lambda )}\]
C) \[\frac{11}{(10\lambda )}\]
D) \[\frac{1}{(9\lambda )}\]
Correct Answer: D
Solution :
\[{{N}_{1}}={{N}_{0}}{{e}^{-10\lambda t}}\]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 that \[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{e},\]\[\therefore \]\[\frac{1}{{{e}^{9\lambda t}}}=\frac{1}{e}\]or \[9\lambda t=1\] or \[t=\frac{1}{9\lambda }\] Hence, the correction option is [d].
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