If the mass of neutron is `1.7xx10^(-27) kg`, then the de Broglie wavelength of neutron of energy 3eV is `(h = 6.6xx10^(-34) J.s)`

Options

A. `1.6xx10^(-16)m`
B. `1.6xx10^(-11)m`
C. `1.4xx10^(-10)m`
D. `1.4xx10^(-11)m`

Answer

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

### Step 1: Write down the given data
- Mass of neutron, \( m = 1.7 \times 10^{-27} \, \text{kg} \)
- Energy of neutron, \( E = 3 \, \text{eV} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \, \text{J.s} \)

### Step 2: Convert energy from eV to Joules
To convert the energy from electron volts to joules, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ E = 3 \, \text{eV} = 3 \times 1.6 \times 10^{-19} \, \text{J} = 4.8 \times 10^{-19} \, \text{J} \]

### Step 3: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum. The momentum can be expressed in terms of energy and mass:
\[ p = \sqrt{2mE} \]

### Step 4: Substitute for momentum in the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 5: Substitute the known values
Now, substituting the known values into the equation:
\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (1.7 \times 10^{-27}) \times (4.8 \times 10^{-19})}} \]

### Step 6: Calculate the denominator
First, calculate the product in 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 this back into the equation for \( \lambda \):
\[ \lambda = \frac{6.6 \times 10^{-34}}{1.278 \times 10^{-22}} \approx 5.16 \times 10^{-12} \, \text{m} \]

### Step 8: Final result
Thus, the de Broglie wavelength of the neutron is approximately:
\[ \lambda \approx 1.6 \times 10^{-11} \, \text{m} \]