A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 `muF` 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`.

Answer

To solve the problem, we need to find the value of the resistance \( R \) such that the ratio of the charge on the capacitor \( Q \) to the activity of the radioactive sample \( N \) remains constant over time.

### Step-by-step Solution:

1. **Understanding the parameters:**
- Average life of the radioactive sample, \( T_m = 30 \, \text{ms} = 30 \times 10^{-3} \, \text{s} \).
- Capacitance of the capacitor, \( C = 200 \, \mu\text{F} = 200 \times 10^{-6} \, \text{F} \).

2. **Radioactive decay and capacitor discharge:**
- The activity \( N \) of the radioactive sample at time \( t \) is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial activity and \( \lambda \) is the decay constant.

- The charge \( Q \) on the capacitor at time \( t \) when discharging through a resistor \( R \) is given by:
\[ Q(t) = Q_0 e^{-\frac{t}{RC}} \]
where \( Q_0 \) is the initial charge on the capacitor.

3. **Setting up the ratio:**
- We are given that the ratio \( \frac{Q}{N} \) is constant:
\[ \frac{Q(t)}{N(t)} = \text{constant} \]
- Substituting the expressions for \( Q(t) \) and \( N(t) \):
\[ \frac{Q_0 e^{-\frac{t}{RC}}}{N_0 e^{-\lambda t}} = \text{constant} \]
- This simplifies to:
\[ \frac{Q_0}{N_0} e^{\left(\lambda - \frac{1}{RC}\right)t} = \text{constant} \]

4. **Condition for constant ratio:**
- For the ratio to remain constant, the exponent must be zero:
\[ \lambda - \frac{1}{RC} = 0 \]
- Rearranging gives:
\[ \lambda = \frac{1}{RC} \]

5. **Finding the decay constant \( \lambda \):**
- The decay constant \( \lambda \) is related to the average life \( T_m \) by:
\[ \lambda = \frac{1}{T_m} \]
- Substituting \( T_m = 30 \times 10^{-3} \, \text{s} \):
\[ \lambda = \frac{1}{30 \times 10^{-3}} = \frac{1000}{30} \approx 33.33 \, \text{s}^{-1} \]

6. **Substituting \( \lambda \) into the equation:**
- From \( \lambda = \frac{1}{RC} \):
\[ 33.33 = \frac{1}{R \cdot 200 \times 10^{-6}} \]
- Rearranging gives:
\[ R = \frac{1}{33.33 \cdot 200 \times 10^{-6}} \]

7. **Calculating \( R \):**
- Performing the calculation:
\[ R = \frac{1}{33.33 \cdot 200 \times 10^{-6}} = \frac{1}{0.006666} \approx 150 \, \Omega \]

### Final Answer:
The value of \( R \) should be \( 150 \, \Omega \).

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