# Continuous X-Ray spectrum

I'm reading Eisberg's Quantum Physics book and came across the following graph

I know very little about the subject, but from that figure I understand that, depending on the energy of the incident electron, the amplitude of the frequencies we get change. That seems pretty logical to me. The problem is that then in the book is stated that:

Electrons in the incident beam can lose different amounts of energy in such encounters and typically a single electron will be brought to rest only after many encounters. The x rays thus produced by many electrons make up the continuous spectrum of Figure 2-10 and are very many discrete photons whose wavelengths vary from $\lambda _{min}$ to $\lambda \to \infty$, corresponding to the different energy losses in the individual encounters.

I don't understand what's being said. What is the relationship between the figure above and the quoted paragraph?

• First let me say that the x-axis could just as easily be the energy of the emitted x-ray rather than the wavelength. What that quote says is the reason that you don't put in an electron with 50 keV and always get back a photon with, for example, 20 keV, is because the electron can lose energy a bunch of times. It's why when you put in a 50 keV electron you get photons with energy 2 keV, 10 keV, 25 keV, etc. It's the reason you have continuous lines on the graph rather than a single point for each electron energy. – pentane Mar 25 '16 at 19:47
• It's called braking radiation, or Bremsstrahlung – Peter Diehr Mar 25 '16 at 21:00
• @pentane Thanks for the answer. Just to check if I got this right... This spectrum is literally continuous or is it discrete but with values too close to be distinguishable? I mean, every time an electron loses energy, it gives away $h\nu$, right? So strictly speaking, the spectrum is discrete but the values the x-axis can take are so close to each other that, practically speaking, it ends up being continuous. Is this reasoning right? – Tendero Mar 26 '16 at 0:09
• Peter Diehr, is this breaking radiation The same as synchrotron radiation? Also I have a question about a double slit experiment using electrons. As an electron travels through a slit experiment can it produce synchrotron radiation and when the radiation reflects back will it affect the path of the electron as it travels alone toward the detection screen? – Bill Alsept Mar 26 '16 at 0:57
• @M.S. Yes I think your reasoning is more or less correct. If you shoot a single electron at the target, maybe you produce one, two, or a handful of photons so then yes an empirical graph of the x-ray output would clearly have discrete points. In a practical x-ray source, you of course have a huge number of electrons so a graph of the data would look continuous. Floris framed this a different (and probably better, from a theoretical standpoint) way by saying it is a graph of relative probability of the x-ray energies that are output. – pentane Mar 27 '16 at 4:05

When electrons hit a solid, they are abruptly decelerated. They don't lose all their energy at once - they lose a certain amount with each "collision" or near-collision with particles (mostly other electrons) in the solid.

We know that a decelerating particle emits electromagnetic radiation; in this case, each electron undergoes multiple decelerations, and emits multiple photons during the process.

The spectrum you see represents the probability that a given electron, starting with a certain energy, will emit a certain photon. The graph you reproduced has the wavelength of radiation along the X axis; I find it more intuitive to use energy (which is related to wavelength by $E=\frac{c}{\lambda}$) because that allows us better to see how energy is lost by an incident electron.

For example, from Kieranmaher - Own work, Public Domain:

You see here the spectrum of an X-ray tube with two different voltages - 60 kVp and 100 kVp. The spectrum for 60 kV is approximately continuous, while the spectrum for 100 kV has "characteristic radiation" peaks on top of the continuous background. There is also a dotted line pointing to "unfiltered 100 kV".

This teaches us a number of things:

1. The probability of an electron decelerating so rapidly that it emits all its energy in a single photon is vanishingly small
2. The probability of emitting lower energy photons is higher
3. As electrons penetrate into the target, the emitted photons have further to go to escape the target; this makes it less likely for low energy photons to be emitted (although they are more likely to be generated inside the material, they don't come out)
4. In real X-ray tubes, external filters (e.g. Aluminum) are used to further limit the amount of low-energy radiation emitted, since most of that radiation will be absorbed by the body: that means it results in "radiation dose" to the patient, but does not contribute much to the image quality.
5. When electrons that were in a particular orbital are knocked out by the incoming electron, the void created may be filled by another electron from a higher orbital "dropping down" into the lower orbital; in the process, it emits a photon with a very specific energy. This is not "bremsstrahlung", but "characteristic radiation" - but it is a real effect that changes the spectrum observed from X-ray tubes*
6. A single electron will give rise to a few emitted photons; the sum of the energy of these photons is no greater than the energy of the incident electron.
7. Different electrons will undergo different collisions, and lead to different energies
8. Taken over a large number of electrons, the total emission spectrum will have the shape given in your graph.

Interpreting this graph, we can estimate the chance of the emission of a 25 keV and a 75 keV photon to be smaller than the chance of emitting two 50 keV photons - mostly because the 25 keV photon is likely to be absorbed before emission (it's in the "filtered" part of the curve)

*) In fact for certain low-energy applications such as mammography, the target material is chosen such that the characteristic radiation is a significant fraction of the emitted X-rays; in essence this creates a more monochromatic beam, which results in a lower absorbed dose for a given image quality (SNR).