Is whole Carbon 13 atom (not just the nucleus) a fermion or boson?

  1. This ought to be simple, I think. But I haven't found a consistent way to think about things yet.

    Is it as simple as adding up all the spins of the elementary particles in the particle and checking whether the total has inter or half-integer spin?

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

    State whether the following is a fermion or a boson:
    (a) an electron
    (b) a proton
    (c) a neutron
    (d) a photon
    (e) a carbon 12 nucleus
    (f) a carbon 13 nucleus
    (g) a carbon 12 atom
    (h) a carbon 13 atom
    (i) a Nitrogen 14 atom
    (j) a Nitrogen 15 atom

    2. Relevant equations

    So I was never taught the basics, I am taking a graduate physical chemistry class coming from an electrical engineering background. Most of the things are rather easy, but sometimes I don't know the theory involved.

    So I formed 2 possible ways to approach this (one simple, the other more involved):

    1) Just add up the spins of the elementary particles and check if the result is half integer or integer.

    2) Based on some other reading I did on the web, it seems like:
    (a) For nuclear spin, if Z is even and A is even, I=0. For other cases, I need to look it up.
    (b) For electronic spin, pair up electrons using Hund's rule, Pauli Exclusion principle, etc. Then S=0.5*(# of unpaired electrons)
    (c) In an atom, electron spin dominates (because the electronic magentic moment is 100x larger)

    Of course, if I ignore (c), and the nuclear spin is half-integer only when A is odd, these methods will give the same answers.

    3. The attempt at a solution

    The following give the same answers either way:
    (a) fermion
    (b) fermion
    (c) fermion
    (d) boson
    (e) boson
    (f) fermion
    (g) boson

    (i) fermion

    The following, however, disagree depending on which way I interpret things.

    (h) 1) The sum of spins of elementary particles is an odd multiple of 1/2. So from this I would say fermion.
    2) But the ground state electronic configuration has two unpaired electrons, and if the electronic spins dominate, this would be a boson

    (j) 1) The sum of spins of elementary particles is an even multiple of 1/2. So from this I would say boson.
    2) But the ground state electronic configuration has three unpaired electrons, and if the electronic spin dominates, this would say fermion.
     
  2. jcsd
  3. If you mean "adding the spins of the electrons, protons and neutrons in an atom", then yes. For example, Helium-3 is a fermion but Helium-4 is a boson. (Bare in mind that protons and neutrons are not elementary particles.)

    If you want to get into all the complications of nuclear shell structure, then you'd need to take a course on nuclear physics, but I doubt it is necessary for your problem.
     
    Last edited: Mar 6, 2012
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