The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, is

Options

A. `h/(sqrt(mkT)`
B. `h/sqrt(3mkT)`
C. `2h/sqrt(3mkT)`
D. `2h/sqrt(mkT)`

Answer

To find the de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature \( T \) (in Kelvin) and mass \( m \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
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.

2. **Relating Momentum to Kinetic Energy**:
The momentum \( p \) can also be expressed in terms of kinetic energy \( K \) as:
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the neutron.

3. **Kinetic Energy in Thermal Equilibrium**:
In thermal equilibrium at temperature \( T \), the average kinetic energy \( K \) of a particle is given by:
\[ K = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant.

4. **Substituting Kinetic Energy into Momentum**:
Now, substituting the expression for kinetic energy into the momentum formula:
\[ p = \sqrt{2m \left(\frac{3}{2} k T\right)} = \sqrt{3mkT} \]

5. **Substituting Momentum into the de Broglie Wavelength Formula**:
Now, substituting this expression for momentum back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

6. **Final Expression**:
Thus, the de Broglie wavelength of the neutron in thermal equilibrium with heavy water at temperature \( T \) is:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]