How do you deduce that an atom is a fermion or a boson? Do you determine it from the number of neutrons because "electrons and protons cancel out each other in a neutral atom"? What does this have to do with spin? Somebody please help.I am really confused here.
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$\begingroup$ You mean experimentally? $\endgroup$– ShingCommented Mar 30, 2018 at 8:30
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$\begingroup$ See this: physics.stackexchange.com/questions/75352/… $\endgroup$– CAFCommented Mar 30, 2018 at 8:50
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$\begingroup$ No,not experimentally.Just theoretically! $\endgroup$– AnayaCommented Mar 30, 2018 at 10:06
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2 Answers
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It has to do with the overall spin. Bosons have integer spin $(0, 1, 2, \dots)$ and fermions have half-integer spin $(n+\tfrac{1}{2})$.
They can be either elementary or composite. Fundamental fermions that we discovered so far are the quarks and the leptons of the Standard Model. Fundamental bosons that we discovered so far are the gauge bosons (gluons, photon, $W^{\pm}$, $Z^{0}$) and the Higgs boson.
Composite particles like baryons are fermions because they are made of three quarks. Mesons are bosons because they are made of two quarks. Protons and neutrons are baryons and therefore are fermions. A nucleus composed of an odd number of nucleons is a fermion, and if it is composed of an even number of nucleons it is a boson. For example, Helium-3 is a fermion and Helium-4 is a boson.
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You just have to see the total spin, if your atom contains an even number of fermions it is a bosons, so since protons/neutrons/electrons are all fermions, take the sum of these if it is even your atom is a boson, if not it is a fermion
