An electron is accelerated from rest through a potential difference of 250 V. Calculate its de Broglie wavelength in nanometres.

Step-by-step solution

1. Use kinetic energy formula: $K=eV$, where $e=1.602\times 10^{-19}$ C and $V=250$ V, giving $K=4.005\times 10^{-17}$ J.
2. Calculate the electron momentum: $p=\sqrt{2mK}$ with $m=9.109\times 10^{-31}$ kg. This gives $p=8.54\times 10^{-24}$ kg m s⁻¹.
3. Apply de Broglie relation: $\lambda =h/p$ with $h=6.626\times 10^{-34}$ J s. Thus $\lambda =6.626\times 10^{-34}/8.54\times 10^{-24}=7.76\times 10^{-11}$ m.
4. Convert metres to nanometres: $7.76\times 10^{-11}$ m = 0.0776 nm.

Concept tested: de Broglie wavelength of accelerated electron

Published 18 September 2026 · How this solution was prepared and checked