Electron Accelerated Potential

Question

Electron accelerated through potential V has de Broglie wavelength λ = ?

RankGuru · Accepted Answer

1. **Kinetic Energy ($K$)**: An electron accelerated through potential $V$ gains energy $K = eV$.
2. **Momentum-Energy Relation**: $K = \frac{p^2}{2m} \implies p = \sqrt{2mK} = \sqrt{2meV}$.
3. **Substitution into de Broglie formula**: $\lambda = \frac{h}{p}$.
4. **Final Expression**: $\lambda = \frac{h}{\sqrt{2meV}}$.
5. **Standard constant**: For an electron, plugging in values gives $\lambda \approx \frac{1.227}{\sqrt{V}} 	ext{ nm}$ or $\frac{12.27}{\sqrt{V}} 	ext{ \AA}$.

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