A radioactive substance emits n beat particles in the first 2 s and `0.5` n beta particles in the next 2 s. The mean life of the sample is

Options

A. 4 s
B. 2 s
C. `(2)/((1n 2))`
D. 2 (1n 2) s

Explanation

To find the mean life of the radioactive substance, we can follow these steps:

### Step 1: Understand the emission of beta particles
In the first 2 seconds, the substance emits \( n \) beta particles. In the next 2 seconds, it emits \( 0.5n \) beta particles.

### Step 2: Determine the remaining particles
After the first 2 seconds, the number of remaining beta particles is:
\[ N = n - n = 0 \]
After the next 2 seconds, the number of remaining particles is:
\[ N = 0 - 0.5n = -0.5n \]
However, since the number of particles cannot be negative, we can interpret this as the substance has decayed significantly.

### Step 3: Relate the decay to half-life
The half-life (\( T_{1/2} \)) can be determined from the decay process. The first emission of \( n \) particles indicates that the substance has undergone a significant decay in the first 2 seconds. The remaining particles after the first 2 seconds can be considered as half of the original amount.

### Step 4: Calculate the half-life
From the problem, we can assume that the half-life is the time taken for the number of particles to reduce to half its initial value. Since we see that the number of particles reduces from \( n \) to \( 0.5n \) in 2 seconds, we can say:
\[ T_{1/2} = 2 \text{ seconds} \]

### Step 5: Relate half-life to mean life
The mean life (\( \tau \)) is related to half-life by the formula:
\[ \tau = \frac{T_{1/2}}{\ln 2} \]
Substituting the value of \( T_{1/2} \):
\[ \tau = \frac{2}{\ln 2} \]

### Step 6: Final answer
Thus, the mean life of the sample is:
\[ \tau = \frac{2}{\ln 2} \]

### Conclusion
The mean life of the radioactive substance is \( \frac{2}{\ln 2} \).

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