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The spin of fundamental particles determines if they are bosons or fermions. The atoms also have bosonic or fermionic behavior, for example $\require{mhchem}\ce{^4He}$ has bosonic and $\ce{^3He}$ has fermionic statistics. Which quantity of atom determines its statistics?

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    $\begingroup$ For a true atom, which is electrically neutral, it's the number of neutrons. Even number of neutrons means bosons, odd means fermions. The proton-electron pairs "cancel" and behave like bosons. $\endgroup$ – Luboš Motl Aug 26 '13 at 18:37
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    $\begingroup$ possible duplicate of Huge confusion with Fermions and Bosons and how they relate to total spin of atom $\endgroup$ – user4552 Aug 26 '13 at 19:32
  • $\begingroup$ @LubošMotl: I think this is somewhat of an oversimplification, for the reasons explained in my answer to the question that this one duplicates. $\endgroup$ – user4552 Aug 26 '13 at 20:25
  • $\begingroup$ Thanks, Ben, for having pointed out the previous question. Both answers to that older question, not just yours, are just plain wrong. There doesn't exist any complication of the sort that would obscure the boson/fermion identity of a particle, whether it's elementary or composite. The signs added under exchange are always defined for a particle with a finite angular momentum and they're always tightly linked to the integrality/half-integrality of the angular momentum. This behavior is never obscured; only your answers are obscuring a very clear issue. See my answer over there. $\endgroup$ – Luboš Motl Aug 27 '13 at 1:14
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A composite particle consisting of an even (odd) number of fermions behaves like a boson (fermion).

Swapping $\require{mhchem}\ce{^3He}$ atoms involves swapping an odd number of fermions (3 nucleons and 2 electrons): an odd number of sign changes of the wave function corresponds to no sign change or bosonic symmetry. Swapping $\ce{^4He}$ atoms involves swapping an even number of fermions (4 nucleons and 2 electrons): an even number of sign changes of the wave function corresponds to a sign change or fermionic behavior.

The same applies to other atoms: $\ce H$ = 1 proton and 1 electron = 2 (even) fermions: bosonic, $\ce{D}$ = 2 nucleons and 1 electron = 3 (odd) fermions: fermionic, etc.

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  • $\begingroup$ I think this is somewhat of an oversimplification, for the reasons explained in my answer to the question that this one duplicates. $\endgroup$ – user4552 Aug 26 '13 at 20:45
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    $\begingroup$ Thanks, @Ben, for having pointed out the previous question. Both answers to that older question, not just yours, are just plain wrong. There doesn't exist any complication of the sort that would obscure the boson/fermion identity of a particle, whether it's elementary or composite. The signs added under exchange are always defined for a particle with a finite angular momentum and they're always tightly linked to the integrality/half-integrality of the angular momentum. This behavior is never obscured; only your answers are obscuring a very clear issue. See my answer over there. $\endgroup$ – Luboš Motl Aug 27 '13 at 1:15

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