Question

Verify that, as stated in Module 43-1, neutrons in equilibrium with matter at room temperature, 300 K, have an average kinetic energy of about 0.04 eV.

Short answer

It is proved that the average kinetic energy is 0.04 eV.

Step-by-step solution

Describe the expression for average kinetic energy

The expression for the average kinetic energy is given by,

${\mathit{K}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathit{K}\mathit{T}$ …… (1)

Here, T is temperature in kelvin.

Find the average kinetic energy

Substitute all the known values in equation (1).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{avg}}=\frac{3}{2}\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(300\mathrm{K}\right) \\ =0.04\mathrm{eV} \end{gathered}$$

Therefore, it is proved that the average kinetic energy is 0.04 eV.