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  • A radioactive sample decays with an average life of . A capacitor of capacitance <img al

A radioactive sample decays with an average life of 20 \mathrm{~ms}. A capacitor of capacitance 100 \mu \mathrm{F} is charged to some potential and then the plates are connected through a resistance R. What should be the value of R so that the ratio of the charge on the capacitor to the activity sample remains constant in time?
 

Option: 1

400 \Omega


Option: 2

100 \Omega


Option: 3

200 \Omega


Option: 4

300 \Omega


Answers (1)

best_answer

The activity of the sample at any time t is given by - \mathrm{A=A_0 e^{-\lambda t}}

The charge on the capacitor \mathrm{Q=Q e^{-t / C R}}

Thus \mathrm{\frac{Q}{A}=\frac{Q_0 e^{-t / C R}}{A_0 e^{-\lambda t}}}
It is independent of t if, \lambda= \mathrm{\frac{1}{C_R} \quad\left[\frac{1}{\lambda}=T\right]}

\begin{aligned} \mathrm{R=\frac{1}{\lambda c} }&\mathrm{ =\frac{T}{c} }\\ &\mathrm{ =\frac{20 \times 10^{-3}}{100 \times 10^{-6}}=200 \Omega }\end{aligned}

Posted by

Deependra Verma

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