An electron is accelerated from rest through a potential difference of 300 V. Calculate its de Broglie wavelength in nanometres.

Step-by-step solution

1. Compute the electron’s momentum: $p=\sqrt{2m_{e}eV}$ where $m_e=9.11\times 10^{-31},$kg, $e=1.6\times 10^{-19},$C and $V=300,$V.
2. Insert the values: $p=\sqrt{2\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}\times 300}=9.36\times 10^{-24},$kg·m/s.
3. Use the de Broglie relation $\lambda =\frac{h}{p}$ with $h=6.626\times 10^{-34},$J·s.
4. Calculate $\lambda =\frac{6.626\times 10^{-34}}{9.36\times 10^{-24}}=7.08\times 10^{-11},$m.
5. Convert to nanometres: $7.08\times 10^{-11},$m = $0.0708,$nm ≈ $0.071,$nm.

Concept tested: de Broglie wavelength of accelerated electron

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked