Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Free neutron vs bound neutron decay

  1. Jul 12, 2014 #1
    I was curious as to the difference in free neutron decay and the lack of decay when bound in a nucleus. I found the following statement here http://en.m.wikipedia.org/wiki/Neutron#Free_neutron_decay

    I don't understand that. Is the wave function of the neutron different when free or bound? The phrase "is balanced" sounds wrong; balanced by who or what?

    Perhaps they mean that instead of a Feynman diagram with a neutron in and proton/electron/anti neutrino out, we should use a diagram with a nucleus in and a nucleus(with higher atomic number)/electron/anti neutrino out. That would suggest to me the wave function of the nucleus supersedes the wave functions of the constituent nucleons.

    Would someone please explain those two Wikipedia sentences, or point me to a more understandable source?

    Follow up question: Since the binding energy per nucleon is different in different elements, is the stability of a bound neutron a function of which element/isotope it is bound in?
  2. jcsd
  3. Jul 12, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For example, let's say we have a 16O nucleus. It's not possible for one of its neutrons to undergo beta decay because the resulting system would be 16F (plus an electron and a neutrino). But 16F has a higher mass-energy than 16O, so this would violate conservation of mass-energy.

    Similarly, a free proton can't beta decay because of conservation of mass-energy.
  4. Jul 12, 2014 #3


    User Avatar
    2015 Award

    Staff: Mentor

    Sure, it is bound, so it has a different energy. Even more important, a produced proton would have a different energy as well - too high to be accessible in the decay.

    That's another (more general) way to see it, yes.

  5. Jul 12, 2014 #4
    I wish Wikipedia would explain as simply as you guys did. Thanks.
  6. Jul 13, 2014 #5
    I've been thinking more on this and I would like a follow up question please.

    I've taught that when two electrons bind in a singlet pair, the identity and individual quantum states of the constituents disappear. Only the quantum states of the singlet may be observed.

    Should I think of a multi nucleon nucleus in a way analogous to the singlet pair?
  7. Jul 14, 2014 #6


    User Avatar
    2015 Award

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Free neutron vs bound neutron decay
  1. Neutron decay? (Replies: 9)

  2. Neutron Decay (Replies: 2)

  3. Neutron decay (Replies: 7)

  4. Neutron decay (Replies: 1)