# I X-ray tube to gamma tube question

1. Mar 13, 2018

### girts

Now a x ray tube working principle is basically the opposite of the photoelectric effect, here instead of photons electrons strike a metal target to deposit their energy which is released as heat (IR radiation) and photons.
So the out coming photon wavelength (energy) is proportional to the kinetic energy of the electron which in turn is directly proportional to the potential difference that accelerated the electron between cathode and anode.

Now what happens when we increase the potential difference between cathode and anode - the electrons have higher kinetic energy which means that the resulting photons will have a shorter wavelength correct?
Also I assume that due to the way an x ray tube works a higher voltage across will give electrons higher kinetic energy but it will also increase the electron current so not only the individual photon wavelength will decrease but the total number of photons will increase correct?

If so far my assumptions are correct, then a question arises, if we keep on increasing voltage which results in higher energy/shorter wavelength photons why can't we get into the gamma region of photon wavelengths with very high voltages? I obviously realize that some physical constraints limit that possibility like the anode target metal endurance and heat removal etc, but theoretically is that possible?

2. Mar 13, 2018

Staff Emeritus
Not the case.

The difference between gamma rays and x-rays is more their source than their energy. Gammas arise through nuclear processes.

3. Mar 13, 2018

### girts

Ok I think I have been wrong about the proportional relationship because as I refreshed my knowledge it seems that light above the threshold frequency required for photoelectron emission doesn't increase linearly with respect to photon wavelength=electron KE. So I assume that light from the threshold wavelength up to a certain smaller wavelength all release electrons with similar kinetic energy.

Is this why you said "Not the case" ? Is the relationship in the x ray tube linear or not between the KE of accelerated electrons to the wavelength of emitted photons?
Because at least the potential difference between cathode and anode with respect to electron KE is directly proportional and linear to the best of my knowledge.
Also why you say that the x ray tube doesn't work similarly in principle but in reverse as the photoelectric effect? It does seem that for one you get electrons from photons and in the other you get photons from electrons, what are the differences (excluding frequencies involved) ?

Also yes I can understand that gammas usually come from nuclear reactions but so do x rays, I would like a more elaborate answer here, because the only difference that a gamma photon has versus an x ray one is their frequency, as well as the difference between a beta particle from a fission decay process and electron from a vacuum tube filament is in the kinetic energy only correct?
Is this because gamma photons have wavelengths so short that in order to produce them the process must involve reactions that are too energetic to be powered efficiently by power sources other than nuclear reactions?

So in theory is it even possibly to achieve the energy threshold for a electron beam to produce gamma spectrum photons from a target?

4. Mar 13, 2018

### Staff: Mentor

The photoeffect releases electrons that were bound to the metal before. The x-ray production does not involve capturing electrons.
By the usual definition x-rays cannot come from nuclear reactions.

You can get photons of arbitrary energy if your electron energy is high enough, but the process that produces a lot of x-rays is limited by the maximal electric field strength in the material. If you shoot in electrons at very high energy other photon production processes dominate at very high energies (e.g. electron-quark interactions).

5. Mar 13, 2018

### girts

Ok I see the "devil in the detail" , so for x rays the electrons that bombard the tungsten target for example hit and accelerate not only electrons but also ions in the target material and as those accelerated particles "brake" they emit a photon of characteristic frequency, at least that's what I got from reading wikipedia?
So photons get emitted as the particles in the target deaccelerate/brake or both when they accelerate and brake? guess the original German word for that would be bremsstrahlung radiation.

now if the photoeffect photons "kick" out the electrons from the target then I assume with time and sufficient photon intensity the plate becomes positively charged due to lack of electrons? For visible light I assume they are the lower energy state electrons that get emitted?

oh another question, since x rays have a window of wavelengths or spectra within that range then what determines the specific x ray frequency of the x ray tube, the accelerating voltage or target material characteristics (resonant frequency for such electron bombardment or what it is called?) or do both of these contribute ?

so what you are saying is that the very physics itself limits the maximum useful photon frequency that we can obtain using an electron beam and a target, and since the energy of a single gamma photon is very high the process that created the photon must also be very energetic and we don't have other efficient means to create such a process other than nuclear reactions themselves?

6. Mar 13, 2018

### Staff: Mentor

No, the radiation is emitted as the incoming electrons brake.
Just Bremsstrahlung. "Strahlung" is German for radiation, and "bremsen" is "to brake".
If you magically isolate the plate perfectly you can accumulate a small charge.
No, the electrons in the highest state are more likely to get emitted. The electrons in lower states don't have enough energy to get freed.
The target material. See the Wikipedia article.
No one said that. There are other methods to create intense high-energy photon beams.

7. Mar 13, 2018

### girts

ok so the target atoms don't get influenced much despite the mention in the wiki article, simply the incoming electrons interact via the electrostatic force and de-accelerate but since they had higher energy and now they have lower energy energy must be conserved so a photon is emitted, much like when a rock is thrown parallel to the surface of water as it touches the water it slows down but its energy is then released as ripples in the water.?
so it is the positive charge of the nucleus within the material that brakes the electron or is it also the repulsion from other nearby electrons?

as for the photoeffect, the highest energy state electrons were the outermost ones ? if yes then the lower energy state electrons would first have to travel to higher orbitals in order to be then released by visible light shining on a target?
I hope I got the higher, lower electron energy levels with respect to the nucleus correct.

the wiki article had a picture of a rare material and the overall x ray response is low but there is a high a sharp peak at a specific energy is this peak ina given material the one that determines the specific x ray frequency a target tube of such sorts will emit?

well what are those other methods of creating a gamma ray beam artificially apart from fission released gammas and fusion ones for some of the fusion reactions?

thank you .

8. Mar 13, 2018

### Vagn

Bremsstrahlung is primarily an electron-nucleus process. The electrons in the target would tend to absorb the energy directly, producing secondary electrons (and heat) instead of x-rays.
No, inner (core) level electrons can be liberated directly if the photon has sufficient energy, this is the foundation of x-ray photoelectron spectroscopy.
The emitted spectrum is dependent on the differences in the electronic energy levels of the material. When a core electron is liberated from the material, it leaves an unoccupied state. An electron in a higher energy state can then decay into the unoccupied state, emitting a photon with energy equal to the difference between the initial and final state.
Note that MFB said high-energy photon beams, rather than gamma rays. This is because the following techniques generate photons with energies greater than typical for gamma rays, but are strictly x-rays as they are generated by external electrons rather than nuclear processes.

Synchrotron insertion devices can obtain energies up to a few hundred keV reasonably easily. An example would be beamline ID15 at ESRF.
Other options include
Free-electron lasers
Laser-accelerated electron sources

Linacs are commonly used for radiotherapy into the MeV range.