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Suppression of beta decay

  1. Jul 4, 2004 #1

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    A free neutron is said to have a half life of something around 15 minutes, or maybe it's 18 minutes, I can't remember exactly. This brings up a couple of questions.

    (1) How does the experimenter maintain a neutron in a free state? Presumably maintaining a high vacuum is one requirement, but given a vacuum chamber, how does one keep a neutron from touching a wall of the chamber? Is the nonzero magnetic moment of the neutron somehow utilized?

    (2) Why is beta decay typically suppressed when the neutron is inside of a nucleus? Once the neutron is part of a nucleus, depending on the particular isotope, it may last billions of years. What is going on? I know that once in a nucleus, the neutrons and protons interconvert due to exchange of charged pions. But one might naively assume that the nucleons would then beta decay in about twice the half life of a free neutron, because of the thought that any given nucleon is spending half of its time as a stable proton. (Actually, I think protons can also decay in certain highly-energetic nuclear states, but that's a topic for another discussion.)
     
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  3. Jul 4, 2004 #2

    arivero

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    (2) Because the neutron is *part* of the nucleus, it is not only that it interconverts, it is that it is part of a system having a collective energy, so decay is not favoured. See it from this way: if it decays to a proton, due to electric repulsion the nucleus will have too many protons, so the energy of the nucleus must be higher than having a neutron and some proton should decay to a neutron again. So to avoid this vicious circle, the neutron does not decay in first place.

    Ah, please note the neutral pion :-) In fact, most of the exchanges in nucleus do not need to carry electrical charge.
     
  4. Jul 4, 2004 #3

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    Arivero, you may very well be right in your explanation; I read it several times trying to get it to sink in.

    Another naive thought is that about 2/3 of the pions being exchanged in the nucleus would be the charged pions, since there are twice as many pions of the charged type as of the uncharged type. Can you offer a reason why "most of the exchanges in the nucleus do not need to carry electrical charge"?
     
  5. Jul 4, 2004 #4

    arivero

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    Olias, you should read "Space-Time Code", by David Filkenstein :-)

    It seems that the more I use the internet, the more I forget my English :-(
    I apologise.

    My internal reasoning here was that neutrons and protons are filling shells with different energy levels, so if they exchange, they are probably to be forced to jump to other shell. On the other hand, exchanges with the same kind particle in the same shell seemed to me less problematic. Hmm I see, it is very weak as an argument. Perhaps I am wrong here.

    On other hand, the people doing nuclear models use a set of four particles whose correspondence with the pions is not straightforward. One of these, the sigma particle, is supposed to be a pair of pions (a friend likes to call it a tetraquark).
     
  6. Jul 4, 2004 #5

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    Actually I was not criticizing your English writing skill. I was just having trouble figuring out how your explanation works is all.
     
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