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.