Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

To find the de Broglie wavelength of a neutron in thermal equilibrium with matter, we first need to understand the relationship between kinetic energy and the de Broglie wavelength. The de Broglie wavelength (\( \lambda \)) is given by the formula:

Understanding the de Broglie Wavelength

The de Broglie wavelength can be calculated using the equation:

\( \lambda = \frac{h}{p} \)

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 (\( KE \)) as follows:

\( p = \sqrt{2m \cdot KE} \)

Here, \( m \) is the mass of the neutron, and \( KE \) is the average kinetic energy of the neutron. In thermal equilibrium at a temperature \( T \), the average kinetic energy of a particle is given by:

\( KE = \frac{3}{2} k T \)

where \( k \) is the Boltzmann constant. Now, let's break this down step by step.

Step 1: Calculate the Average Kinetic Energy

At a temperature of 300 K, we can calculate the average kinetic energy of the neutron:

\( KE = \frac{3}{2} k T \)

Using the value of Boltzmann's constant \( k \approx 1.38 \times 10^{-23} \, \text{J/K} \), we have:

\( KE = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times 300 \, \text{K} \)

Calculating this gives:

\( KE \approx 6.21 \times 10^{-21} \, \text{J} \)

Step 2: Determine the Momentum

Next, we need to find the momentum of the neutron. The mass of a neutron is approximately \( m \approx 1.675 \times 10^{-27} \, \text{kg} \). Using the kinetic energy we just calculated:

\( p = \sqrt{2m \cdot KE} \)

Substituting the values:

\( p = \sqrt{2 \times (1.675 \times 10^{-27} \, \text{kg}) \times (6.21 \times 10^{-21} \, \text{J})} \)

Calculating this gives:

\( p \approx 1.15 \times 10^{-23} \, \text{kg m/s} \)

Step 3: Calculate the de Broglie Wavelength

Now we can find the de Broglie wavelength using the momentum we just calculated:

\( \lambda = \frac{h}{p} \)

Substituting in Planck's constant \( h \approx 6.626 \times 10^{-34} \, \text{J s} \):

\( \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{1.15 \times 10^{-23} \, \text{kg m/s}} \)

Calculating this yields:

\( \lambda \approx 5.77 \times 10^{-11} \, \text{m} \)

Final Result

Thus, the de Broglie wavelength of a neutron in thermal equilibrium at 300 K, with an average kinetic energy of \( \frac{3}{2} k T \), is approximately \( 5.77 \times 10^{-11} \, \text{m} \) or 57.7 picometers. This wavelength is on the order of atomic dimensions, illustrating the wave-particle duality of matter at the quantum level.