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Conversion of mass to energy

  1. Jan 16, 2005 #1
    I have a fundamental question. I am reading this modern physics book, and it says that an electron and a proton that are released and come together into a bound state release a photon. Fine. But the explanation given is that the sum of the individual masses of the proton and electron is greater than the mass of the bound system, and that the extra rest mass was converted to energy in the release of the photon. I am confused because it seems that an equally good explanation is that the electron and proton have a potential energy when they are apart, and it is this energy that is released as a photon. Here there would be no conversion of mass into energy needed to save conservation of energy. Does the bound system behave as though it has less mass than the constituent particles, or is the second explanation just as good? Please help. :confused:
  2. jcsd
  3. Jan 18, 2005 #2
    Start with the electron and proton infinitely far apart. Call this 0 potential energy. Now let them fall together under their electrostatic attraction. They feel an attractive force, and the energy of the final configuration is less than it was at infinity--i.e., work must be done to pull them apart, i.e., you must add energy to the system to make them both "free" again. Now, whether the electron and proton spit out a photon to get rid of this energy, or whether the energy is removed in some other way, there is less interaction energy between them when they are bound into an atom than when they were at infinity. The rest mass is the same in both cases, just the rest mass of the electron and the rest mass of the proton. But when they are bound, their potential energy is less. Since energy has mass (E=mc^2), the mass of the bound system is m_proton + m_electron - E_binding/c^2. To find the binding energy, integrate the coulomb force between the particles from infinity to the electronic radius of the atom.

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