The de - Broglie wavelength of a neutron at `27 ^(@) C` is `lambda`. What will be its wavelength at `927^(@) C`?

Options

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

Explanation

To find the de Broglie wavelength of a neutron at a temperature of \(927^\circ C\), given that its wavelength at \(27^\circ C\) is \(\lambda\), we can follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
- For \(27^\circ C\):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \(927^\circ C\):
\[ T_2 = 927 + 273 = 1200 \, K \]

### Step 2: Understand the relationship between de Broglie wavelength and temperature
- The de Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{p} \]
where \(p\) is the momentum.
- For a particle in thermal equilibrium, the momentum can be related to temperature. The root mean square speed \(v_{rms}\) of the particles is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{m}} \]
where \(R\) is the gas constant and \(m\) is the mass of the neutron.

### Step 3: Relate the wavelengths at different temperatures
- The de Broglie wavelength is inversely proportional to the momentum:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
- Therefore, we can write the ratio of wavelengths at two different temperatures:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}} \]

### Step 4: Substitute the values of temperatures
- Substitute \(T_1 = 300 \, K\) and \(T_2 = 1200 \, K\):
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2 \]
- This implies:
\[ \lambda_2 = \frac{\lambda_1}{2} \]

### Step 5: Conclusion
- Therefore, the wavelength at \(927^\circ C\) is:
\[ \lambda_2 = \frac{\lambda}{2} \]

### Final Answer
The de Broglie wavelength of the neutron at \(927^\circ C\) is \(\frac{\lambda}{2}\).

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