X-ray tube cutoff (${\lambda }_{\min }=hc/(eV)$)

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: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 $V=hc/(e\lambda )\simeq 62.0\mathrm{kV}$ in this model. A continuous frequency spectrum spans lower photon energies; the cutoff calculation predicts neither its shape nor the efficiency of photon emission.