The de-broglie wavelength of neutron at `27^(@)C` is `lambda` . The wavelength at `927^(@)C` will be

Options

A. `(lambda)/(9)`
B. `(lambda)/(4)`
C. `(lambda)/(2)`
D. `(lambda)/(3)`

Explanation

To find the de Broglie wavelength of a neutron at a temperature of 927°C given that its wavelength at 27°C is λ, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

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

3. **Kinetic Energy of a Gas Molecule**:
For a gas molecule, the kinetic energy can be expressed as:
\[ KE = \frac{f}{2} k T \]
where \( f \) is the degrees of freedom, \( k \) is Boltzmann's constant, and \( T \) is the absolute temperature in Kelvin.

4. **Substituting Kinetic Energy into the Wavelength Formula**:
Substitute the expression for kinetic energy into the wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot \frac{f}{2} k T}} = \frac{h}{\sqrt{fmkT}} \]

5. **Calculate the Wavelength at Different Temperatures**:
- For 27°C (300 K):
\[ \lambda_{27} = \frac{h}{\sqrt{fmk \cdot 300}} \]
- For 927°C (1200 K):
\[ \lambda_{927} = \frac{h}{\sqrt{fmk \cdot 1200}} \]

6. **Express the Wavelength Ratio**:
To find the relationship between the two wavelengths:
\[ \frac{\lambda_{927}}{\lambda_{27}} = \frac{\sqrt{300}}{\sqrt{1200}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

7. **Final Result**:
Therefore, if the wavelength at 27°C is λ, then the wavelength at 927°C will be:
\[ \lambda_{927} = \frac{\lambda}{2} \]

### Conclusion:
The wavelength at 927°C will be \( \frac{\lambda}{2} \).