Neutrons in thermal equilibrium with matter have an average kinetic energy of $\left(3/2\right)\mathbf{kT}$, where is the Boltzmann constant and T, which may be taken to be $\mathbf{300}\mathbf{K}$, is the temperature of the environment of the neutrons. (a) What is the average kinetic energy of such a neutron? (b) What is the corresponding de Broglie wavelength?

The given data can be listed below as,

• The average kinetic energy is,$KE=\left(\frac{3}{2}\right)kT$ .
• The value of temperature is, $T=300\text{K}$.

In this question, the de Broglie wavelength of the neutron can be obtained with the help of the value of the average kinetic energy. The relationship between the de Broglie wavelength and average kinetic energy is an inverse one.

The expression to calculate the average kinetic energy of a neutron is expressed as,

$KE=\frac{3}{2}kT$

Here,KE is the average kinetic energy of a neutron and k is the Boltzmann constant whose value is $1.38\times {10}^{-23}\text{J/K}$.

Substitute all the known values in the above equation.

$\begin{array}{rcl}KE& =& \frac{3}{2}\left(1.38\times {10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\\ & =& 6.21\times {10}^{-21}\text{J}\\ & =& \left(6.21\times {10}^{-21}\text{J}\right)\times \left(\frac{\frac{1}{1.6\times {10}^{-19}}\text{eV}}{1\text{J}}\right)\left(\frac{{10}^3\text{meV}}{1\text{eV}}\right)\\ & \approx & 38.8\text{meV}\end{array}$

Thus, the average kinetic energy of a neutron is $38.8\text{meV}$.

The expression to calculate the corresponding de Broglie wavelengthis expressed as,

$\lambda =\frac{h}{\sqrt{2m_n\left(KE\right)}}$

Here, $\lambda$is the de Broglie wavelength, h is the Plank’s constant whose value is $6.63\times {10}^{-34}\text{J}·\text{s}$and $m_n$is the mass of a neutron whose value is $1.66\times {10}^{-27}\text{kg}$.

Substitute all the known values in the above equation.

$\begin{array}{rcl}\lambda & =& \frac{\left(6.63\times {10}^{-34}\text{J}·\text{s}\right)\left(\frac{1\text{N}·\text{m}·\text{s}}{1\text{J}·\text{s}}\right)}{\sqrt{2\left(1.66\times {10}^{-27}\text{kg}\right)\left(6.21\times {10}^{-21}\text{J}\right)\left(\frac{1{\text{N}}^{\text{2}}·{\text{s}}^{\text{2}}}{1\text{kg}·\text{J}}\right)}}\\ & \approx & 1.46\times {10}^{-10}\text{m}\\ & \approx & \left(1.46\times {10}^{-10}\text{m}\right)\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ & \approx & 146\text{pm}\end{array}$

Thus, the corresponding de-Broglie wavelength is $146\text{pm}$.