Average life ofa radioactive sample is 4 ms Initially the total number of nuclie is `N_(0)` A charged capacitor of capacity `20 mu f` is connected across a resistor R. The value of R such that ratio of number of nuclei remaining to the charge on capacitor remains constant with time is

Options

A. `100 Omega`
B. `200 Omega`
C. `300 Omega`
D. `400 Omega`

Answer

To solve the problem, we need to find the value of the resistor \( R \) such that the ratio of the number of radioactive nuclei remaining to the charge on the capacitor remains constant over time.

### Step-by-Step Solution:

1. **Understanding the Decay of Radioactive Nuclei:**
The number of radioactive nuclei remaining at time \( t \) can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei and \( \lambda \) is the decay constant.

2. **Charge on the Capacitor:**
The charge \( Q(t) \) on the capacitor at time \( t \) can be expressed as:
\[ Q(t) = Q_0 e^{-\frac{t}{RC}} \]
where \( Q_0 \) is the initial charge on the capacitor, \( R \) is the resistance, and \( C \) is the capacitance.

3. **Setting Up the Ratio:**
We need to find \( R \) such that the ratio \( \frac{N(t)}{Q(t)} \) remains constant over time:
\[ \frac{N(t)}{Q(t)} = \frac{N_0 e^{-\lambda t}}{Q_0 e^{-\frac{t}{RC}}} \]

4. **Simplifying the Ratio:**
This simplifies to:
\[ \frac{N(t)}{Q(t)} = \frac{N_0}{Q_0} e^{-\left(\lambda - \frac{1}{RC}\right) t} \]
For the ratio to remain constant, the exponent must equal zero:
\[ \lambda - \frac{1}{RC} = 0 \]
Thus, we have:
\[ \lambda = \frac{1}{RC} \]

5. **Finding the Decay Constant \( \lambda \):**
The average life \( T \) of the radioactive sample is related to the decay constant by:
\[ T = \frac{1}{\lambda} \]
Given that the average life \( T = 4 \, \text{ms} = 4 \times 10^{-3} \, \text{s} \), we can find \( \lambda \):
\[ \lambda = \frac{1}{T} = \frac{1}{4 \times 10^{-3}} = 250 \, \text{s}^{-1} \]

6. **Substituting Values:**
Now substituting \( \lambda \) into the equation \( \lambda = \frac{1}{RC} \):
\[ 250 = \frac{1}{R \times 20 \times 10^{-6}} \]

7. **Solving for \( R \):**
Rearranging gives:
\[ R = \frac{1}{250 \times 20 \times 10^{-6}} = \frac{1}{5 \times 10^{-3}} = 200 \, \Omega \]

### Final Answer:
The value of \( R \) such that the ratio of the number of nuclei remaining to the charge on the capacitor remains constant with time is:
\[ R = 200 \, \Omega \]