X-ray tube and photon energy question?

[NocMedic]

Hey all, I am just studying for an upcoming medical imaging exam however I am having some trouble understanding a concept:

With regards to X-ray beam production, the effective photon energy of a beam is roughly 60-70% of the maximum photon energy (Emax) at a particular Tube voltage in the range of 40-150KV. For example a 100KV tube voltage produces an effective photon energy of approx. 60keV.

My questions are:

1) Why does this occur, what is the physics behind the value being roughly 60% of Emax?
2) How is the effective photon energy different to the Average photon energy (the most frequent beam) or is this the same thing?

Any help is greatly appreciated!

 

[bpsbps]

Most of the x-rays are coming from "bremsstrahlung" radiation. When the accelerated electron bangs around the metal target is loses energy and this is given off as radiation.

The highest possible x-ray energy is equal to the initial kinetic energy of the incoming electron this would happen if the electron went from it's full speed to completely stopped. Most likely the electron is going to undergo several collisions, each time releasing part of it's energy.

A curve showing a typical x-ray spectrum is shown in the below site:

http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/xrayc.html

Note the abrupt edge at the smallest frequency -- this is the highest possible energy. The maximum of the bremsstrahlung curve is related to the most likely deceleration and I *ASSUME* is the 60-70% that is called the "effective photon energy". This value will vary depending on the initial voltage and the metal that the target is made of.

The "average photon energy" would be if you added up all the x-ray energies and divided by the number of x-rays. Notice that the bremsstrahlung curve is not symmetric therefore the "average photon energy" is not necessarily at the maximum of the bremsstrahlung curve.

You should double check on the meaning of Effective Photon Energy, but to me this is what I'd guess.

 

[charng]

Hello! I can provide some insight into your questions about X-ray tube and photon energy.

1) The reason for the effective photon energy being roughly 60-70% of the maximum photon energy is due to the process of X-ray beam production. When high-energy electrons from the cathode strike the anode, they lose energy and produce X-ray photons with a range of energies. However, the anode material and the tube voltage determine the maximum energy (Emax) that can be produced. This maximum energy is typically higher than the effective photon energy because some of the electrons do not lose all of their energy in the process, resulting in a range of energies being produced. This is known as the bremsstrahlung process. The effective photon energy represents the average energy of the X-ray photons produced, which is typically around 60-70% of the maximum energy due to the range of energies produced.

2) The effective photon energy is different from the average photon energy. The average photon energy refers to the average energy of all the photons in the X-ray beam, while the effective photon energy specifically represents the average energy of the X-ray photons that contribute to the image formation. This is because not all photons in the X-ray beam are able to pass through the patient and contribute to the image formation, so the effective photon energy takes into account only those photons that actually contribute.

I hope this helps clarify the concept for you. Good luck on your exam!