An electron is accelerated from rest through a potential difference of $100\text{V}$. Calculate its de Broglie wavelength in metres.

Step-by-step solution

1. The kinetic energy gained by the electron is $K=eV$, where $e=1.602\times 10^{-19}\text{C}$ and $V=100\text{V}$.
2. Thus $K=1.602\times 10^{-19}\times 100=1.602\times 10^{-17}\text{J}$.
3. For a non‑relativistic electron, $K=\frac{p^2}{2m}$, so the momentum $p=\sqrt{2mK}$, with $m=9.11\times 10^{-31}\text{kg}$.
4. Calculate $p=\sqrt{2\times 9.11\times 10^{-31}\times 1.602\times 10^{-17}}=5.40\times 10^{-24}{\text{kgms}}^{-1}$.
5. The de Broglie wavelength is $\lambda =\frac{h}{p}$, where $h=6.626\times 10^{-34}\text{Js}$.
6. Therefore $\lambda =\frac{6.626\times 10^{-34}}{5.40\times 10^{-24}}=1.23\times 10^{-10}\text{m}$.

Concept tested: de Broglie wavelength of accelerated electron

Published 11 September 2026 · How this solution was prepared and checked