Answer to Question #203290 in Atomic and Nuclear Physics for Yash sharma

Question #203290

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 ??


1
Expert's answer
2021-06-07T09:32:19-0400

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

dsinϕ=nλBd \sin \phi = n \lambda_B

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

d=λB=hp\displaystyle d = \lambda_B = \frac{h}{p}

p=hd=6.63103441010=1.6581024  [kgm/s]\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

Ek=p22m=2.75104821.671027=0.8231021J=0.514102eV=5.1meV\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


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