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.

Question

An ideal inductor, a resistor of resistance $R$ Ohms and a capacitor with adjustable capacitance are connected in series to an alternating voltage with an effective value of $V$ Volts and frequency of $f$ Hz. The current flowing through the circuit when the capacitance of the capacitor is set to $C_1$ is the same as when the capacitance of the capacitor is set to $C_2$, $C_2 > C_1$. The inductance of the inductor $L$ is given by:

Answer 1

Idea: Match the two time constants so both quantities fall off at the same rate.

Radioactive side: Activity is $A = \lambda N = \lambda N_0 e^{-\lambda t}$. The mean (average) life is $\tau = 1/\lambda = 20\ \text{ms}$, so $A \propto e^{-t/\tau}$ with $\tau = 20\times 10^{-3}\ \text{s}$.

Capacitor side: A charged capacitor released across a resistor loses charge as $Q \propto e^{-t/(RC)}$, so its characteristic time is $RC$.

Constant ratio condition: The quotient $Q/A$ carries the factor $e^{-t/(RC)}/e^{-t/\tau} = e^{-t\left(1/RC - 1/\tau\right)}$. It is time independent only when the decay rates are identical:
\[ \frac{1}{RC} = \frac{1}{\tau} \;\Rightarrow\; RC = \tau. \]
Plugging in the numbers,
\[ R = \frac{\tau}{C} = \frac{0.020}{100\times 10^{-6}} = 200\ \Omega. \]
Hence the resistance must be 200 ohm.
\[ \boxed{R = 200\ \Omega} \]

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The Correct Option is B

Solution and Explanation

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