If the mass of neutron = `1.7 xx 10^(-27)` kg, then the de-Broglie wavelength of neutron of energy 3 eV is `(h = 6.6 xx 10^(-34) J s)`

Options

A. `1.6 xx 10^(-16)m`
B. `1.6 xx 10^(-11)m`
C. `1.4 xx 10^(-10)m`
D. `1.4 xx 10^(-11)m`

Answer

To find the de-Broglie wavelength of a neutron with a given energy, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) can be expressed in terms of kinetic energy (\( K \)):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the neutron and \( K \) is its kinetic energy.

### Step 3: Substitute the expression for momentum into the de-Broglie wavelength formula
Substituting \( p \) into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 4: Convert energy from eV to Joules
Given that the energy \( K = 3 \, \text{eV} \), we need to convert this to Joules:
\[ K = 3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.8 \times 10^{-19} \, \text{J} \]

### Step 5: Substitute known values into the formula
Now we can substitute \( h = 6.6 \times 10^{-34} \, \text{J s} \), \( m = 1.7 \times 10^{-27} \, \text{kg} \), and \( K = 4.8 \times 10^{-19} \, \text{J} \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.7 \times 10^{-27} \times 4.8 \times 10^{-19}}} \]

### Step 6: Calculate the denominator
First, calculate the expression under the square root:
\[ 2 \times 1.7 \times 10^{-27} \times 4.8 \times 10^{-19} = 1.632 \times 10^{-45} \]
Now take the square root:
\[ \sqrt{1.632 \times 10^{-45}} \approx 1.278 \times 10^{-22} \]

### Step 7: Calculate the de-Broglie wavelength
Now substitute back into the equation for \( \lambda \):
\[ \lambda = \frac{6.6 \times 10^{-34}}{1.278 \times 10^{-22}} \approx 5.16 \times 10^{-12} \, \text{m} \]

### Final Answer
Thus, the de-Broglie wavelength of the neutron is approximately:
\[ \lambda \approx 5.16 \times 10^{-12} \, \text{m} \]