Question

A beam of high energy neutrons is scattered from a metal lattice, where the space between the nuclei is around 0.4nm. To see the quantum diffraction effect, the kinetic energy of electron must be around ??

Answer 1

To see that neutron is a particle with de Broglie wavelength, one needs to see the interference of diffracted waves.

$d \sin \phi = n \lambda_B$

To see the interference at least in the lowest order ($n=1$),

$\displaystyle d = \lambda_B = \frac{h}{p}$

$\displaystyle p = \frac{h}{d} = \frac{6.63 \cdot 10^{-34}}{4 \cdot 10^{-10}} =1.658 \cdot 10^{-24} \; [kg \cdot m/s]$

So the kinetic energy of neutron is

$\displaystyle E_k = \frac{p^2}{2m} = \frac{2.75 \cdot 10^{-48}}{2 \cdot 1.67 \cdot 10^{-27}} = 0.823 \cdot 10^{-21} J = 0.514 \cdot 10^{-2} eV = 5.1\, meV$