A) \[\lambda \]
B) \[\frac{1}{2}\lambda \]
C) \[\frac{1}{4\lambda }\]
D) \[\frac{e}{\lambda }\]
Correct Answer: C
Solution :
If N is the number of radioactive nuclei present at some instant, then\[N={{N}_{0}}{{e}^{-\lambda t}}\] The constant \[{{N}_{0}}\] represents the number of radioactive nuclei at t = 0 Now,\[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{e}^{-{{\lambda }_{1}}t}}}{{{e}^{-{{\lambda }_{2}}t}}}\]or\[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{e}^{-5\lambda t}}}{{{e}^{-\lambda t}}}={{e}^{-4\lambda t}}\] but\[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{e}\](as provided) Therefore,\[\frac{1}{e}=\frac{1}{{{e}^{4\lambda t}}}\]or\[4\lambda t=1\]or\[t=\frac{1}{4\lambda }\]You need to login to perform this action.
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