= X-ray tube cutoff
{c}
{title2=$\lambda_{\min}=hc/(eV)$}
For an <Electron> starting with negligible <kinetic energy> and accelerated through a constant positive <electric potential difference> $V$, electrical work supplies $K=eV$. In the ideal single-<Electron> <bremsstrahlung> model, <conservation of energy> bounds any one emitted <photon>:
$$
E_\gamma=h\nu=\frac{hc}{\lambda}\le eV,\qquad
\nu_{\max}=\frac{eV}{h},\qquad
\lambda_{\min}=\frac{hc}{eV}.
$$
The endpoint corresponds to essentially all available <kinetic energy> going into one <photon>. It is a limiting bound, not a claim that every <Electron> produces an endpoint <photon>. Neglect target recoil, initial <kinetic energy>, and losses before the emitting interaction. If the accelerating <voltage> varies, its largest value sets the corresponding ideal upper endpoint; a mean <voltage> does not substitute for that value.
For a calculated constant-$V$ example, $50\,\mathrm{kV}$ gives $E_{\max}=50\,\mathrm{keV}$ and $\lambda_{\min}=0.0248\,\mathrm{nm}$. Doubling $V$ halves $\lambda_{\min}$. Producing a $0.0200\,\mathrm{nm}$ <photon> requires at least $V=hc/(e\lambda)\simeq62.0\,\mathrm{kV}$ in this model. A continuous <frequency spectrum> spans lower <photon> <energies>; the cutoff calculation predicts neither its shape nor the efficiency of <photon emission>.
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