An electron is accelerated from rest through a potential difference of 400 V. What is its de Broglie wavelength in nanometers?

Step-by-step solution

1. Compute the kinetic energy of the electron: $E_{\text{kin}}=eV=400\text{ eV}=400\times 1.602\times 10^{-19},\text{J}=6.408\times 10^{-17},\text{J}$.
2. Compute its momentum using $p=\sqrt{2m{E}_{\text{kin}}}$: $p=\sqrt{2(9.11\times 10^{-31},\text{kg})(6.408\times 10^{-17},\text{J})}=1.080\times 10^{-23},\text{kg⋅m/s}$.
3. Use the de Broglie relation $\lambda =h/p$: $\lambda =\frac{6.626\times 10^{-34},\text{J⋅s}}{1.080\times 10^{-23},\text{kg⋅m/s}}=6.13\times 10^{-11},\text{m}=0.0613,\text{nm}$.

Concept tested: de Broglie wavelength of accelerated electron

Published 20 September 2026 · How this solution was prepared and checked