A radioactive element decays by `beta-`emission. A detector racords `n`-beta particles in `2` sec and in next `2` sec it records `0.65n`-beta particles. Find mean life.

Options

A. 4 s
B. 2 s
C. `(2)/((In 2))s`
D. 2 (In 2)s

Explanation

To solve the problem, we need to determine the mean life of a radioactive element that decays by beta emission. We are given the number of beta particles detected in two different time intervals.

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- The decay of a radioactive substance can be described by the exponential decay law. The number of particles detected in a given time interval can be used to find the decay constant (λ).

2. **Setting Up the Problem**:
- In the first 2 seconds, the detector records `n` beta particles.
- In the next 2 seconds (from 2 to 4 seconds), it records `0.65n` beta particles.

3. **Using the Decay Formula**:
- The number of particles remaining after a time \( t \) can be expressed as:
\[ N(t) = N_0 e^{-\lambda t} \]
- Here, \( N_0 \) is the initial number of particles, \( N(t) \) is the number of particles remaining after time \( t \), and \( \lambda \) is the decay constant.

4. **Calculating the Decay Constant**:
- From the first interval (0 to 2 seconds):
\[ N(2) = N_0 - n \]
- From the second interval (2 to 4 seconds):
\[ N(4) = N(2) - 0.65n \]
- We can express \( N(4) \) in terms of \( N_0 \):
\[ N(4) = N_0 e^{-\lambda \cdot 4} \]
- Setting the two expressions for \( N(4) \) equal gives:
\[ N_0 e^{-\lambda \cdot 4} = (N_0 - n) - 0.65n \]
- Simplifying this:
\[ N_0 e^{-\lambda \cdot 4} = N_0 - n - 0.65n = N_0 - 1.65n \]

5. **Finding the Ratio**:
- From the first interval:
\[ N(2) = N_0 e^{-\lambda \cdot 2} = N_0 - n \]
- Dividing the two equations:
\[ \frac{N_0 - 1.65n}{N_0 - n} = e^{-\lambda \cdot 2} \]
- This gives us a way to relate \( n \) and \( \lambda \).

6. **Solving for the Decay Constant**:
- Rearranging and solving for \( \lambda \) will yield the decay constant.

7. **Calculating Mean Life**:
- The mean life \( \tau \) is related to the decay constant by:
\[ \tau = \frac{1}{\lambda} \]
- Since we can express \( \lambda \) in terms of the half-life \( t_{1/2} \):
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
- Given that the half-life is 2 seconds, we can substitute:
\[ \lambda = \frac{\ln 2}{2} \]
- Therefore, the mean life is:
\[ \tau = \frac{1}{\lambda} = \frac{2}{\ln 2} \]

### Final Answer:
The mean life of the radioactive element is \( \frac{2}{\ln 2} \) seconds.