Electrons accelerated through a potential difference of 500 V are incident on a copper crystal whose lattice planes have spacing 0.208 nm. Using Bragg’s law for electron diffraction, determine the angle at which the first‑order diffraction maximum is observed.

Step-by-step solution

1. Find the electron’s de Broglie wavelength: $\lambda =\frac{h}{\sqrt{2m_{e}eV}}$ with $V=500,$V.
2. Compute $p=\sqrt{2\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}\times 500}=1.207\times 10^{-23},$kg·m/s.
3. Calculate $\lambda =\frac{6.626\times 10^{-34}}{1.207\times 10^{-23}}=5.49\times 10^{-11},$m = $0.0549,$nm.
4. Apply Bragg’s law for first order ($n=1$): $2dsin\theta =\lambda$, with $d=0.208,$nm.
5. Solve for $\theta$: $sin\theta =\frac{0.0549}{2\times 0.208}=0.132$.
6. Take $\theta =\arcsin (0.132)\approx 7.58^{\circ }\approx 7.6^{\circ }$.

Concept tested: Bragg diffraction of electrons

Published 15 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked