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 relationship between de-Broglie wavelength and temperature
The de-Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the neutron,
- \( v \) is the velocity of the neutron.

### Step 2: Relate velocity to temperature
The root mean square velocity (\( v_{rms} \)) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass.

From this, we can see that the velocity \( v \) is directly proportional to the square root of the temperature:
\[ v \propto \sqrt{T} \]

### Step 3: Relate wavelength to temperature
Since \( v \) is directly proportional to \( \sqrt{T} \), we can express the wavelength as inversely proportional to the square root of temperature:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]

### Step 4: Set up the equation for the two temperatures
Let \( \lambda_1 \) be the wavelength at 1327°C and \( \lambda_2 \) be the wavelength at 127°C. We can write:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]
where \( T_1 \) and \( T_2 \) are the absolute temperatures corresponding to 1327°C and 127°C, respectively.

### Step 5: Convert temperatures to Kelvin
Convert the given temperatures from Celsius to Kelvin:
- \( T_1 = 1327 + 273 = 1600 \, K \)
- \( T_2 = 127 + 273 = 400 \, K \)

### Step 6: Substitute the values into the equation
Now substituting the temperatures into the equation:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{1600}}{\sqrt{400}} \]

### Step 7: Simplify the equation
Calculating the square roots:
\[ \sqrt{1600} = 40 \]
\[ \sqrt{400} = 20 \]
Thus,
\[ \frac{\lambda_2}{\lambda_1} = \frac{40}{20} = 2 \]

### Step 8: Solve for \( \lambda_2 \)
This means:
\[ \lambda_2 = 2 \lambda_1 \]

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

### Final Answer
If the wavelength at 1327°C is \( \lambda \), then at 127°C, the wavelength will be:
\[ \lambda_2 = 2\lambda \]