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Problem on tritons and deuterons

  1. Dec 2, 2006 #1
    Suppose we have a beam balance capable of measuring even smallest differences in weights.
    Now on one side we have a Triton and Deuteron.
    And on the other side we have He-nucleus and a neutron

    I want to know which will weigh heavier and why?

    I saw on a book cover with balance tilting towards the former one thus i concluded that the deuterons and tritons will weigh more but my explanation on that conclusion don't seem to be logical.

    Is that book cover right or I have been just fooled and my fumndamentals are just checked?
    Last edited: Dec 2, 2006
  2. jcsd
  3. Dec 2, 2006 #2
  4. Dec 2, 2006 #3
    This means helium nucleus is more stable as it do have such even arrangement of both protons and neutrons.
  5. Dec 6, 2006 #4
    Why would even=stable? Yes, He4 is stable AND seems to be a fundamental building block for heavier elements as He4 is often a decay component. I believe there is a fundamental "structure" to He4 that gives it its stability, though I will admit that the "even" P/N count may provide for this stable structure.
  6. Dec 6, 2006 #5
    Having an equal number of protons and neutrons does not always result in stability--for example see this isotope chart for Beryllium:
    Note that Beryllium-8, which has 4 P and 4 N [NNNNPPPP] is not stable.
  7. Dec 7, 2006 #6

    Meir Achuz

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    2p + 2n is the smallest number of nucleons that can fit into the same ground S-state. At the same time, each of these nucleons feels the attractive force of the nearby other three. That is why He^4 is the stablest nucleus.
    Be^8 is not stable because it decays ilnto a pair of these most stable He^4.
  8. Dec 7, 2006 #7
    I wouldn't categorize that explanations as "why", but it's the current leading theory. "feels attractive force" is a concept with no structure, but I don't disagree. True, He^4 is the most stable known nuclei, but there may still be baryonic nuclear structures "larger" but more stable. One possible candidate would be a baryonic neutrolino, but the current theory gives a huge range of energy values and no composition or structure. Another could be n^4 - no charge, no magnetic moment, very stable and very dense. The later is predicted by W.F. Hagen's second paper on Energiewirbel.

    Hmm.. How about Be^8 is not stable AND it decays into a pair of He^4. Why it is unstable remains unexplained. This decay example does however reinforce one of my first points that He^4 is a building block of larger nuclei.
  9. Dec 7, 2006 #8
    One hypothesis as to why Be-8 is unstable derives from the Close-Packed Spheron Model of Linus Pauling. Here is quote from Pauling: "I assert that nuclei beyond A = 4 are stable only if two or more resonating structures involving hydrogen-3, helium-3, and helium-4 can be written". Dr. Linus Pauling, Research Notebook, #26 . 13 February 1966 :http://osulibrary.oregonstate.edu/specialcollections/rnb/26/rnb26.html

    According to this Pauling model there "must" be present a three nucleon cluster, either triton [NPN] or [PNP], rotating against either (1) another three nucleon cluster or (2) a helium-4 cluster (the alpha)for any isotope with A > 4 to be "stable". Hence, Be-8, being composed of two helium-4 rotating clusters [PNPN]~[PNPN] violates this basic "rule of assembly" dynamic and is thus "unstable". Note the same explanation holds for He-5, in fact no mass 5 or 8 isotopes are stable. Now, before anyone starts throwing darts, Linus Pauling was a genius, holds two Nobel Prizes-- and is it not interesting that his model that explains why no stable mass 5 or 8 isotopes exist, published in Science and Proc. National. Acad. Sci. in 1965, has never been falsified.
    Last edited by a moderator: Dec 7, 2006
  10. Dec 7, 2006 #9
    Great link, thanks!!
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