The de Broglie wavelength of an neutron corresponding to root mean square speed at `927^(@)` is `lambda`. What will be the de Broglie wavelength of the neutron corresponding to root mean square speed at `27^(@)C`?

Options

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

Answer

To solve the problem, we need to find the de Broglie wavelength of a neutron at two different temperatures and relate them. Let's break it down step by step.

### 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 Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The kinetic energy (E) of a particle is related to its momentum by:
\[ E = \frac{p^2}{2m} \]
From this, we can express momentum as:
\[ p = \sqrt{2mE} \]

### Step 3: Substitute kinetic energy in terms of temperature
For a neutron, the kinetic energy can also be expressed in terms of temperature (T) as:
\[ E = \frac{3}{2} k T \]
where \( k \) is Boltzmann's constant.

### Step 4: Substitute this back into the de Broglie wavelength equation
Substituting the expression for kinetic energy into the momentum equation gives:
\[ p = \sqrt{2m \left(\frac{3}{2} k T\right)} = \sqrt{3mkT} \]
Now substituting this into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

### Step 5: Determine the relationship between wavelength and temperature
From the equation:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
This indicates that the de Broglie wavelength is inversely proportional to the square root of the temperature.

### Step 6: Convert temperatures from Celsius to Kelvin
1. Convert 927°C to Kelvin:
\[ T_1 = 927 + 273 = 1200 \, K \]
2. Convert 27°C to Kelvin:
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 7: Use the relationship to find the new wavelength
Using the relationship derived:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} \]
Substituting the values of \( T_1 \) and \( T_2 \):
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2 \]
Thus, we have:
\[ \lambda_2 = 2 \lambda_1 \]

### Conclusion
The de Broglie wavelength of the neutron corresponding to the root mean square speed at 27°C is:
\[ \lambda_2 = 2 \lambda \]

### Final Answer
The correct option is **C) 2λ**.

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