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B Why is Carbon-14 Unstable?

  1. Nov 28, 2017 #1
    Hi pf,

    We are currently learning about nuclear stability in class by looking at the nuclear stability graph when you plot the proton number against the neutron number. I understand that if an isotope is on the stability line then it is not radioactive. Therefore unstable nuclei have too many protons or too many neutrons causing it to sit off the line. In my textbook it says too many protons is unstable because it increases the electrostatic repulsion in the nucleus. It goes on to say that by undergoing beta decay it turns a proton into a neutron which reduces the electrostatic repulsion as extra neutrons decrease the electrostatic force. However, it gives no explanation of why too many neutrons causes an instability. If neutrons decrease the electrostatic repulsion then how is having slightly too many unstable?

    Thanks
     
  2. jcsd
  3. Nov 28, 2017 #2

    Vanadium 50

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    Because the neutrons have to go into higher energy shells. Eventually you get to a point where it is energetically favorable to turn a neutron in a high energy level intp a proton at a low energy level.
     
  4. Nov 29, 2017 #3
    Thank you. So it is energetically more favourable for the extra two neutrons that carbon-14 has over carbon-12 to decay into protons than to move into higher energy shells? So even in carbon-14 which doesn't have many neutrons - they still occupy high energy levels?
     
  5. Nov 29, 2017 #4
    According to the so-called nuclear shell model the lowest levels are ##0s_{1/2}##, ##0p_{3/2}## and ##0p_{1/2}##. In these levels you can put 2, 2 and 4 neutrons respectively. The lowest "magic" numbers are 2 and 8, since it is a large energy splitting between the ##0s## and the two ##0p## levels. However, for light nuclei the splitting between ##0p_{3/2}## and ##0p_{1/2}##. Therefore, one has a semi-magic number of 6, i.e. 2 neutrons in ##0s_{1/2}## and 4 in ##0p_{3/2}##.
    For carbon-14 you need to put 2 neutrons in ##0p_{1/2}## also. It is thus energetically more favorable to have a carbon-12 nucleus + 2 free neutrons instead of a bound carbon-14.
     
  6. Nov 29, 2017 #5

    DrClaude

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    But a carbon-14 nucleus doesn't decay to carbon-12, but to nitrogen-14:
     
  7. Nov 29, 2017 #6
    Yes, it is correct! Thanks DrClaude for pointing my mistake. I was writing a bit too fast.
     
  8. Dec 1, 2017 #7
    It's all about balance of force.
    Think of protons as two pieces of paper and the neutrons as glue holding them together.

    Not enough glue and they won't stick together. Too much glue and it never dries so they don't stick together.
     
  9. Dec 2, 2017 #8

    mfb

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    I don't think that is a useful analogy.
     
  10. May 22, 2018 #9
    So I heard that one of the neutrons will decay into a proton because a nutrino comes along and gives the neutron a W boson. This turns one of the down quarks in the neutron into an up quark and additionally the nutrino becomes an electron (because it lost the W boson). So would it be right to conclude that the reason nutrinos can interact with the neutrons in an unstable nucleus, and not a stable nucleus, is because only the neutrons in the unstable nucleus have enough energy for the interaction to happen?
     
  11. May 22, 2018 #10

    jtbell

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    Where did you hear that? No neutrino comes in before the decay. The down-quark changes into an up-quark by "emitting" a virtual W, which "decays" into an electron and an antineutrino.

    (The W is virtual because we can't detect it directly. It's basically a calculational tool that allows us to arrive at the final result.)
     
  12. May 22, 2018 #11

    mfb

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    Neutrinos can induce reactions (inverse beta decay), but that is a very rare process, and it is not relevant for Carbon-14. Unlike radioactive decays this reaction gets additional energy from the neutrino, it can also happen in stable nuclei if the neutrino energy is sufficient.
    It does not "lose" a W boson. It didn't contain a W boson before.
     
  13. May 24, 2018 #12

    dlgoff

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    14C is unstable, ... barely. :olduhh:

     
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