A radioactive sample decays with an average - life of `20 ms`. A capacitor of capacitance `100 muF` is changed to a potential V and then a resistance R is connected across the capacitor. What should be value of R so that the ratio of charge on the capacitor to the activity of the radioactive sample remains constant in time.

Explanation

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`.