Question

A radioactive sample decays with an average life of $20ms$. A capacitor of $100\mu F$ is charged to some potential and then the plates are connected through a resistance $R$ . What should be the value of '$20\times R$' (in $k\Omega$), so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time ?

Answer 1

Activity of sample at time 't' is given by
$A=A_0{e}^{-\lambda t}$.
The charge on capacitor at time 't' is given by
$Q=Q_0{e}^{-\frac{t}{RC}}$.
$\frac{Q}{A}=\frac{Q_0{e}^{-\frac{t}{RC}}}{A_0{e}^{-\lambda t}}$ should be independent of time, therefore
$\lambda =\frac{1}{RC}$
$\Rightarrow R=\frac{1}{\lambda C}=\frac{t_{avg}}{C}=\frac{20\times {10}^{-3}}{100\times {10}^{-6}}=200\Omega$
($\because t_{avg}=\frac{1}{\lambda }$)
So, '$20\times R$' (in $k\Omega$) is 4.