Question

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

Hint:

This problem cleverly links two different exponential decay processes: capacitor discharge and radioactive decay. For their ratio to be time-independent, their decay rates must be identical, which means their time constants (or mean lifetimes) must be equal. Here, $\tau_{RC} = RC$ and $\tau_{radioactive} = 1/\lambda$.

Answer 1

Correct Answer: 150

The charge q on a capacitor discharging through a resistor R is given by:
$q(t) = q_0 e^{-t/RC}$, where $q_0$ is the initial charge and RC is the time constant.
The activity A of a radioactive sample is given by:
$A(t) = A_0 e^{-\lambda t}$, where $A_0$ is the initial activity and $\lambda$ is the decay constant.
We are given that the ratio $\frac{q(t)}{A(t)}$ is fixed with respect to time (i.e., it is constant).
$\frac{q(t)}{A(t)} = \frac{q_0 e^{-t/RC}}{A_0 e^{-\lambda t}} = \frac{q_0}{A_0} e^{(-\frac{t}{RC} + \lambda t)} = \frac{q_0}{A_0} e^{t(\lambda - \frac{1}{RC})}$
For this ratio to be constant for all t, the exponent of the exponential term must be zero.
$\lambda - \frac{1}{RC} = 0 \implies \lambda = \frac{1}{RC}$
This means the time constant of the RC circuit must be equal to the mean life of the radioactive sample, since the mean life $\tau_{avg} = 1/\lambda$.
So, $RC = \tau_{avg}$.
We are given:
Average life, $\tau_{avg} = 30$ ms = $30 \times 10^{-3}$ s.
Capacitance, $C = 200 \, \mu\text{F} = 200 \times 10^{-6}$ F.
Now we can solve for R:
$R = \frac{\tau_{avg}}{C} = \frac{30 \times 10^{-3} \text{ s}}{200 \times 10^{-6} \text{ F}}$
$R = \frac{30}{200} \times 10^3 = \frac{3}{20} \times 1000 = 3 \times 50 = 150 \, \Omega$.