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Binding energy and radioactive decay

  1. Jul 10, 2008 #1
    As I understand it, in all forms of radioactive decay, constrained by E=mc^2, a spontaneous breakdown of an element or isotope occurs: that is, a massive element (parent) decays to a less massive element, isotope or leptons [daughter(s)], plus a release of energy. However, in alpha decay, the energy release, unless understood as binding or potential energy, is not apparent in the end product accounting (see example below). Whereas in gamma decay, the energy release in the form of a high-energy photon is. But, unless I consider gamma decay as the end stage of alpha or beta decay, where the nucleus is often left in an excited state, I don't see the little bit of matter in the break away end products. Only in beta decay do I explicitly see both the break away of a small bit of matter and energy release; a neutron turns into a proton or a proton turns into a neutron, plus an electron and other leptons with an energy release in the form of kinetic energy associated with the movement of the electron.

    Is it correct to say that binding energy is the small bit of missing mass of the bounded parent (mass deficit) as compared to the total of the unbounded masses of the parent's constituents (nucleons) free from the nucleus? While I think this is correct, I found an explanation of both alpha and beta decay that says the opposite: "Energy is released in the process of alpha [and beta] decay. Careful measurements show that the sum of the masses of the daughter nucleus and the alpha [beta particle] is a bit less than the mass of the parent isotope. ...the mass that is lost in such decay is converted into kinetic energy and carried away by the decay products."

    Moreover, given that protons and neutrons are also in motion all the time, kinetic energy must be somehow involved in all nuclear reactions not just in radioactive decay. And since nuclear potential energy is stored energy inside nuclei which is a measure of the work done to bind protons to protons, neutrons to neutrons?, and protons to neutrons, it too must be somehow involved in all nuclear reactions.

    Example: In alpha decay, the element uranium (92U238) decays to an isotope of thorium (90Th234) + 2He4 (alpha particle). A little bit of matter (the tightly bound helium nucleus breaks away from the unstable uranium element , it decays to a thorium isotope which itself is unstable so decay continues until a stable isotope is produced. Unless the binding energy is fleeting existing only during the process or lost in the rounding, I don't find it when during an accounting of the mass in alpha decay. In the decay of uranium (92U238 --> 90Th234 + 2He4), all the mass is accounted for without allocating any mass to energy in the end products; if binding energy is involved, it seems only so in terms of its mass equivalent but not explicitly as energy per se.
     
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  3. Jul 10, 2008 #2

    jtbell

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    The masses of the three atoms are 238.050788, 234.043601, and 4.002603 u respectively. The total mass of the products is less than the original atom by 0.004584 u, which corresponds to 4.27 MeV of energy.
     
  4. Jul 10, 2008 #3
    Thanks for your response! It agrees with information I found at the website: ABCs of Nuclear Science but it does not seem to agree with this posting taken from Wikipedia on Binding Energy. In the example below the parent mass is less than the constituent masses. Is the Wikipedia article wrong? What am I missing?

    "Specific quantitative example:
    A deuteron is the nucleus of a deuterium atom, and consists of one proton and one neutron. The experimentally-measured masses of the constituents as free particles are mproton = 1.007825 u; mneutron= 1.008665 u; mproton + mneutron = 1.007825 + 1.008665 = 2.01649 u. The mass of the deuteron (also an experimentally measured quantity) is atomic mass 2H = 2.014102 u. The mass difference = 2.01649−2.014102 u = 0.002388 u. Since the conversion between rest mass and energy is 931.494MeV/u, a deuteron's binding energy is calculated to be 0.002388 u × 931.494 MeV/u = 2.224 MeV."


    Also, how would you characterize the energy equivalent of the missing mass, is it kinetic energy or nuclear potential energy, or is it both?

    Dave
     
  5. Jul 10, 2008 #4

    Vanadium 50

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    That means the parent is bound. The deuteron does not decay.
     
  6. Jul 11, 2008 #5

    jtbell

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    This means that you must supply at least 2.224 MeV of energy in order to separate a deuteron into a proton and neutron. Think of this energy as going into the "extra" mass of the separate particles. If you supply more energy, it becomes kinetic energy of the proton and neutron.

    A deuteron doesn't spontaneously separate into proton and neutron, whereas U238 does decay spontaneously into Th234 and 4He.
     
    Last edited: Jul 11, 2008
  7. Jul 11, 2008 #6
    Yes, I think this is getting at the core of my confusion but I can't help thinking that a bit of circularity or at least a difference in focus or emphasis lies beneath the contradictions referred to earlier. Let me try to explain.

    It seems that whether we are focusing on stable elements or unstable isotopes, the "originals" must contain binding energy in the form of nuclear potential energy mediated by the strong force in addition to mass, otherwise the protons/neutrons would fly away. In the case of radioactivity, internal forces are sufficient to break away small bits of matter (e.g. the alpha or beta particles) but with stable elements, an external source of energy, that is at least equal to the binding engery, is needed to bring about a split. With this in mind, perhaps a more complete discussion of deuterium fission would involve the following:

    deuterium (2.014102 u) which surely must include the mass equvalent of its binding energy (proton to neutron); to which we add 2.224 MeV yielding certain constituents having, in total, a greater mass than the original. If the additional mass is solely explained in terms of the mass equivalent of the disassembly energy (2.2224 MeV) externally supplied, what happens to the nuclear potential energy that bound the proton and neutron in deuterium? Either it flys away, thus fleeting, or it is not accounted for because it is presumed inconsequential. It can't, however, be explained as constituent binding energy, in the case of deuterium, as its constituents are an unbounded proton and neutron. And explaining the additional constituent mass in terms of the external energy supplied is surely part of the explanation but I can't help think there is a bit more to it.​

    Thanks for your patience in helping me work through this confusion. I'm a 70 year old financial guy who is just now trying to understand chemical and nuclear reactions as preperation for a foray into theoretical physics.

    Dave
     
  8. Jul 12, 2008 #7
    Do you (or anybody else) agree with this statement? If not, please explain. I think a response would go a long way toward eliminating my confusion or perhaps, if you agree, provide confirmation. Thanks!

    Do you (or anybody else) know the answer to this question and do you agree with the premise on which it is based? If not, please explain. Again, an answer would go a long way toward eliminating my confusion or perhaps, if you agree, provide confirmation. Thanks!

    Dave
     
    Last edited: Jul 12, 2008
  9. Jul 12, 2008 #8

    jtbell

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    The potential energy of a bound system is less than the PE of the same system after it is separated. By convention, we set PE = 0 when the system is completely separated, which makes it negative when the system is bound.

    This is true for gravitational potential energy (earth + moon) and electrostatic potential energy (electron + nucleus in an atom), as well as for nuclear potential energy (proton + neutron in a deuteron). When you separate any of these bound systems, the potential energy increases. This increase is what we call the binding energy.

    Binding energy is usually spoken of as positive, so it really should be called "un-binding energy" or "separation energy."
     
  10. Jul 12, 2008 #9
    Thank you jtbell, your responses have been extremely helpful.

    Dave
     
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