Recently I went to the dentist and had a full set of routine X-rays taken. The technology has changed a lot since my childhood. Mostly because now my dentist hooks up his laptop to an electronic device that is put in my mouth and the X-rays are ready within fractions of seconds after exposure. This way I don’t have to wait as much for the X-ray films to be developed and in the end this is a cheaper way to get the information. I’m also assured that the new electronic readers require less exposure than film so that my total radiation exposure is reduced by quite a bit. I also get to wear the lead apron to prevent some of the soft tissues in my chest cavity and lower body from being exposed to unnecessary radiation.
So what do X-rays actually measure? This is what I thought I would ask and answer today for my first post in a long time. Part of this question is about how the photons in the X-rays interact with matter and why is lead used to stop X-rays.
This is very different than the superhero X-ray vision. Surprisingly, for most superheroes it is still true that lead will stop them from seeing an object.
For the first part, the theory of how electrons and photons interact is called quantum electrodynamics. With very little extra effort one can also describe how photons and protons interact. The important aspect you need to know is that photons can be absorbed and emitted by charged particles, so photons and electrons and photons and protons can scatter from each other.
This process is called Compton scattering. And the Compton experiments were some of the key experiments that prove the quantization of the electromagnetic field. This is because without photons, the resulting scattered radiation would not change color. This happens in the quantum theory because energy and momentum are conserved, so when the charged particle receives a kick and radiates a new photon in a different direction, the conservation of energy typically requires the outgoing photon to have less energy that the original photon. Here below you can see the Feynman diagrams that lead to Compton scattering.
The two diagrams that lead to Compton scattering
The diagram indicates a process where an electron absorbs a photon and emits a photon, or where the electron first emits a photon and then absorbs a photon. The red line indicates the electron line, and the wavy lines indicate the photons.
The same diagrams can be used for protons, if we change the electron parameters for proton parameters instead. Indeed, for this process, the main thing we need is the charge and mass of the particle that we are scattering from.
Because there are two places where a photon is absorbed or emitted in each diagram, each of them measures the charge of the particle, and this has to be squared to obtain a probability.
In quantum electrodynamics, each square of the charge is usually accompanied by a factors of 4 π in the denominator. This combination is called the fine structure constant, and this is measured in the laboratories.
We get that the probability of scattering from a charged particle is proportional to 
If we have such a scattering process, it is typically measured by a cross section. A cross section indicates the size of the area that a photon needs to hit in order to scatter off the charged particle. This is controlled by the mass of the particle (the only other scale in the system). In natural units a unit of mass is equivalent to a unit of inverse length (or time). This relation for a particle of some mass is called the Compton wavelength of the particle.
This is given by
The way you put this together is that you need an area (controlled by the square of the Compton wavelength) and
a probability (controlled by 
If you take into account that a proton weighs about two thousand times the weight of the electron, you realize that in the case of small atomic number the cross section to hit the heavy particle is typically small compared to the cross section to hit an electron. When the atomic number is high, the enhancement to the cross section goes only like 
Also, the electron is typically kicked out of the orbit of the atom and it is freed, so an interaction with X-rays ionizes the material. This is dangerous in high doses as it damages the molecular structures in cell tissues.
So X-rays usually measure the density of light charged particles in matter. Given a density of electrons ρ and a cross section, we get a penetration length of order
For most matter, the density of atoms per unit volume is very similar, so the density of electrons is proportional to the atomic number of the material 
When you take X-rays, Calcium stops them pretty well (atomic number 20), compared to water, nitrogen or hidrocarbons ,which is why X-rays are good for diagnosing bones. To read more about this, go here.
Incidentally, both Compton and Roentgen (who discovered how to make and record X-rays) were awarded the Nobel prize in physics for their studies of X-rays.
Hi,
I don’t get why $\sigma$ is constant, you wrote earlier that it goes like $Z^2$ (from what you write it seems that $l \sim 1/Z^3$).
Hi Spino:
Sigma is the same for all electrons and there are Z electrons per atom. The scattering of light from nuclei involves a much smaller cross section and can be ignored.
Superman’s X-ray vision works by backscattering, which is why he doesn’t have to be on both sides of the object he’s looking at (as with a conventional X-ray scanner).
So the flip-side of your dental X-ray are the “low Z” backscatter devices that the DHS uses to look at your naughty bits.
Sidenote: I took a trip to Point Roberts, WA on the weekend and saw their new radiation detectors. They look fairly ominous, but being bright yellow monoliths is better than being hidden behind a tree.
Friday, I was a chess player who was a little confused about the concept that X-rays are also photons, after a short explanation, asked me something like “okay, then X-rays also must come in different frequencies, like the colors of the rainbow”. I said, “yes”. He continued, “how do they choose what frequency to use for the x-rays at the dentist’s office? Is that the same frequency as used for chest x-rays?”
Carl:
The usual way of producing X-rays is not all that monochromatic: you dump a beam of electrons into tungsten and you get a wide spectrum of X-ray photons, many of them come from Brehmstralung (the electrons are stopped by
scattering with nuclei, and there is radiation from acceleration). Since in dentist and chest X-rays all you are looking for is absorption of -rays, the frequency does not alter all that much the end result. Soft X-rays are more readily absorbed (the scattering process has more complications due to atomic structure), so they are usually blocked with some metal plate, leaving only the hard X-rays.
Usually these are the same X-rays, but depending on the amount of tissue that they have to go through one uses higher intensities.
In the same system occasionally an electron collision kicks out another electron from the lowest shell of the tungsten atom, and a third electron from the next shell falls in to the lowest shell replacing the missing electron and emitting a single X-ray photon with a well defined frequency (these are K-shell x-rays). This way you can get a high intensity approximately monochromatic beam. These are the ones that are used for X-ray crystallography. Different materials have different K-shell X-ray emission lines, so that is how one can dial the color of the X-ray. However, there is no knob: you have to change the material that the electron beam collides with to produce X-rays at different frequencies.
Actually, the energies that diagnostic x-ray machines use is just below the energy needed for Compton effects to dominate. The photoelectric effect plays a large role–different interaction, but at the end you mostly end up with a $Z^3$ instead of a $Z^2$.
The typical system for describing the energy of the spectrum is the peak voltage across the x-ray tube; the x-rays in the spectrum (mostly) have lower energy than that. The mean energy is around half (or is it a third, I never can recall…) that of the peak.
I don’t have a good source for the energies used for dental x-rays handy, but they tend to be a lower energy (lower frequency) than chest x-rays. I think that for dental systems the kVp is around 60-70 kVp, while for a chest x-ray it will be about twice that.
Now, we would actually rather see an image acquired at a lower kVp, usually. There is contrast loss at higher kVp. However, you also have to consider penetration (higher energy photons are more likely to pass through the body), radiation exposure to the patient (the more photons you send through the patient, the more dose), and noise in the image (the more photons, the less noise).
Hi Dustin:
Thanks for the tip. It seems that I inadvertently described higher energy processes than what actually happens: the gamma ray scattering off electrons, rather than the X-ray energies.
It is still true that the softer parts of X-rays are filtered before irradiating a patient, and that higher Z atoms filter more than lower Z atoms.
I’ll try to figure out exactly how the photoelectric calculation works: particularly the factor of the cube power of Z seems intriguing.
I think calorimetric processes would help you greatly. These would be seen in LHC and Fermi(formerly Glast).
Best,
Also,
Experimentalists probe the structure of the proton by scattering electrons (white line) off quarks which interact by exchanging a quantum of light (wavy line) known as a photon.
Best,