The wavelength of de-Brogile waves associted with neutrons at room temperature T K is

Options

A. `(1.82)/(T) Å`
B. `(1.82)/(sqrt(T))Å`
C. `(25.15)/(sqrt(T))Å`
D. `(30.7)/(T) Å`

Explanation

To find the wavelength of the de Broglie waves associated with neutrons at room temperature, we can follow these steps:

### Step-by-Step Solution

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda_d \)) is given by the formula:
\[ \lambda_d = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Relate Kinetic Energy to Velocity**:
The kinetic energy (\( KE \)) of a particle can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging this gives us:
\[ v = \sqrt{\frac{2 KE}{m}} \]

3. **Substitute Kinetic Energy in Terms of Temperature**:
For a particle at temperature \( T \), the average kinetic energy can be expressed using the Boltzmann constant (\( k \)):
\[ KE = \frac{3}{2} kT \]
Substituting this into the expression for velocity gives:
\[ v = \sqrt{\frac{2 \cdot \frac{3}{2} kT}{m}} = \sqrt{\frac{3kT}{m}} \]

4. **Substitute Velocity Back into the de Broglie Wavelength Formula**:
Now, substituting \( v \) back into the de Broglie wavelength formula:
\[ \lambda_d = \frac{h}{m \sqrt{\frac{3kT}{m}}} \]
Simplifying this gives:
\[ \lambda_d = \frac{h}{\sqrt{3mkT}} \]

5. **Insert Known Values**:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of neutron \( m = 1.675 \times 10^{-27} \, \text{kg} \)
- Boltzmann constant \( k = 1.38 \times 10^{-23} \, \text{J/K} \)

Thus, we have:
\[ \lambda_d = \frac{6.626 \times 10^{-34}}{\sqrt{3 \cdot (1.675 \times 10^{-27}) \cdot (1.38 \times 10^{-23}) \cdot T}} \]

6. **Final Expression**:
The final expression for the de Broglie wavelength becomes:
\[ \lambda_d \approx \frac{6.626 \times 10^{-34}}{\sqrt{3 \cdot 1.675 \times 10^{-27} \cdot 1.38 \times 10^{-23} \cdot T}} \]
This can be further simplified to yield a numerical value when \( T \) is specified.