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Atomic Physics - Radioactive Decay and Stability

  1. Nov 13, 2011 #1
    Hi,

    Explain in terms of the number of nucleons and the forces between them, why argon-36 is stable and argon-39 is radioactive.

    My first doubt regards the number of nucleons. If a nucleon is the collective number of neutrons and protons, if we take carbon 12 for example, does it have 6 or 12 nucleons?

    So, from what I understand, the greater the binding energy, the greater the stability. Unstable nuclei will decay, that is, give out energy (radioactively for example) in order to approach a more stable arrangement.

    Argon 39 would therefore be less stable than Argon 36 due to the fact it has three more neutrons and, hence, since neutrons weigh more than protons, have more mass, which translates into less binding energy?

    Does that explain why each is stable or radioactive?

    Can someone please guide me with this?

    Thanks!
     
  2. jcsd
  3. Nov 13, 2011 #2

    SteamKing

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    The atomic number of carbon is 6. For all isotopes of carbon, there are 6 protons in the nucleus. There are different isotopes of carbon, for example, carbon 12 and carbon 14.
    In the carbon 12 nucleus, there are 6 protons and 6 neutrons. In the carbon 14 nucleus, there are 6 protons and 8 neutrons.
     
  4. Nov 13, 2011 #3
    Ok, but the y-axis of a Binding Energy curve is Binding Energy/nucleon, that is, we would get the Binding Energy of Carbon 12 and divide it by 12 or by 6?
     
  5. Nov 13, 2011 #4

    SteamKing

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    You're not paying attention. Carbon 12 has 12 nucleons (= protons + neutrons)
    Carbon 14 has 14 nucleons. Argon 39 has 39 nucleons
     
  6. Nov 13, 2011 #5
    Ok, thanks got it now.
     
  7. Nov 14, 2011 #6
    You also can't just look at the binding energy per nucleon to determine it's stability. The neutron to proton ratio also plays a role. If this ratio is too high or too low, the isotope is unstable.
     
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