There is a stream of neutrons with a kinetic energy of 0.0327eV. If the half-life of neutrons is 700 seconds, the fraction of neutrons that will decay before they travel a distance of 10 Km . (Given mass of neutron `m=1.6758xx10^(-27)` kg )

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the kinetic energy from eV to Joules
The kinetic energy (KE) of the neutrons is given as 0.0327 eV. To convert this to Joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\).

\[ KE = 0.0327 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 5.232 \times 10^{-21} \, \text{J} \]

### Step 2: Use the kinetic energy to find the speed of the neutrons
The kinetic energy can also be expressed using the formula:

\[ KE = \frac{1}{2} mv^2 \]

Where \(m\) is the mass of the neutron. Rearranging the formula to find \(v\):

\[ v^2 = \frac{2 \times KE}{m} \]

Substituting the values:

\[ v^2 = \frac{2 \times 5.232 \times 10^{-21} \, \text{J}}{1.6758 \times 10^{-27} \, \text{kg}} = 625 \times 10^{4} \, \text{m}^2/\text{s}^2 \]

Taking the square root to find \(v\):

\[ v = \sqrt{625 \times 10^{4}} = 2500 \, \text{m/s} \]

### Step 3: Calculate the time taken to travel 10 km
The distance to be traveled is 10 km, which is \(10^4\) meters. The time \(t\) taken to travel this distance is given by:

\[ t = \frac{\text{distance}}{\text{speed}} = \frac{10^4 \, \text{m}}{2500 \, \text{m/s}} = 4 \, \text{s} \]

### Step 4: Calculate the number of half-lives in the time period
The half-life of the neutron is given as 700 seconds. We can find the number of half-lives \(n\) that occur in 4 seconds:

\[ n = \frac{t}{T_{1/2}} = \frac{4 \, \text{s}}{700 \, \text{s}} = \frac{4}{700} = \frac{1}{175} \]

### Step 5: Calculate the fraction of neutrons that decay
The fraction of neutrons remaining after \(n\) half-lives is given by:

\[ \frac{N}{N_0} = 2^{-n} = 2^{-\frac{1}{175}} \]

Using the approximation \(2^{-x} \approx 1 - \frac{x \ln(2)}{1}\) for small \(x\):

\[ \frac{N}{N_0} \approx 1 - \frac{1}{175} \ln(2) \approx 1 - 0.004 \]

Thus,

\[ \frac{N}{N_0} \approx 0.996 \]

The fraction of neutrons that decay is:

\[ f = 1 - \frac{N}{N_0} = 1 - 0.996 = 0.004 \]

### Final Answer
The fraction of neutrons that will decay before they travel a distance of 10 km is **0.004**.

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