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

0

Explanation

The activity of the sample at time t is given by `A=A_(0)e^(-lambda t)` where `lambda` is the decay constant and `A_0` is the activity at time t =0 when the capacitor paltes 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 mu F` Thus, `(Q)/(A)=(Q_(0))/(A_(0)) (e^(-t//CR))/(e^(-lambdat))`
It is independent of `t` is `lambda=(1)/(CR)`
or ``R=(1)/(lambdaC)=(t_(av))/(C)=(20xx10^(-3)s)/(100xx10^(-6)F)=200Omega`.