Photons and Bremsstrahlung radiation

Question

When fast electrons goes into a target, part of their kinetic energy is converted into electromagnetic radiation, that we call Bremsstrahlung radiation, as they change their velocity. The energy spectrum is smooth because electron can either lose all their energy and stop inside the material or lose only a fraction of its kinetic energy.

But what can we say about the number of photons? Does it emits just one photon with frequency given by the loss of energy for each interaction? If yes, can I conclude that the number of photons generated by Bremsstrahlung radiation is proportional to the density of the target, while the average energy of each photon is proportional to the $Z$ of the target?

My reasoning is that an electron that goes into a nucleus will lose energy (either a fraction or all its energy) and emit just one photon. If the nucleus has a high charge, then it will consistently lose a large part of its energy than a small part, resulting in a higher average energy of each photon.

Instead if the electron interact with a lot of nucleus (high density target), then there will be a larger number of photons.

Answer 1

Classically, each electron makes an electromagnetic pulse. This isn't a bad model. Quantum mechanically, the Fourier power of the pulse in a band $\Delta\omega$ at frequency $\omega$ gives you the expected number of emitted photons of energy $\hbar\omega$. As @Triatticus points out, the bremsstrahlung power spectrum is flat enough versus frequency that at low frequencies the expected number of photons diverges.