An electron is accelerated through a potential difference of 200 V. Calculate its de Broglie wavelength.

Step-by-step solution

1. Convert the kinetic energy from electron‐volts to joules: $200\text{ eV}=200\times 1.602\times 10^{-19}\text{ J}=3.204\times 10^{-17}\text{ J}$.
2. Use the de Broglie relation $lambda=h/p$, with $p=\sqrt{2mK}$ for a non‑relativistic electron.
3. Substitute $m=9.11\times 10^{-31}\text{ kg}$ and $K=3.204\times 10^{-17}\text{ J}$ to find $p=\sqrt{2(9.11\times 10^{-31})(3.204\times 10^{-17})}=7.642\times 10^{-24}\text{ kg⋅m/s}$.
4. Finally compute $\lambda =6.626\times 10^{-34}/7.642\times 10^{-24}=8.67\times 10^{-11}\text{ m}=0.0867\text{ nm}\approx 0.087\text{ nm}$.

Concept tested: de Broglie wavelength of a moving particle

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked