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?

Answer

`200 omega`

Explanation

Let at `t=0`,
Capacitor starts discharging then at time t, activity of radioactive sample `=R=R_(0)e^(-lambdat)`
charge on capacitor `=Q=Q_(0)e^(-t//RC)`
Now `R/Q=(R_(0)e^(-lambdat))/(Q_(0)e^(-t//RC))=R_(0)/Q_(0) e^(-t(lambda-1//RC))`
It is independent of time if `lambda=1/(RC)`
`R=1/(lambdaC)=(t_(avg))/C=(20xx10^(-3))/(100xx10^(-6))=200 Omega`