Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Gamma radiation, photon energies and wavelength question

  1. Jul 13, 2017 #1
    I haven't though about this from such a perspective but today while reading wikipedia (yes yes not the best source) I got confused, now the "eV" is said to measure the energy gained by an electron between a potential difference of 1V.
    I assume particle physicists use this measurement because its quite handy for particle accelerator where if you have say 250kV between cathode and anode and you use an electron beam you can then say the electrons got a final energy of 250 keV, correct?

    but here's the part that confuses me, in terms of gamma radiation which is among the highest frequency EM radiation and so ionizing, I see a chart where frequency, wavelength and the corresponding energy is given and it is written in eV.
    it is written that frequencies from the Ghz region up towards the Phz and even EHz, have corresponding energies of only μeV to keV, and only from gamma the energy starts in the MeV region, does that mean that photon Em radiation is not very powerful except starting from X ray to gamma region? And even then I see X rays have on average only about 100KeV, although I assume its enough to damage a cell because the binding energy between an electron and the nucleus is far far less than that between protons and neutrons inside the nucleus correct? What is the typical binding energy between electrons and nucleus in eV for organic materials like flesh?

    If I for example accelerate a beam of electrons between a cathode and anode to a potential of 100KeV, do they then reach an energy equivalent to the 100KeV X ray range photons for example? I have a feeling this might be the way an X ray tube works by accelerating electrons which are then
    directed into the anode which serves also as the target creating photon emission of an energy which is proportional to the applied potential difference?
    I read also that most of the energy of the electron beam gets absorbed in the anode and simply manifests itself as heat, yet the photons that are emitted by bremsstrahlung type of radiation from the target are of the corresponding eV energy of the PD between cathode and anode does this simply mean that of all the electrons from the beam (intensity) most don't make for a photon emission event and get trapped or otherwise end up as heat but those few that get their job done then emit a photon of their corresponding energy?
    If my assumptions are correct seems like the opposite of the photoelectric effect only with bigger losses in the conversion?

    Oh and one final question if I may, what determines the wavelength of an EM wave, I have always wondered why the wavelengths of even such high frequencies as Mhz and up to Ghz have such large wavelengths, is the visual picture of a photon flying through space and time making up and down motions much like a dot that would ride a sine wave a correct way of thinking about this? and then between each full up-down-up moment there is some distance that the photon has traveled in the horizontal frame which then become the wavelength?

    thank you all for taking time to answer.
  2. jcsd
  3. Jul 13, 2017 #2


    User Avatar
    2017 Award

    Staff: Mentor

    It is also a more convenient unit in terms of its size - you don't need to talk about zeptojoule or other prefixes no one ever heard about.
    Define "very powerful". In addition to the photon energy, the number of photons can matter as well.
    Typically a few eV for the outermost electrons (=the electrons that matter for chemical bonds), this applies to all chemical bonds.
    Sure. It also means the maximal photon energy you can get out of this setup is about 100 keV.
    I'm not sure how useful that comparison is.

    wavelength*frequency = propagation speed (phase velocity, but let's skip technical details). This is true for waves in general. For light, the propagation speed is the speed of light, which is typically very fast.
    That is completely wrong. A photon doesn't even have a position, and there is no up and down motion of anything. The wavelength is the distance between two subsequent maxima of the electric field strength, for example.
  4. Jul 13, 2017 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    That's right. The amount of energy per photon is not very large until you get to very high frequencies.

    Typically a few eV at most.

    The electric and magnetic fields at any point in space can be represented by vectors (arrows) whose magnitude (length) represents the strength of the electric or magnetic force a charged particle would feel. The direction the arrows point represent the direction of the field lines, which can be thought of as the direction of the force on a charged particle. In the case of the electric field at least. The magnetic field is a bit different since the force acts perpendicular to the field line, but the idea is the same.

    An EM wave is a fluctuation of the EM field that propagates outwards from a source. These fluctuations can be thought of as all of these vectors rapidly oscillating directions and magnitudes as the wave passes. This rapid oscillation creates an alternating force on charged particles as it passes, and, for example, causes electrons in an antenna to oscillate back and forth, which can be detected by a radio receiver or other equipment attached to the antenna.

    If you draw a line running perpendicular to the wavefront (so in a spherical wave the line would run straight out from the source), the tips of the field vectors will form the squiggly line you always see representing light. The wavelength of the EM wave is the distance between the nearest vectors pointing in the same direction. The frequency is how quickly these oscillations occur.

    Note that a photon is not a classical particle and does not travel up and down as it moves through space. An EM wave interacts with matter in such a way as to transfer discrete "chunks" of energy to the matter. This interaction, this quanta of energy, is a photon.
  5. Jul 15, 2017 #4
    Right my fault, I had read about the photon earlier but somehow though about it still like a classical particle,

    well to rephrase it would it then be fair to say that the photon is simply a phenomenon which we have observed and so have given a name to it, and the phenomenon in question being that certain spectrum light (EM radiation) gives off certain energy electrons aka the now famous photoelectric experiment and we then simply concluded that if there are electrons emitted with specific energy and intensity which somehow manages to be proportional to the incoming light (EM radiation) so it must be that light also consists of discrete quanta only apart from electrons we cannot directly measure or see them so we observe them indirectly through their interaction with a metal target for example, is this a correct way of approaching this question?

    Now I personally can understand the wavelength in terms of a generator rotor magnet passing by a coil as it approaches the induced field going up in strength then reaching its maxima and then as it goes away decreasing etc, I can also understand it when a semiconductor or a spark gap switches current at a fast pace creating a high frequency EM wave and then it can be seen for example in a microwave oven if something other than food is put in there and some parts of it become burned fast while other stay relatively cool because they were inbetween the maximum points of the EM sine wave,

    though it becomes harder for me to understand why for example a decaying nucleus emits a very high frequency photon/s that we call gamma rays, since all EM waves travel at c in vacuum is the frequency sort of an approximation of the energy of the event that created the "light" photons in the first place so for example nuclear fission releases very high frequency photons simply because the event itself release alot of energy?

    Sorry if this seems obvious but one more question, I can understand when a nucleus breaking apart or fusing creates particles that are ejected like alpha ones for fusion or beta electrons from fission or neutrons for that matter but it is a bit harder to understand why it emits photons since they are not originally part of the atomic structure, is it simply a mechanism for which we have no deeper explanation than the one which simply states that during certain transitions in nuclear reactions a photon is created etc? or is there more detail to it ?
  6. Jul 15, 2017 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    It's a good start. We have many more observations than just the photoelectric effect that supports the idea that EM radiation is quantized. See the bottom part of the first reply to this reddit question for an example.

    Pretty much. Decaying nuclei release a lot of energy, which either goes into the kinetic energy of the decay products, the creation of new particles (such as neutrinos), or EM radiation. Since the energy released is much larger than chemical reactions, nuclear reactions release radiation with much more energy.

    The full picture would require an understanding of quantum electrodynamics, but we can say that we have a bunch of interacting, electrically charged particles and that when electrically charged particles interact they tend to release EM radiation. Classically this happens when a charged particle is accelerated, but the nucleus can release radiation even when no fusion or fission events occur and no particles are accelerated. Like I said, we'd need to get into quantum theory to fully explain it.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Gamma radiation, photon energies and wavelength question
  1. Gamma and photons (Replies: 7)

  2. Gamma radiation (Replies: 13)