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Nuclear Fusion vs. Photons

  1. Aug 25, 2008 #1
    Hello. Please let me start off by saying that I'm not very smart. I know I am most likely going to embarrass myself, but I hope that I do learn a lot, and maybe can one day be smart enough to not have to ask such obvious questions. Until then I'm truly sorry.

    Ok. Recently I have been greatly interested in the process of nuclear fusion. It is my understanding that all matter in the universe other than hydrogen is the result of nuclear fusion. I guess I consider everything in the Universe to be star dust.

    So protium is made up of one electron, and one proton. The mass of each protium being 1.007825 u.

    So with luck, and enough energy; one of the protium are transformed.through inverse beta decay into a neutron with an electron neutrino, and we can't forget the anti-electron..

    So now we have an electron and proton, and a neutron, neutrino and anti-electron from the two protium

    So then the anti-electron hits the electron which turns them into photons.

    So far two protium transform into one deuterium while simultaneously giving off 2 photons, and one neutrino.

    So now a deuterium combines with a proton from a protium while the electron from the protium finds an anti electron, and forms a photon.. With one neutron, and two proton, we now have He3.

    With little luck, the He3 hits another He3. They combine, and lose two protons. These protons find an electron, and are now protium. The two neutron, and two proton form He4.

    In conclusion, the photons released throughout the process is the energy, and I'm guessing light that is given off from the sun. Neutrinos are shot out from the sun also, but they are a particle I think, and move at a speed slightly less that of light.

    My ultimate question is this:

    I am under the impression that all light is made up of photons. What I can't understand is how can a photon be made with something as simple as fire? Even electricity.

    Electrons bump into each other, but where does the random positron come from that is needed to create a photon?

    I have been told that all mass can ve converted to energy. E = mc2 right? It seems to me that only electrons and anti electrons can be transformed into a photon. Does this not imply that only electrons can transform into the pure form of energy known as light?

    I was told by my father that the light we see from fire is because oxygen in the air is pulling carbon from the wood. The ionization is the light that I was seeing. Is there maybe different kinds of light?

    I know this question must come off as obvious, but I am really lost. I'm getting to the point where it's either too easy, and I already know, or it's way over my head. "Kinda like relativity."

    Anyways, it's 3am so I should get to bed. I wanted to add up all the mass of the particles involved in nuclear fusion to see exactly how much mass is lost into nutrinos, and photons, but I'm on my blackberry, and the calculator on here is not big enough. Maybe another time. :)

    Goodnight, and thank you for your patience.

  2. jcsd
  3. Aug 25, 2008 #2
    For anyone confused about the term protium, like I was, this is from http://www.protiumenergy.com/home.html

    "The scientific name for what we commonly refer to as Hydrogen is Protium. It is the most common isotope of the element hydrogen and has one proton, one electron but no neutrons. It is denoted by the symbol ¹H. Other isotopes of hydrogen are the rarer Deuterium which has one proton, one neutron and one electron, and the radioactive isotope Tritium which has one proton, two neutrons and one electron. You may be interested to know that Hydrogen is the only element that has unique names for each of its isotopes."
  4. Aug 25, 2008 #3
    Anyway, to help answer your question(I'm no expert), when a bound electron goes from a higher energy state to a lower one, a photon is released... it's frequency dependent on that energy state change relationship.
  5. Aug 25, 2008 #4


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    The fusion process takes place in the core of a star (or Sun) where temperatures and densities are high enough. The principal process is the proton-proton cycle.


    Two protons fuse to form a deuteron (and positron (anti-electron) and neutrino). That's the origin of the positron which will find an electron and annihilate producing two 0.511 MeV gamma rays.

    The deuteron fuses with a proton to form He-3 and another gamma ray is released.

    Two He-3 nuclei then fuse to form He-4 and 2 protons, or a proton may fuse with He-3 to form He-4, another positron and neutrino.

    In addition, electrons will scatter off the larger nuclei and produce bremsstrahlung radiation.

    Also, the sun does contain other elements and a small fraction of fusion is due to the CNO cycle, http://csep10.phys.utk.edu/astr162/lect/energy/cno.html - and that also produces gammas and positrons. In addition, the ionization and recombination process will produce X-rays.

    Fire is a combustion (chemical) process, which release the atomic bond energies as fuel molecules combine with oxygen molecules (exchange electrons), as opposed to fusion which is a nuclear process. Fire also produces light, but no X-rays or gamma rays. The heat both fire and fusion are manifest in the kinetic energy of the products, which for fire is on the order of 0.1-0.3 eV, but for fusion is on the order of MeV.
  6. Aug 25, 2008 #5
    Dear Astronuc

    Your response was VERY informative. Thank you very much for your explanation. I must ask though. The light that the fire produces. Are those photons? I have a feeling that they are not due to the fact that no positron exists to combine with an electron. This leads me to believe that the light produced by a fire is not composed of photons, which in-turn makes me question what light is made of. I am under the impression that light is composed of photons. False?

    Thanks again everyone for your help. I know I must seem clueless.


  7. Aug 25, 2008 #6


    Staff: Mentor

    Photons are produced in many other interactions besides a nuclear fusion reaction! Essentially, any time a charged particle changes its state a photon will be released or absorbed. Photons are released when a charged particle accelerates, vibrates, changes quantum state, in many nuclear reactions, in blackbody thermal radiation, etc.

    In a fire I believe that the visible flame is caused by small pieces of material (soot) which are hot enough to emit blackbody radiation in the visible spectrum. The heat is from a chemical reaction, not a nuclear reaction.
  8. Aug 26, 2008 #7
    Dear DaleSpam

    I have been doing some research on charged particles, and although I feel a little more comfortable; I am, for the most part, lost.

    To my understanding, a charged particle can be any type of matter in the universe with a charge. Any matter can produce photons as long as it comes into contact with its anti-particle. This does make sense though I will have to look into it at an atomic level deeper at another time to get a full understanding. I think this is off subject though.

    I also did some reading on blackbody radiation. Can we conclude that this is the same as thermal radiation? In any case, I feel a little more comfortable with thermal radiation also.

    I guess my next question is how exactly is a photon produced other then coming into contact with its anti-particle? I understand that a photon can be produced when a charged particle changges its state, but is this due to that fact that a proton turns into a neutron emiting a positron which comes into contact with an electron? Am I close?

    Thank you very much for teaching me, and i'm learning alot. I really do appreciate all of your help on this extremely confusing matter.



    P.S. Sorry about the poor spelling. I am typing this on my computer which hasn't a spell check, unlike my cell phone in which I normally use.


    I was just talking to my father, and I am not sure if he is correct, or if I understand so one more question.

    Is inverse beta decay "electron capture" the only way a photon can be formed? Is a proton being converted into a neuton the only way a photon can be produced?
    Last edited: Aug 26, 2008
  9. Aug 26, 2008 #8


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    Light is basically photons. Light (visible light) is that group of photons in the small part of the electromagnetic spectrum, which is visible to our eyes (wavelengths in the range of ~400-700 nm).


    But in general light (photons) have a broad range of energies from very high to low: gammas, X-rays, UV, visible light, infrared, microwave, radiowave

    Gammas are those photons which result from nuclear or subatomic reactions, e.g. gamma decay of a nucleus, or annihilation, as in the annihilation of positron+electron.

    X-rays are photons arising from electrons falling in the K or L shells of atoms which have had at least one K or L electron ejected by a collision.

    Brehmsstrahlung (breaking) radiation is cause by acceleration of electrons near nuclei, where the electrons have too high an energy to be captured.

    Fire produces visible light and infrared because the chemical reactions have very low energy per reaction compared to nuclear reactions, which I mentioned earlier.

    Microwave and radiowaves are caused by electrons oscillating in a conductor (antenna). They have very long wavelengths and consequently relative low frequency and low energy, compare to visible light.

    With regard to matter-antimatter, the electron+positron annihilation produces two gamma rays of ~ 0.511 MeV each, which is equivalent to the rest mass of an electron.

    A proton-antiproton annihilation produces pions, which decay to muons (or in some cases gammas), which decay to electrons (and neutrinos and anti-neutrinos)

    This might help

  10. Aug 26, 2008 #9
    Dear Astronuc

    Thank you for your explanation, but can you please specify how, exactly a photon is produced in each of these situations?


  11. Aug 26, 2008 #10


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    Well to answer 'how' a photon is produced there, requires quite much Quantum Electrodynamics (which is graduate level class in physics).

    But I can give you a classical 'analogy': A charge that accelerates (changes is velocity vector) will emit EM radiation (photons in quantum physics), this is covered in standard electrodynamics courses in school. Now this can be transfered to gamma rays from radioactive nuclei. The configuration of protons in the nucleus can change, since there are many different possible configurations. But only one configuration is the lowest one (the ground state) and physical systems strives to reach the lowest energy possible. So in order to that in this situation is to change its proton distribution in 6dim-phase-space (space and momentum) - changing this leads to changes in velocities of charges, and photon(s) will be emitted.

    Same 'argument' can be used to 'Brehmsstrahlung'.
  12. Aug 26, 2008 #11


    Staff: Mentor

    No, this is not what I meant at all. I meant, for example, the change in state when an electron jumps from one orbital to another. There is no anti-particle involved, simply the movement of an electron from one shell to another. Similarly for the electrons that move back and forth in an antenna. In each case they change their state (orbital position in the first case, state of motion for the second case), and with that change in state there is a change in energy. In order to conserve energy a photon is either absorbed or released.

    Think of photons primarily as the universal way to conserve energy. When a particle collides with its anti-particle there is a lot of energy so a high-energy photon is produced. When an electron goes to a lower orbital there is a medium amount of energy, so a medium-energy photon is produced. When an electron accelerates in an antenna there is a very small change in energy, so very low energy photons are produced.
  13. Aug 26, 2008 #12
    Dear DaleSpam

    Your explanation is definitely informative. I feel like I'm starting to understand. I must say though that I have a lot of questions, so for this; I'm sorry.

    After reading your response, my next question is; when an electron moves from one level around the atom to another, and produces a photon, there has to be mass lost. Please confirm there is mass lost. If not, then please explain, because for a photon to be produced, there has to be mass lost. I hope. If there isn't, then I'm WAY off.

    Thanks for your patcience.


  14. Aug 26, 2008 #13
    I just read that photons and gravitons are massless. I don't believe that an electron loses mass when disturbed...If I am wrong, the mass will return the next time the electron is bumped,,,Comments..right? wrong?
  15. Aug 26, 2008 #14
    Dear D. Richmond

    You are correct in the sense that photons havn't any mass. This, I believe, is because there are pure energy. Photons move at the speed of light. Something no matter can do. Even neutrinos which have mass move at slightly less then the speed of light.

    I think you forget Einsteins famous equation E = MC^2, where E is energy, M is mass, and C is the speed of light in a vacuum. As you can see, you need mass to get energy.

    This is why I'm a little confused about this issue also.


  16. Aug 26, 2008 #15


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    Well when an electron in an atom goes into a lower state and emits a photon, the atom will be more bound and hence have lower mass that it had before, so you are correct here.
  17. Aug 26, 2008 #16
    Dear Malawi_Glenn

    Thank you for your imput. Can you please be a little bit more specific as to where in the atom this mass is lost? I know that inverse beta decay an electron, and positron are lose. In this case, an electron is simply shifted from one electron cloud level to another. So where in the atom does the mass come from to produce this low energy photon.


  18. Aug 26, 2008 #17


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    The constituents in a bound system has lower mass then the free constituents, see:


    That thread is about the atomic nucleus, but the same physics is applicable to atoms aswell. Also this is valid for the solar system, so it is a general physical feature.
  19. Aug 26, 2008 #18


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    I also am confused on this matter - say an electron goes from one energy level to a lower one and emits a red photon:

    An electron mass of 9.109 382 15(45) × 10–31 kg translates to about 511,000 Electron Volts.

    A typical red light photon is 1 Electron Volt.

    Is it proper to visualize the electron as giving up a photon as it falls from one energy level to another, and that the Electron would now be 510,999 Electron volts. If not this, then where did the extra energy come from?

    Is it a given that the electron mass does not change or can this be viewed a different way?
  20. Aug 26, 2008 #19
    Dear Malawi_Glenn

    Thank you for your answer, and attempt to help me understand, but your response is telling me nothing. I'm guessing I am not smart enough to understand what it is your trying to get across. Sorry.

    In any note, can someone please explain to me where the mass comes from in a way I can understand. Please remember that I"m in 8th grade, and havn't any scientific classes to assist me. I'm trying my best to understand though, and I do appreciate the help.


  21. Aug 26, 2008 #20


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    The electron mass you are refering to is when the electron is free.

    In bound systems you have binding energy, which is negative, and makes the system as a whole to get mass lower than the mass of its constutients. You can not say that ' a proton in a deutron has mass 0.98proton_free" and similar.
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