Nuclear Physics And Radioactivity Question 442
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:
[d] $ 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 $
BETA
