Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly : (Take half - life of the neutron = 12.8 min)

Options

A. 6000 km
B. 12000 km
C. 18000 km
D. 24000 km

Answer

To solve the problem of finding the distance in free space at which the intensity of a 5 eV neutron beam reduces to half, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life:**
The half-life of a neutron is the time it takes for half of the neutrons in a beam to decay or lose their intensity. In this case, the half-life is given as 12.8 minutes.

2. **Convert Half-Life to Seconds:**
Since we will be calculating distance in terms of kilometers and time in seconds, we need to convert the half-life from minutes to seconds.
\[ \text{Half-life in seconds} = 12.8 \text{ minutes} \times 60 \text{ seconds/minute} = 768 \text{ seconds} \]

3. **Calculate the Kinetic Energy in Joules:**
The kinetic energy (KE) of the neutron is given as 5 eV. We need to convert this energy into Joules.
\[ \text{KE} = 5 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 8 \times 10^{-19} \text{ J} \]

4. **Use the Kinetic Energy Formula to Find Velocity:**
The kinetic energy of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the neutron (approximately \( 1.67 \times 10^{-27} \) kg). Rearranging this formula to find the velocity \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 8 \times 10^{-19}}{1.67 \times 10^{-27}}} \]

5. **Calculate the Velocity:**
Plugging in the values:
\[ v = \sqrt{\frac{16 \times 10^{-19}}{1.67 \times 10^{-27}}} \approx \sqrt{9.58 \times 10^{8}} \approx 3.09 \times 10^{4} \text{ m/s} \approx 31 \text{ km/s} \]

6. **Calculate the Distance:**
Now that we have the velocity, we can calculate the distance traveled in the time equal to the half-life:
\[ \text{Distance} = \text{Velocity} \times \text{Time} = 31 \text{ km/s} \times 768 \text{ s} \approx 23748 \text{ km} \]

7. **Final Result:**
The distance at which the intensity of the neutron beam reduces to half is approximately:
\[ \text{Distance} \approx 24000 \text{ km} \]

### Conclusion:
The distance in free space at which the intensity of a 5 eV neutron beam reduces to half is nearly **24000 km**.