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

To solve this problem, let us first understand the scenario: We have a capacitor discharging through a resistor and a radioactive sample decaying simultaneously. The given condition is that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant over time.

The relevant equations are as follows:

1. The charge on a capacitor at any time, t, is given by: 

\( Q(t) = Q_0 e^{-\frac{t}{RC}} \)

Where Q₀ is the initial charge, R is the resistance, and C is the capacitance.

2. The activity of a radioactive sample, A(t), is given by:

\( A(t) = A_0 e^{-\frac{t}{\tau}} \)

Where A₀ is the initial activity and τ is the average life of the sample.

We know from the problem statement:

\( \frac{Q(t)}{A(t)} = \frac{Q_0}{A_0} \)

Substitute the expressions for \( Q(t) \) and \( A(t) \):

\( \frac{Q_0 e^{-\frac{t}{RC}}}{A_0 e^{-\frac{t}{\tau}}} = \frac{Q_0}{A_0} \)

Simplifying gives:

\( e^{-\frac{t}{RC}} e^{\frac{t}{\tau}} = 1 \)

Thus:

\( -\frac{t}{RC} + \frac{t}{\tau} = 0 \)

This simplifies to:

\( \frac{1}{RC} = \frac{1}{\tau} \)

Hence:

\( RC = \tau \)

Given \( τ = 30 \, ms = 30 \times 10^{-3} \, s \) and \( C = 200 \, \mu F = 200 \times 10^{-6} \, F \)

Solve for R:

\( R \cdot 200 \times 10^{-6} = 30 \times 10^{-3} \)

\( R = \frac{30 \times 10^{-3}}{200 \times 10^{-6}} \)

\( R = 150 \, \Omega \)

The calculated resistance \( R \) is 150 Ω. This falls within the specified range (150, 150).

Therefore, the value of 'R' should be 150 Ω.