The de Broglie wavelength associated with a neutron at room temperature is 1.5 Å. The de Broglie wavelength of same at `327^@` C will be

Options

A. 0.75 Å
B. 1.06 Å
C. 0.375 Å
D. 2.1 Å

Explanation

To find the de Broglie wavelength of a neutron at 327°C, we can follow these steps:

### Step 1: Understand the relationship between de Broglie wavelength and kinetic energy.
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. The momentum \(p\) can be expressed in terms of kinetic energy (KE):
\[ p = \sqrt{2m \cdot KE} \]
Thus, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 2: Determine the kinetic energy at room temperature (27°C).
At room temperature (27°C), we convert this temperature to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
The kinetic energy (KE) of a particle at temperature \(T\) is given by:
\[ KE = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant. Therefore, the kinetic energy at room temperature is:
\[ KE_1 = \frac{3}{2} k \cdot 300 \]

### Step 3: Determine the kinetic energy at 327°C.
Now, convert 327°C to Kelvin:
\[ T_2 = 327 + 273 = 600 \, \text{K} \]
The kinetic energy at this temperature is:
\[ KE_2 = \frac{3}{2} k \cdot 600 \]

### Step 4: Establish the relationship between the two wavelengths.
From the earlier expression for de Broglie wavelength, we can see that:
\[ \lambda_1 \propto \frac{1}{\sqrt{KE_1}} \quad \text{and} \quad \lambda_2 \propto \frac{1}{\sqrt{KE_2}} \]
Thus, we can write the ratio of the two wavelengths as:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{KE_1}{KE_2}} \]

### Step 5: Substitute the kinetic energy values.
Substituting the expressions for kinetic energy:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{\frac{3}{2} k \cdot 300}{\frac{3}{2} k \cdot 600}} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Calculate λ₂ using λ₁.
Given that \(\lambda_1 = 1.5 \, \text{Å}\):
\[ \lambda_2 = \lambda_1 \cdot \frac{1}{\sqrt{2}} = 1.5 \, \text{Å} \cdot \frac{1}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.06 \, \text{Å} \]

### Final Answer:
The de Broglie wavelength of the neutron at 327°C is approximately **1.06 Å**.

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