The continuous part of the x ray spectrum is due the deceleration of electrons. I know that a decelerating charged particle emits a braking radiation according the EM theory. However, what's in the photon picture of light that says a decelerating charged particle must emit photons continuously? Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?
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2 Answers
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Classical models suffice for many cases of electromagnetic radiation. Bremsstrahlung is a phenomenon that's very difficult to approach from a quantum viewpoint, so we usually model it classically. Empirically, this is usually adequate. The classical theory predicts the intensity spectrum, and the intensity divided by the photon energy is the expected rate of detection of photons of that energy.
Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?
They do. Most electrons colliding with the target in an x-ray tube don't lose much energy to bremsstrahlung. The majority of energy is lost to atomic excitations that cascade down into heat. Some atomic excitations radiate photons at characteristic x-ray energies, so there is generally a line spectrum added to the bremsstrahlung spectrum.
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$\begingroup$ Is it fair to say that Quantum Mech primarily models the interaction between light and matter, and not the behavior of light or EM fields alone? $\endgroup$– RC_23Commented Mar 1, 2023 at 23:18
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$\begingroup$ Me asking why electrons don't lose their energies to atoms residing in the anode was a follow up to the original question, that is I meant to say why there is a braking radiation at all if electrons can lose their energies to residing atoms? So instead of getting continuous spectrum embedded in it some characteristic ones, why not they aren't all characteristics? $\endgroup$– JackCommented Mar 2, 2023 at 0:25
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1$\begingroup$ @Jack Why do your tires squeal if you press the brake pedal too hard? Most of the kinetic energy of your car heats the brakes and tires, but some goes into sound. In an x-ray tube, all the processes of energy loss are happening at the same time, so you get heat, bremsstrahlung, and characteristic x-rays. $\endgroup$ Commented Mar 2, 2023 at 0:33
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$\begingroup$ @RC_23 For electromagnetic wave propagation and other reversible processes below ~100 EHz, Maxwell and QED models agree, so we don't need to consider quantum phenomena. But, when the electromagnetic field interacts irreversibly with something and we look at the interaction on a small enough scale, we observe wave function collapse, photons, etc. $\endgroup$ Commented Mar 2, 2023 at 14:42
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1$\begingroup$ @Jack Bremsstrahlung comes from the close encounters of electrons with atomic nuclei. Accelerated charge produces radiation, and the nucleus can cause a large change in the electron's trajectory over a picometer distance. On the other hand, the acceleration in an x-ray tube occurs over a centimeter or so, relatively gentle. That produces very little radiation. $\endgroup$ Commented Mar 3, 2023 at 13:42
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However, what's in the photon picture of light that says a decelerating charged particle must emit photons continuously?
I don't think the usual "photon picture" is compatible with "emitting continuously". The emission should be a series of well separable countable events to conform to the "photon picture". Emitting continuously evokes emission of a classical EM field wave that does not manifest any photon behaviour.
You may have meant "emit photons with high cadency". Then the photon picture would be as follows: from time to time, the electron bathing in external electric field and moving rapidly emits a photon; and energy and momentum for this newly created particle is taken from energy of the combined system (moving electron + EM field in its vicinity). In a sense, electron acts as a "condensation center" on which EM energy and EM momentum concentrates and gets moving as a particle. This EM energy and EM momentum come from EM energy and momentum in the space all around the electron in its vicinity. This energy and momentum is zero or unable to leave the electron when the electron is alone (free electron does not emit photons), but they are non-zero when the electron is in external field (this can be seen already in classical EM theory).
If instead of electron in static electric field, we have electron moving rapidly through a region filled with thermal EM radiation, then there can be a different kind of braking radiation: it is possible that the fast electron scatters a low energy photon into a high energy photon, and thus the electron loses kinetic energy while it produces "more visible" photons. This is called the inverse Compton effect.
Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?
Electrons do not interact directly with atoms or other electrons, they interact with the immediate EM field where they are, and they interact with atoms only through this EM field. So naturally when electrons lose energy, this goes first into EM energy of radiation, and this spread in all directions, thus not all of it can be absorbed by the matter in the anode. Some of it may reach electrons in the atoms and produce characteristic radiation from there.
answered Mar 2, 2023 at 1:12