An electron is accelerated from rest through a potential difference of $250\text{V}$. Calculate its de Broglie wavelength in picometres. (Use $h=6.626\times 10^{-34}\text{J⋅s}$, $m_e=9.11\times 10^{-31}\text{kg}$, $e=1.602\times 10^{-19}\text{C}$.)

Step-by-step solution

1. The kinetic energy gained equals the electric work: $K=eV=1.602\times 10^{-19}\times 250\text{J}=4.005\times 10^{-17}\text{J}$.
2. For a non‑relativistic electron, $K=\frac{p^2}{2m_e}$, so momentum $p=\sqrt{2m_{e}K}$.
3. Substituting values: $p=\sqrt{2\times 9.11\times 10^{-31}\times 4.005\times 10^{-17}}=8.55\times 10^{-24}\text{kg⋅m/s}$.
4. De Broglie wavelength $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{8.55\times 10^{-24}}=7.75\times 10^{-11}\text{m}$.
5. Convert to picometres: $1\text{pm}=10^{-12}\text{m}$, so $\lambda =7.75\times 10^{-11}\text{m}=7.75\text{pm}\approx 7.8\text{pm}$.
6. Thus the wavelength is $7.8\text{pm}$.

Concept tested: de Broglie wavelength of accelerated electron

Published 21 September 2026 · How this solution was prepared and checked