The de-Broglie wavelength of neutron in thermal equilibrium at temperature T is

Options

A. `(30.8)/sqrtTÅ`
B. `(0.308)/TÅ`
C. `(0.025)/sqrtTÅ`
D. `(0.25)/sqrtTÅ`

Answer

To find the de-Broglie wavelength of a neutron in thermal equilibrium at temperature \( T \), 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 of the particle.

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

### Step 3: Express kinetic energy in terms of temperature
In thermal equilibrium, the average kinetic energy of a particle is given by:
\[ K = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature.

### Step 4: Substitute kinetic energy into the momentum formula
Substituting the expression for kinetic energy into the momentum formula, we get:
\[ p = \sqrt{2m \left(\frac{3}{2} k T\right)} = \sqrt{3mkT} \]

### Step 5: Substitute momentum into the de-Broglie wavelength formula
Now, substituting \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

### Step 6: Simplify the expression
To express the wavelength in terms of the mass of the neutron, we use:
- Mass of neutron \( m \approx 1.67 \times 10^{-27} \) kg
- Planck's constant \( h \approx 6.63 \times 10^{-34} \) J·s
- Boltzmann constant \( k \approx 1.38 \times 10^{-23} \) J/K

Substituting these values into the equation gives:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T}} \]

### Step 7: Calculate the numerical value
Calculating the denominator:
\[ \sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T} \]
This will yield a numerical value that can be simplified further.

### Step 8: Final expression for de-Broglie wavelength
After simplification, the expression can be approximated to:
\[ \lambda \approx \frac{30.8}{\sqrt{T}} \text{ angstroms} \]

### Conclusion
Thus, the de-Broglie wavelength of a neutron in thermal equilibrium at temperature \( T \) is:
\[ \lambda = \frac{30.8}{\sqrt{T}} \text{ angstroms} \]