Electrons are accelerated through a potential difference of 600 V and then incident on a nickel crystal whose lattice planes have spacing $d=0.190$ nm. Using Bragg’s law for electron diffraction, determine the angle at which the first‑order diffraction maximum occurs.

Step-by-step solution

1. Compute the de Broglie wavelength of the accelerated electrons: $E_{\text{kin}}=eV=600,\text{eV}=600\times 1.602\times 10^{-19},\text{J}=9.612\times 10^{-17},\text{J}$, $p=\sqrt{2m{E}_{\text{kin}}}=1.323\times 10^{-23},\text{kg⋅m/s}$, $\lambda =h/p=6.626\times 10^{-34}/1.323\times 10^{-23}=5.009\times 10^{-11},\text{m}=0.0501,\text{nm}$.
2. Apply Bragg’s law for first‑order ($n=1$): $nlambda=2dsin\theta$. Solve for $\theta$: $sin\theta =\frac{\lambda }{2d}=\frac{0.0501}{2\times 0.190}=0.1318$. $\theta =\arcsin (0.1318)=7.57^{\circ }$.

Concept tested: Electron diffraction, Bragg’s law

Published 20 September 2026 · How this solution was prepared and checked