This ought to be simple, I think. But I haven't found a consistent way to think about things yet.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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