The de Broglie wavelength of a thermal neutron at `927^circC` is `lamda`. Its wavelength at `27^circC` will be

Options

A. `(lamda)/(2)`
B. `(lamda)/(sqrt2)`
C. `lamdasqrt2`
D. `2lamda`

Explanation

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

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

### Step 2: Use the relationship between de Broglie wavelength and temperature
The de Broglie wavelength (\(\lambda\)) is inversely proportional to the square root of the temperature:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
This implies:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}} \]

### Step 3: Substitute the known values into the equation
Substituting \(T_1\) and \(T_2\) into the equation:
\[ \frac{\lambda}{\lambda_2} = \sqrt{\frac{300}{1200}} \]

### Step 4: Simplify the fraction under the square root
\[ \frac{300}{1200} = \frac{1}{4} \]
Thus,
\[ \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 5: Solve for \(\lambda_2\)
From the equation:
\[ \frac{\lambda}{\lambda_2} = \frac{1}{2} \]
We can rearrange this to find \(\lambda_2\):
\[ \lambda_2 = 2\lambda \]

### Conclusion
The de Broglie wavelength of the thermal neutron at \(27^\circ C\) is:
\[ \lambda_2 = \frac{\lambda}{2} \]

### Final Answer
The wavelength at \(27^\circ C\) will be \(\frac{\lambda}{2}\).

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