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I'm studying for my radiology exams and I don't understand the answer to this question.

It states that the following sentence is true:

Reducing the filament voltage has no effect on the maximum photon energy produced by characteristic or Bremsstrahlung radiation.

I get that the filament voltage won't affect characteristic radiation because that is based on the target material but I don't understand how the filament voltage won't affect the maximum photo energy of Bremsstrahlung radiation. I thought that the kVp was the only thing that changed the maximum photo energy and that is related to filament voltage, or am I getting confused?

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  • $\begingroup$ For the outsider: “kVp” is the peak voltage across the x-ray tube, measured in kilovolts. For the asker: the first image at this link suggests you are correct, that the maximum bremsstrahlung energy is for a single photon to carry away the electron’s entire kinetic energy, and that this bremsstrahlung endpoint is therefore equal to the peak voltage across the filament. I’ll hold off on a proper answer in case a real expert comes along. $\endgroup$
    – rob ♦
    Commented May 8, 2022 at 17:59

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The filament voltage controls the filament temperature, which controls the electron current, but not the electron energy. Thus, increasing the filament voltage increases the flux of x-ray photons, but not their energy. The Bremsstrahlung cutoff remains the same.

The anode voltage controls the electron energy, and that controls the Bremsstrahlung cutoff.

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  • $\begingroup$ Thanks for this, this makes more sense now. $\endgroup$
    – Hollie Campbell
    Commented May 22, 2022 at 12:39

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