A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life `tau`.Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

Options

A. `(tau)/(C )`
B. `(2tau)/(C )`
C. `(tau)/(2C)`
D. `(3tau)/(2C)`

Answer

To solve the problem, we need to find the value of resistance \( R \) such that the ratio of the electrostatic energy stored in a capacitor to the activity of a radioactive sample remains constant over time.

### Step-by-Step Solution:

1. **Understand the Discharge of a Capacitor**:
The charge \( Q(t) \) remaining on a capacitor after time \( t \) when discharged through a resistance \( R \) is given by:
\[ Q(t) = Q_0 e^{-t/(CR)} \]
where \( Q_0 \) is the initial charge, \( C \) is the capacitance, and \( R \) is the resistance.

2. **Calculate the Energy Stored in the Capacitor**:
The energy \( U \) stored in the capacitor can be expressed as:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
Substituting \( Q(t) \) into this equation gives:
\[ U(t) = \frac{1}{2} \frac{(Q_0 e^{-t/(CR)})^2}{C} = \frac{Q_0^2 e^{-2t/(CR)}}{2C} \]

3. **Determine the Activity of the Radioactive Sample**:
The activity \( A(t) \) of a radioactive sample decaying over time is given by:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial activity and \( \lambda \) is the decay constant. The relationship between \( \lambda \) and the average life \( \tau \) is:
\[ \lambda = \frac{1}{\tau} \]

4. **Formulate the Ratio of Energy to Activity**:
We need to find the ratio \( \frac{U(t)}{A(t)} \):
\[ \frac{U(t)}{A(t)} = \frac{\frac{Q_0^2 e^{-2t/(CR)}}{2C}}{A_0 e^{-\lambda t}} = \frac{Q_0^2 e^{-2t/(CR)}}{2C A_0 e^{-\lambda t}} \]
This simplifies to:
\[ \frac{U(t)}{A(t)} = \frac{Q_0^2}{2C A_0} e^{-\left(2/(CR) - \lambda\right)t} \]

5. **Set the Exponent to Zero for Constant Ratio**:
For the ratio \( \frac{U(t)}{A(t)} \) to remain constant over time, the exponent of \( e \) must be zero:
\[ \frac{2}{CR} - \lambda = 0 \]
Rearranging gives:
\[ \frac{2}{CR} = \lambda \]

6. **Substituting for Decay Constant**:
Since \( \lambda = \frac{1}{\tau} \), we can substitute this into our equation:
\[ \frac{2}{CR} = \frac{1}{\tau} \]

7. **Solving for Resistance \( R \)**:
Rearranging the equation to solve for \( R \):
\[ R = \frac{2\tau}{C} \]

### Final Answer:
The value of resistance \( R \) for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time is:
\[ R = \frac{2\tau}{C} \]