A radioactive sample decays with a mean life of 20ms. A capacitor of capacitance `100muF` is charged to some potential and then the plates are connected to a resistance .R.. If the ratio of the charge on the capacitance to the activity of radioactive sample remains constant in time is `R=100(x)Omega`, then x=

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to analyze the relationship between the charge on the capacitor and the activity of the radioactive sample over time. We are given the mean life of the radioactive sample and the capacitance of the capacitor.

### Step-by-Step Solution:

1. **Understand the Mean Life and Decay Constant**:
- The mean life (τ) of the radioactive sample is given as 20 ms.
- The decay constant (λ) is related to the mean life by the formula:
\[ \lambda = \frac{1}{\tau} \]
- Substituting the value of τ:
\[ \lambda = \frac{1}{20 \times 10^{-3}} = 50 \, \text{ms}^{-1} \]

2. **Activity of the Radioactive Sample**:
- The activity (A) of the radioactive sample at time t is given by:
\[ A = A_0 e^{-\lambda t} \]
- Where \( A_0 \) is the initial activity.

3. **Charge on the Capacitor**:
- The charge (Q) on the capacitor at time t is given by:
\[ Q = Q_0 e^{-\frac{t}{RC}} \]
- Where \( Q_0 \) is the initial charge, R is the resistance, and C is the capacitance.

4. **Setting Up the Ratio**:
- According to the problem, the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant:
\[ \frac{Q}{A} = \text{constant} \]
- Substituting the expressions for Q and A:
\[ \frac{Q_0 e^{-\frac{t}{RC}}}{A_0 e^{-\lambda t}} = \text{constant} \]

5. **Simplifying the Equation**:
- This can be simplified to:
\[ \frac{Q_0}{A_0} e^{-\left(\frac{1}{RC} - \lambda\right)t} = \text{constant} \]
- For this ratio to remain constant over time, the exponent must equal zero:
\[ \frac{1}{RC} - \lambda = 0 \]
- Thus, we have:
\[ \frac{1}{RC} = \lambda \]

6. **Substituting Values**:
- We know \( \lambda = 50 \, \text{ms}^{-1} \) and \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \).
- Rearranging gives:
\[ R = \frac{1}{\lambda C} = \frac{1}{50 \times 10^{-3} \times 100 \times 10^{-6}} \]
- Calculating R:
\[ R = \frac{1}{5 \times 10^{-9}} = 200 \, \Omega \]

7. **Finding x**:
- The problem states that \( R = 100x \, \Omega \).
- Therefore:
\[ 200 = 100x \implies x = \frac{200}{100} = 2 \]

### Final Answer:
Thus, the value of \( x \) is \( 2 \).

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