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VITEEE 2018 Paper

Section: Physics

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Question : 9 of 125

Marks: +1, -0

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

0.922 per minute
0.691 per minute
0.461 per minute
0.230 per minute

Solution:

We know that

\;\frac{dN}{dt}=\lambda N

Now

\;\frac{d N_0}{dt}=\lambda N_0 \;\frac{d N_t}{dt}=\lambda N_t

9750=\lambda N_0 ; 975=\lambda N_t

\;\frac{N_0}{N_t}=\;\frac{9750}{975}=\;\frac{10}{1} \Rightarrow N_0=10 N_t

We know that

N_t = N_0 e^{-\lambda t}

\;\frac{N_t}{N_0}=e^{-\lambda t} \Rightarrow \;\frac{1}{10}=e^{-\lambda 5}

10^{-1}=e^{-5\lambda}

Taking log on both sides,

-1=-5\lambda \times \;\frac{1}{2.303}

\lambda=\;\frac{1}{5} \times 2.303=0.461

per minute.

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