In the Davisson–Germer experiment electrons are accelerated through a potential difference of $54\text{V}$ and a first‑order diffraction maximum is observed at a scattering angle of $50^{\circ }$. Using Bragg's law for electron diffraction, calculate the interplanar spacing $d$ of the crystal.

Step-by-step solution

1. First find the de Broglie wavelength of the electrons: kinetic energy $K=eV=54\times 1.6\times 10^{-19}=8.64\times 10^{-18}\text{J}$.
2. Momentum $p=\sqrt{2mK}=\sqrt{2\times 9.11\times 10^{-31}\times 8.64\times 10^{-18}}=1.255\times 10^{-23}\text{kg⋅m/s}$.
3. Wavelength $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{1.255\times 10^{-23}}=5.28\times 10^{-11}\text{m}=0.0528\text{nm}$.
4. For first‑order diffraction ($n=1$) Bragg's law is $2d\sin \theta =n\lambda$.
5. Insert $\theta =50^{\circ }$: $\sin 50^{\circ }=0.7660$.
6. Solve for $d$: $d=\frac{\lambda }{2\sin \theta }=\frac{5.28\times 10^{-11}}{2\times 0.7660}=3.45\times 10^{-11}\text{m}=0.0345\text{nm}$.

Concept tested: Electron diffraction and Bragg’s law

Published 7 September 2026 · How this solution was prepared and checked