Math for identifying bosonic atoms

1. Jun 11, 2015

itssilva

(And fermionic atoms, as well). Knowing the weird statement that rubidium atoms and such are bosons and reading some of the threads, I've convinced myself that it is a perfectly acceptable possibility; however, I'm not so sure if I learned how to point my finger and say atom X is a boson/fermion.
From what I see, the math is something like this: Wikipedia (BEC article) says 7Li is a boson; it has 3 e, 3 p and 4 n, right? All of which are fermions (s=1/2 for e; for p and n as well?). Now, to obtain the total spin of the ground state, divide each 'class' of fermion by 2; the remainder is the number of unpaired fermions in each class (1 e, 1 p, 0 n). This is the ground state, so the unpaired electron oughta be paired up with the proton (semiclassically: two magnets configured like N-S|N-S have less energy than N-S|S-N). End of the day: no leftover unpaired fermions, S=0. Boson. This is the rule of thumb I've adopted so far.
I say 'rule of thumb' because I'm highly suspicious of this method, since I've learned in my QM that spin addition is a quite fastidious procedure (determining Clebsch-Gordon coefficients and whatnot). Is this the correct math?
Also, I've heard in my QFT class (if I understood correctly) that, if you consider an energy scale less than that which would take to break apart a bound state, it 'looks like a fundamental particle' propagating away; is the same principle at work here? I mean, if the first excitation energy of 7Li is 5 eV and first ionization 11 eV (numbers I made up), does it mean that I oughta keep my lithium at energies lesser than 5 eV (in order to avoid spin-flipping) and/or 11 eV (to avoid ionization) to have it work with B-E statistics?

2. Jun 12, 2015

fzero

Orbital angular momentum is always an integer, so the only way to get a half-integer total angular momentum is to have an odd number of fermions. The rule you have is fine; an atom with an odd number of fermions must obey fermionic statistics.

To answer your second question. First, a spin flip changes the total angular momentum by $\pm 1$, so if the bound state started as a fermion, it will remain one after the spin flip. Second, the statistics of an ion will be determined by the same rule about odd/even number of fermions. If we remove a single electron from a fermionic atom we will have a bosonic ion and vice versa.

3. Jun 12, 2015

DrDu

You don't have to pair up the spins so that S=0 and normally, this won't be the case. However, the total spin will always be an integer number.
Arguing in terms of Clebsch Gordan coefficients: if s1 and s2 are half integer, so will be the z-components of spin. However, the z-components add up directly, so Sz will be integer. This is only possible if S is integer, too, as S_z ranges from S to -S in steps of 1.

4. Jun 12, 2015

mathman

A simple way to tell is by the number of neutrons in the nucleus. If odd it is a fermion, if even it is a boson. This is a result from the obvious fact that the number of protons equals the number of electrons.