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Mass energy conversion

  1. Dec 29, 2009 #1
    I don't know basically anything about nuclear physics, but I do know that mass can be converted to energy. Nuclear reactions, anti-matter, etc. But can energy be converted into mass? If so, where would this happen?
     
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  3. Dec 29, 2009 #2

    Astronuc

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    In particle colliders, the kinetic energy of the particles is converted into matter and anti-matter. Unless stored, the anti-matter usually finds corresponding matter and annihilates.

    In the case of high energy gammas, E > 1.022 MeV, pair production may occur when the gamma interacts with a nucleus. In pair production, an electron-positron pair are formed, and usually the positron finds another electron and they annihilate into 2 gammas of energy ~0.511 MeV.
     
  4. Dec 29, 2009 #3

    mheslep

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    Must the gamma be precisely of that energy for pair production, or would, say, 1.122 MeV also work? In that case, what happens to the remaining 0.1 MeV?
     
  5. Dec 29, 2009 #4

    Astronuc

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    The 1.022 MeV is the minimum energy required to produce the rest mass. AFAIK, the rest of the energy goes into the kinetic energy of the particles. In addition to pair prodcution, a gamma could knock out a neutron, which is know as photoneutron production. This doesn't create mass as much as it liberates a neutron from the nucleus. Basically the gamma energy has to match the binding energy of the last nucleon.
     
  6. Dec 29, 2009 #5

    Andrew Mason

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    Further to everything that Astronuc has stated, energy is converted into mass any time the energy content of a body increases - eg. when a body absorbs radiation. But because of the law of entropy, this process is generally one in which the energy/mass becomes more dispersed over time. So creation of new clumps of matter is not likely to occur. Since the big bang, energy and mass concentration in the universe has been decreasing.

    AM
     
  7. Dec 29, 2009 #6
    The idea of mass–energy equality unites the concepts of conservation of mass and conservation of energy, allowing particles which have rest mass to be converted to other forms of energy which have the same mass but require movement.
     
  8. Dec 30, 2009 #7

    mheslep

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    Thanks for the response.
    Right, just the enough for the rest mass of the two particles.

    What factor determines whether the gamma results in pair production, or in a photoneutron, or just excites the nucleus to a higher energy state? If this were the interaction of a photon and electrons I know where to go for a model, but I don't see the analogy here for the nucleus.
     
  9. Dec 30, 2009 #8

    Astronuc

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    There are two interactions with regard to photons (gamma rays) and atomic electrons: photoelectric effect (photon completely absorbed, and electron ejected from atom), Compton effect (photon scatters and loses energy, and electron is ejected from atom). Pair production involves a photon interaction with the nucleus.

    When a relatively low energy neutron is absorbed by a nucleus, the nucleus emits a gamma ray (which is the binding energy). Conversely, a gamma ray with sufficient energy can cause a neutron to be ejected from the nucleus. In the case of a deuteron, a gamma ray of sufficient energy (~2.2 MeV) can cause separation (dissociation) of the proton and neutron (the process is called 'photodissociation').
     
  10. Dec 30, 2009 #9

    mheslep

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    Or, electron stays in a higher energy orbital.
    My question was, given a incoming gamma of known energy can we say exactly which of the various scenarios will occur? I'm guessing now that the outcome is analogous to an incoming neutron: X% change of absorption, 100-X% chance of fission and so on.
     
  11. Dec 30, 2009 #10

    Astronuc

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    The various reactions compete based on cross-section. For gammas below 1.022 MeV, there will be no pair production, and so only photo-electric effect and Compton scattering occur. There's plenty of experimental data that have been used to develop cross-sections for the different reactions as a function of gamma-ray energy and element.

    See figure 3 in this - http://www.physics.uoguelph.ca/~cschultz/labcourses/outlines/highres.pdf [Broken]

    Note the decrease in cross-sections for the photo-electric and Compton scattering.


    And just for interest - a thesis on photodisintegration of a deuteron
    http://he3.dartmouth.edu/Photodisintegration/AbbyThesis.pdf [Broken]
     
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  12. Dec 30, 2009 #11

    mheslep

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    Thank you
     
    Last edited by a moderator: May 4, 2017
  13. Jan 1, 2010 #12

    epenguin

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    Isn't the result all around us? (Although not a bit self-evident).
    All the nuclei heavier than that of iron have been created in a process involving energy-mass conversion (although there may have been mass-energy conversion on the way).

    A lot of it if I remember happened in supernovae. :blushing:I am a bit vague beyond that, in fact I would be grateful for indications of useful sources of info. books/articles at any level about element creation.
     
  14. Jan 1, 2010 #13

    Astronuc

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    We don't know how the elements were formed in the beginning, but certainly all elements have been formed in stars, but they simply transform H and He into heavier elements.
     
  15. Jan 30, 2010 #14
    Reading some of the responses to energy mass conversion question, has prompted me to ask the experts in this forum the following question:
    What would be the length of time that visible light would be emitting from the process of converting enough energy into a body of mass of 6.0×(10 to the 24th power) KG ?
     
  16. Jan 30, 2010 #15
    You haven't given a rate of energy conversion.
     
  17. Jan 31, 2010 #16
    Nuclear physics is my interest/curiosity but not my educational background,so I have a lot to learn. And being on this forum is an opportunity that I greatly appreciate because it brings me in contact with some great minds in this fascinating field of study of our nature.

    In reference to your question about the rate of conversion I am not sure how to answer it, because I am sure that the rate depends on some other factors that I am not familiar with. Can we put up some scenarios?
     
  18. Jan 31, 2010 #17
    I'm far from being a great mind, but I'll try to explain when I know to you.
    First off, I don't think there are any radioactive processes that give off light in the visible spectrum, I believe all electromagnetic radiation given off is in the gamma ray range. Luminescence from radioactive elements is from interactions of radiation given off, not from the products themselves.
    You can choose any rate you want since you haven't mentioned any process in particular.
     
  19. Jan 31, 2010 #18
    Well... I am thinking of an atomic bomb explosion that emanates a lot of light for a brief moment. that is where E=MC^2 is applicable. Duration of that light is, what... in microseconds... or perhaps milliseconds?
    On the Sun the process is kind of reversed M=E/C^2. Hydrogen ---> Helium ---> to heavier and heavier elements. That process also emanates light, but of course, "zillions" of years.

    Lets imagine we have some source of tremendous amounts of energy and we want to form a mass of Iron of 2.0e+24Kg that eventually cools to lets say room temperature.

    What are the needed variables that we would have to control to cause visible spectrum light emission to last for a duration of time in lets say hours, not milliseconds or seconds or minutes or millions of years.
     
  20. Feb 1, 2010 #19
    Light is not a direct consequence of fission or fusion. It's important for you to understand how processes work before trying calculations of any kind.
    In an atomic explosion and in the sun, surrounding materials are heated and the excitation and returning to lower energy states of the electrons in those materials gives off light in the visible wavelength.

    About your proposed question. What materials are you starting with, pure hydrogen? Should this enormous mass of iron created be cooled by other means than radiating it? Are you assuming iron is removed to some container the instant it is created and that all energy produced by creating it is converted to visible light?

    I'm not trying to be rude or discourage you, but what I'm trying to say is that until you fully know what you are talking about, it's difficult to ask questions that can easily be answered.
     
  21. Feb 1, 2010 #20

    mheslep

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    ^Visible light is not ..., only gamma electromagnetic radiation.
     
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