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Decay Modes

  1. May 16, 2009 #1
    Hello, I am having issues with some nuclear physics revision questions

    1. The problem statement, all variables and given/known data

    The following atomic masses have been determined (in Atomic mass units, u) (A = atomic mass number, Z = proton number)
    Li(A=7,Z=3) - 7.0182 u
    Be(A=7,Z=4) - 7.0192 u
    Indicate the stable nucleus and determine its mode(s) of decay.


    2. Relevant equations



    3. The attempt at a solution

    Ok I have approached this knowing the answer, but not understanding it. The Be nucleus is unstable and decays into the Li, most likely by electron capture.

    My problem is I do not understant how to identify the unstable nucleus without reading the answer. The only thing I have noticed which may or may not be relevant is that both nuclei have the same mass number, A, making them isobars.

    Can someone please explain how to identify the unstable nucleus?

    Many thanks,

    Peter
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 16, 2009 #2

    diazona

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    Homework Helper

    The only thing I notice is that the beryllium atom has a slightly greater mass (by 0.001 u) than the lithium atom. When particles decay they lose mass, they don't gain it.
     
  4. May 17, 2009 #3
    Modern Physics by Tipler/Llewellyn (4th ed.) has a good overview of nuclear physics. On pp. 516-518 they have a good discussion of nuclear stability.

    The short of it is this: for light elements (like Li and Be), the most stable ratio of nucleons is roughly 1:1. Both neutrons and protons are fermions, so the exclusion principle makes it very energetically unfavorable to have a nucleus composed of purely protons or purely neutrons, hence the stability when they're about half and half. With heavier elements, Coulombic interactions among the protons becomes very significant, such that the population of neutrons becomes noticeably larger than that of protons. For example, uranium-238 [the most stable isotope of uranium] has 92 protons and 146 neutrons.

    So in the case of your problem, we have an odd number of nucleons, so we can pick either 3 neutrons/4 protons (Be) or 4 neutrons/3 protons (Li). To minimize Coulombic repulsion, pick the one with fewer protons as the stable nucleus. As for which reaction it undergoes, there are only four common ones (alpha, beta+, beta-, and electron capture). Pick the one that gives you the correct change in A and Z, and if there is more than one that could do so, use conservation laws to figure out if one is more favorable than the other.
     
  5. May 17, 2009 #4
    Thanks to both of you, I shall go in search of that book in the morning.
     
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