Question 11.17:
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x ${10}^{-10}$ m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) KT at 300K.

(a)
Hint: $$K=\frac{p^{2}}{2m}$$

Step 1: Find the momentum of a neutron.
According to De Broglie
$$\lambda=\frac{h}{p}\\p=\frac{h}{\lambda}$$
Step 2: Find the kinetic energy of a neutron.
$$K=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}$$
$\begin{array}{l}=\frac{\left(6.63×{10}^{-34}{\right)}^{2}}{2×\left(1.40×{10}^{-10}{\right)}^{2}×1.66×{10}^{-27}}=6.75×{10}^{-21}J\\ =\frac{6.75×{10}^{-21}}{1.6×{10}^{-19}}=4.219×{10}^{-2}eV\end{array}$
Hence, the kinetic energy of the neutron is 6.75 x 10-21 J or 4.219 x 10-2 eV.

(b)
Hint: $$\lambda^{\prime}=\frac{h}{\sqrt{2 K m }}$$

Step 1: Find the average kinetic energy of the neutron.
The average kinetic energy of the neutron:
$\begin{array}{l}{K}^{\mathrm{}}=\frac{3}{2}kT\\ =\frac{3}{2}×1.38×{10}^{-23}×300=6.21×{10}^{-21}J\end{array}$

Step 2: Find the wavelength of a neutron.
The relation for the de Broglie wavelength is given as ${\lambda }^{\mathrm{\prime }}=\frac{h}{\sqrt{2{K}^{\mathrm{}}{m}_{}}}$
$\begin{array}{l}{\lambda }^{\mathrm{\prime }}=\frac{6.63×{10}^{-34}}{\sqrt{2×6.21×{10}^{-21}×1.66×{10}^{-27}}}\\ =1.46×{10}^{-10}m=0.146nm\end{array}$
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.