Atoms And Nuclei Question 372

Question: The activity of a radioactive sample is measured as $ N_{0} $ counts per minute at t=0 and $ N_{0}/e $ counts per minute at t=5 minutes. The time (in minutes) at which the activity reduces to half its value is

Options:

A) $ {\log_{e}}2/5 $

B) $ \frac{5}{{\log_{e}}2} $

C) $ 5{\log_{10}}2 $

D) $ 5{\log_{e}}2 $

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Answer:

Correct Answer: D

Solution:

  • $ N=N_{0}{{e}^{-\lambda t}} $

    Here, $ t=5 $ minutes $ \frac{N_{0}}{e}=N_{0}.{{e}^{-5\lambda }}\Rightarrow 5\lambda =1,,\lambda =\frac{1}{5}, $

    Now, $ {T_{1/2}}=\frac{\ell n2.}{\lambda }=5\ell n2 $



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