An electron is accelerated through a potential difference of $150\text{V}$. Assuming it starts from rest, calculate the de Broglie wavelength of the electron in nanometres. Use $h=6.626\times 10^{-34}\text{J⋅s}$, $m_e=9.11\times 10^{-31}\text{kg}$ and $e=1.602\times 10^{-19}\text{C}$.

Step-by-step solution

1. The kinetic energy gained by the electron is $K=eV=(1.602\times 10^{-19}\text{C})(150\text{V})=2.403\times 10^{-17}\text{J}$.
2. Relate kinetic energy to momentum: $K=\frac{p^2}{2m_e}\Rightarrow p=\sqrt{2m_{e}K}$.
3. Substitute the numbers: $p=\sqrt{2(9.11\times 10^{-31}\text{kg})(2.403\times 10^{-17}\text{J})}=6.62\times 10^{-24}\text{kg⋅m/s}$.
4. Apply the de Broglie relation $\lambda =\frac{h}{p}$: $\lambda =\frac{6.626\times 10^{-34}}{6.62\times 10^{-24}}=1.00\times 10^{-10}\text{m}$.
5. Convert metres to nanometres: $1.00\times 10^{-10}\text{m}=0.10\text{nm}$.

Concept tested: de Broglie wavelength

Published 15 September 2026 · How this solution was prepared and checked