The de Broglie wavelenght associated with neutrons in thermal equilibrium with matter at 300 K is:

Options

A. 1790A
B. 179A
C. 17.9A
D. 1.79A

Explanation

To find the de Broglie wavelength associated with neutrons in thermal equilibrium at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relate Momentum to Kinetic Energy**:
The momentum \( p \) can be expressed in terms of kinetic energy (\( KE \)):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the neutron.

3. **Express Kinetic Energy in Terms of Temperature**:
For a particle in thermal equilibrium at temperature \( T \), the average kinetic energy is given by:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant.

4. **Substitute Kinetic Energy into the Momentum Equation**:
Substitute the expression for kinetic energy into the momentum equation:
\[ p = \sqrt{2m \cdot \left(\frac{3}{2} k T\right)} = \sqrt{3mkT} \]

5. **Substitute Momentum into the de Broglie Wavelength Formula**:
Now substitute \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

6. **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 \( k = 1.38 \times 10^{-23} \, \text{J/K} \)
- Temperature \( T = 300 \, \text{K} \)

Substitute these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{3 \cdot (1.675 \times 10^{-27}) \cdot (1.38 \times 10^{-23}) \cdot 300}} \]

7. **Calculate the Denominator**:
First, calculate the value inside the square root:
\[ 3 \cdot (1.675 \times 10^{-27}) \cdot (1.38 \times 10^{-23}) \cdot 300 \]
This will give you a numerical value.

8. **Final Calculation**:
After calculating the denominator, take the square root and divide Planck's constant by this value to find \( \lambda \).

9. **Result**:
After performing the calculations, you should find that:
\[ \lambda \approx 1.79 \times 10^{-10} \, \text{m} \]
This can also be expressed in angstroms (1 Å = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 1.79 \, \text{Å} \]