Question

A radioactive sample decays with an average life of 20 ms. A capacitor of capacitance 100 µF is charged to some potential. 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? Choose the correct answer.

• A. 100 Ohm
• B. 200 Ohm
• C. 300 Ohm
• D. 10 Ohm

Hint:

Set the capacitor time constant equal to the nuclear mean life: \(RC = \tau\), then \(R = \tau/C\).

Answer 1

The Correct Option is B

Step 1: The charge on a discharging capacitor decays exponentially with the RC time constant:
\[ Q(t) = Q_0\, e^{-t/RC}. \]
Step 2: The activity of a radioactive sample decays exponentially with its mean life \(\tau\) (where the decay constant is \(\lambda = 1/\tau\)):
\[ A(t) = A_0\, e^{-t/\tau}. \]
Step 3: The ratio is
\[ \frac{Q(t)}{A(t)} = \frac{Q_0}{A_0}\, e^{-t\left(\frac{1}{RC} - \frac{1}{\tau}\right)}. \]
For this ratio to be constant in time, the exponent must vanish, i.e. \(\frac{1}{RC} = \frac{1}{\tau}\), so \(RC = \tau\).
Step 4: Solve for \(R\):
\[ R = \frac{\tau}{C} = \frac{20\times 10^{-3}\ \text{s}}{100\times 10^{-6}\ \text{F}} = 200\ \Omega. \]
\[ \boxed{R = 200\ \Omega} \]