A) \[\lambda \]
B) \[\frac{1}{2}\lambda \]
C) \[\frac{1}{4}\,\lambda \]
D) \[\frac{e}{\lambda }\,\]
Correct Answer: C
Solution :
We know that \[N={{N}_{0}}{{e}^{-\,\lambda t}}\] N = number of radioactive nuclei present at some instant \[{{N}_{0}}=nuclei at\,\,t = 0\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{e}^{-\,{{\lambda }_{1}}t}}}{{{e}^{-\,{{\lambda }_{2}}t}}}=\frac{{{e}^{-\,{{\lambda }_{5}}t}}}{{{e}^{-\,\lambda t}}}={{e}^{-\,4\lambda t}}\] \[\Rightarrow \,\,\,\,\,\,\,As,\,\,\,\,\,\,\,\,\,\,\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{e}\] (given) \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{e}={{e}^{-\,4\lambda t}}\,\,\Rightarrow \,\,\,{{e}^{-\,4\lambda t}}=e\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,-4\lambda t=1\,\,\,\Rightarrow \,\,\,t=\frac{1}{4\lambda }\]You need to login to perform this action.
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