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Why are neutrons in the nucleus stable vs their free counterparts?

  1. Jul 4, 2012 #1
    If the free neutron decays in only 15 minutes, why are neutrons in the nucleus attached to protons stable? This has always bewildered me.


    Just a side question, are 'tetraneutrons' stable or even possible?
     
  2. jcsd
  3. Jul 4, 2012 #2

    Bill_K

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    Because if a neutron inside a nucleus decayed, it would leave a nucleus having one more proton and one less neutron. Whether this can actually happen depends on whether there is enough energy available, i.e. whether the mass of the first nucleus exceeds the mass of the second.
     
  4. Jul 4, 2012 #3
    But still, surely a free neutron requires the same amount of energy to undergo Beta decay?
     
  5. Jul 4, 2012 #4

    Bill_K

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    In order for a decay process to happen there must be at least as much rest mass in the original nucleus as in the end products. Here's an example:

    The 7Li nucleus has a mass of 7.016003 amu. If one of its neutrons beta-decayed, the result would be a 7Be nucleus, which has one less neutron and one more proton. But 7Be has a mass of 7.016926 amu, which is more. So there's not enough energy for this process to happen.
     
  6. Jul 4, 2012 #5

    jtbell

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    In effect, a neutron in a nucleus gets the energy required for the decay, from the difference in binding energy between the initial and final nuclei.
     
  7. Jul 4, 2012 #6
    So is the extra mass from the energy of the newly created proton opposing the nucleus or am i completely missing this?
     
  8. Jul 4, 2012 #7

    jtbell

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    The sum of the masses of the final nucleus and the electron has to be less than the mass of the original nucleus. Changing a neutron into a proton does increase the overall "internal electrical repulsion" of the nucleus, which would tend to increase the mass, but that's far from the whole story because of the strong nuclear force. Nuclear structure is a complicated subject. For a simplified model, look up the "semi-empirical binding energy formula." (Keep in mind that binding energy is negative.)
     
  9. Jul 5, 2012 #8
    Thanks I understand now, didn't think about binding energies but a free neutron has a binding energy of it's 'system' - (which is just itself) of 0 so therefore any other state has a higher binding energy and so is preferred?
     
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