Activity of a radioactive sample is measured as 9750 counts per minute at t = 0 and as 975 counts pe

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13.Nuclei

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The activity of a radioactive sample is measured as $9750$ counts per minute at $t = 0$ and as $975$ counts per minute at $t = 5$ minutes. The decay constant is approximately ............ per minute

A

$0.230$

B

$0.461$

C

$0.691$

D

$0.922$

Solution

$A = {A_0}{e^{ – \lambda t}} \Rightarrow 975 = 9750\;{e^{ – \lambda \times 5}}$ $ \Rightarrow {e^{5\lambda }} = 10$

$ \Rightarrow 5\lambda = {\log _e}10 = 2.3026{\log _{10}}10 = 2.3026$

$ \Rightarrow \lambda = 0.461$

Standard 12

Physics

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