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Calculate the binding energy

  1. May 1, 2013 #1
    To calculate the binding energy released when alpha decay converts radium 226, with atomic mass of 226.025, into radon 222, with atomic mass 222.01757, I understand that we subtract the atomic mass of radium 226 from the sum of the atomic masses of the radon 222 and the alpha particle, which is the helium nucleus (atomic mass 4.002603). My question involves the two electrons that are "left over" and I assume are contained in the radon atom after the transmutation. In other words, during the alpha decay, two protons and two neutrons are emitted as an He 4 nucleus. We are then left with 86 protons and 136 neutrons and to get radon 222. Where are the 88 electrons that radium 226 started with? Since a helium nucleus is emitted, I assume the electrons are in the radon, producing a negatively charged atom. So I would think that to calculate the binding energy, I would have to include the mass of the two extra electrons in addition to the atomic mass of radon. But this seems to give me an incorrect answer. Or is the book mistaken? or is 4.002603 the mass of the helium atom, rather than that of the alpha particle as reported. I am confused about these two electrons! How are they accounted for?
    Last edited: May 1, 2013
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  3. May 2, 2013 #2

    Simon Bridge

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    Where could they be?
    If the only thing that has happened is the ejection of a He nucleus from the radium nucleus then, logically, what happened to the electrons?

    Generally the atomic binding energies of the electrons are so small compared with the nuclear energies that we don't bother to include them.
  4. May 2, 2013 #3


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    The masses that you find in tables are always atomic masses (including enough electrons to make a neutral atom), not nuclear masses. The mass that you give for an alpha particle is actually for a neutral helium atom, so it includes the mass of two electrons.

    For a nuclear decay, strictly speaking you should use nuclear masses when calculating the energy released, but it's OK to use atomic masses in alpha decay because the total number of "extra" electron masses is the same before and after so they cancel out. With beta+ and beta- decay you have to be more careful.
  5. May 5, 2013 #4
    I remember alphas as the worst type of ionizing radiation because of the net positive charge. The 2 He 4 with a +2 charge. An alpha is so ionizing it can't make it through a sheet of paper. So it would reason that after the decay the ejected alpha would quickly strip the electrons from somewhere.
  6. May 5, 2013 #5

    Simon Bridge

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    @JMP1961: any of this of any use?
    We cannot help you if you don't answer questions.
  7. May 5, 2013 #6

    Vanadium 50

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    The OP hasn't been here since he posted that question.
  8. May 5, 2013 #7

    Simon Bridge

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    Has OP turned on email notifications?
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