The de-Broglie wavelength of a neutron at `1327^@C` is `lambda` . What will be its wavelength at `127^@C` ?

Options

A. `lambda`
B. `2 lambda`
C. `lambda/2`
D. `8lambda`

Answer

To solve the problem of finding the de-Broglie wavelength of a neutron at 127°C given its wavelength at 1327°C, we can follow these steps:

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

### Step 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.

### Step 3: Express Kinetic Energy in Terms of Temperature
For an ideal gas, the average kinetic energy is given by:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

### Step 4: Substitute Kinetic Energy into the Wavelength Formula
Substituting the expression for kinetic energy into the momentum formula gives:
\[ p = \sqrt{2m \cdot \frac{3}{2} k T} = \sqrt{3mkT} \]
Now substituting this back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

### Step 5: Determine the Relationship Between Wavelength and Temperature
From the above equation, we can see that:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
This implies that the wavelength is inversely proportional to the square root of the temperature.

### Step 6: Set Up the Ratio for Different Temperatures
Let \( \lambda_1 \) be the wavelength at \( T_1 = 1327°C \) and \( \lambda_2 \) be the wavelength at \( T_2 = 127°C \). We can express this relationship as:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} \]

### Step 7: Convert Temperatures to Kelvin
Convert the temperatures from Celsius to Kelvin:
- \( T_1 = 1327 + 273 = 1600 \, K \)
- \( T_2 = 127 + 273 = 400 \, K \)

### Step 8: Calculate the Ratio of Wavelengths
Now, substituting the values into the ratio:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{1600}{400}} = \sqrt{4} = 2 \]
This means:
\[ \lambda_2 = 2 \lambda_1 \]

### Conclusion
The de-Broglie wavelength of the neutron at 127°C will be twice that at 1327°C.

### Final Answer
The wavelength at 127°C is \( 2\lambda \).

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