Electrons are accelerated through $250\text{V}$ and then incident on a copper crystal whose lattice planes have spacing $0.20\text{nm}$. Using Bragg’s law for electron diffraction, determine the angle at which the first‑order diffraction maximum is observed.

Step-by-step solution

1. Calculate the electron’s speed from kinetic energy $eV=\frac{1}{2}{m}_e{v}^2$.
2. Compute $v=\sqrt{2eV/m_e}=\sqrt{\frac{2\times 1.602\times 10^{-19}\times 250}{9.109\times 10^{-31}}}\approx 9.38\times 10^6{\text{m,s}}^{-1}$.
3. Find the de Broglie wavelength $\lambda =\frac{h}{m_{e}v}=\frac{6.626\times 10^{-34}}{9.109\times 10^{-31}\times 9.38\times 10^6}\approx 7.74\times 10^{-11}\text{m}=0.0774\text{nm}$.
4. Apply Bragg’s condition for first order ($n=1$): $2dsin\theta =\lambda$.
5. Compute $sin\theta =\frac{\lambda }{2d}=\frac{0.0774}{2\times 0.20}=0.1935$.
6. Find $\theta =si{n}^{-1}(0.1935)\approx 11.1^{\circ }$.

Concept tested: Electron diffraction (Bragg's law)

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked