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A radioactive sample decays with an aver...

A radioactive sample decays with an average life of `20 ms`. A capacitor of capacitance `100 muF` 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 of the radioactive sample remains constant in time?

Text Solution

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The activity of the sample at time `t` is given by
`A = A_0 e^(-lambdat)`
where `lambda` is the decay constant and `A_0` is the activity at time `t = 0` when the capacitor plates are connected. The charge on the capacitor at time `t` is given by
`Q= Q_0 e^(-t/CR)`
where `Q_0` is the charge at `t = 0` and `C = 100 muF` is the capacitance. Thus,
`Q/A = Q_0/A_0 (e^(-t/CR)/(e^(-lambdat)`.
It is independent of `t` if `lambda = 1/(CR)`
or, `R = 1/(lambdaC) = t_(av)/C = (20 xx 10^(-3) s)/(100 xx 10^(-6) F) = 200 Omega`.
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