$(a)$ Obtain the de Broglie wavelength of a neutron of kinetic energy $150 \; eV$. An electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. $(m_{n} = 1.675 \times 10^{-27} \; kg)$
$(b)$ Obtain the de Broglie wavelength associated with thermal neutrons at room temperature $(27 \; ^\circ C)$. Hence,explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

(N/A) Kinetic energy $K = 150 \; eV = 150 \times 1.6 \times 10^{-19} \; J = 2.4 \times 10^{-17} \; J$.
Mass of neutron $m_{n} = 1.675 \times 10^{-27} \; kg$.
The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2 m_{n} K}}$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}}} \approx 2.33 \times 10^{-12} \; m$.
Since the inter-atomic spacing in a crystal is $\approx 10^{-10} \; m$,and the wavelength $2.33 \times 10^{-12} \; m$ is much smaller than this,the neutron beam is not suitable for diffraction.
$(b)$ At $T = 300 \; K$,the average kinetic energy $K = \frac{3}{2} k_{B} T$.
$K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \; J$.
$\lambda = \frac{h}{\sqrt{2 m_{n} K}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}} \approx 1.45 \times 10^{-10} \; m$.
This wavelength is comparable to the inter-atomic spacing of crystals $(\approx 10^{-10} \; m)$,making thermal neutrons suitable for diffraction experiments.