Question

A radioactive sample has an average life of $30\text{ms}$ and is decaying. A capacitor of capacitance $200\mu \text{F}$ is first charged and later connected with resistor ${}^{\prime }{R}^{\prime }$. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of ${}^{\prime }{R}^{\prime }$ should be _____$\Omega$.

Answer 1

Given:
$T_m=30\text{ms};C=200\mu \text{F}$

$\frac{Q}{N}=\frac{Q_0{e}^{-t/RC}}{N_0{e}^{-\lambda t}}=\frac{Q_0}{N_0}{e}^{t(\lambda -\frac{1}{RC})}$

Since $Q/N$ is constant hence,

$e^{t(\lambda -\frac{1}{RC})}=1=e^0$

$\lambda =\frac{1}{RC}$

$R=\frac{1}{\lambda C}=\frac{T_m}{C}=\frac{30\times {10}^{-3}}{200\times {10}^{-6}}=150\Omega$