= Electron accelerating-voltage wavelength
{title2=$\lambda=h/\sqrt{2m_e eU}$}
Let an <Electron> start with negligible <kinetic energy> and move through an increase $U>0$ in <electric potential>. Its <electric charge> is $q=-e$, where $e>0$ is the <elementary charge>, so the decrease in <potential energy> is $eU$. In a field-free region after acceleration,
$$
K=eU,\qquad |\mathbf p|=\sqrt{2m_e eU},\qquad \lambda=\frac{h}{\sqrt{2m_e eU}}\simeq\frac{1.2264\,\mathrm{nm}}{\sqrt{U/(1\,\mathrm V)}}.
$$
This is an <Electron> formula, not a universal conversion from <voltage> to <wavelength>. It assumes negligible initial <kinetic energy>, negligible losses, and $eU\ll m_ec^2$, where $c$ is the <speed of light>. It describes the <de Broglie wavelength> outside the crystal; the <potential energy> inside a material can change the local <kinetic momentum>.
For $U=54\,\mathrm V$, take $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$, $e=1.602176634\times10^{-19}\,\mathrm C$, and $m_e\simeq9.1093837\times10^{-31}\,\mathrm{kg}$. The rounded <Electron> <mass> is consistent with the 2022 recommended constants.<CODATA Value: electron mass by National Institute of Standards and Technology>[CODATA electron mass: numerical value and 2022 adjustment label]{ref}
$$
K=8.65175\times10^{-18}\,\mathrm J=54\,\mathrm{eV},\qquad
|\mathbf p|\simeq3.97019\times10^{-24}\,\mathrm{kg\,m\,s^{-1}},\qquad
\lambda\simeq1.66895\times10^{-10}\,\mathrm m=0.166895\,\mathrm{nm}.
$$
The <electronvolt> is an <energy> unit: $54\,\mathrm{eV}$ must be converted to <joules> before combining it with $m_e$ in SI units. <Dimensional analysis> checks both steps:
$$
[eU]=\mathrm{C\,V}=\mathrm J,\qquad [\sqrt{2m_e eU}]=\sqrt{\mathrm{kg\,J}}=\mathrm{kg\,m\,s^{-1}},\qquad [h/p]=\frac{\mathrm{J\,s}}{\mathrm{kg\,m\,s^{-1}}}=\mathrm m.
$$
The resulting <velocity> is about $4.3584\times10^6\,\mathrm{m\,s^{-1}}$, or $0.01454c$, so the nonrelativistic assumption is consistent. For a larger accelerating <voltage>, the <relativistic energy-momentum relation> gives $pc=\sqrt{K(K+2m_ec^2)}$, with $K=eU$, while $\lambda=h/p$ still holds. At $54\,\mathrm V$ the relativistic <wavelength> is shorter by about $0.00264\%$.
\Image[/de-broglie-wavelength-momentum-voltage.png]
{title=<de Broglie wavelength> versus <momentum> and <Electron> accelerating <voltage>}
{description=Original formula plots. Left: $\lambda=h/p$, with $p$ the <momentum> magnitude. Right: the nonrelativistic <electron accelerating-voltage wavelength> for $10$–$500\,\mathrm V$ and negligible starting <kinetic energy>. Labels mark calculated examples, including $54\,\mathrm V$; no points represent experimental measurements. Quadrupling $U$ doubles $p$ and halves $\lambda$.}
{source=de-broglie-wavelength-momentum-voltage.py}
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