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The Orbital Angular momentum of an electron in an atom is given by $\hbar \sqrt{l(l+1)}$ and the Spin Angular Momentum is $\hbar \sqrt{s(s+1)}$. So what is the resultant Angular momentum of an electron in an orbit in an atom?

I came across a question which asked the value of Angular momentum of an electron in its ground state in an atom. Does this refer to resultant of both kinds of Angular momentum, or am I misinterpreting the question?

Also, what exactly does ground state refer to in an atom? Is it the 1s orbital?

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    $\begingroup$ Electrons do not orbit the atom. This is confusing, but try to just accept it: they have angular momentum, but are not "orbiting" as in moving in an orbit around anything. They have spin angular momentum, but are not spinning. $\endgroup$
    – Stian
    Commented Jan 16, 2019 at 20:16
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    $\begingroup$ I would argue that individual electrons do not have well defined angular momentum in an atom, only the many-electron state can be assigned to a well defined angular momentum. Also, not all states are angular momentum eigenstates. $\endgroup$
    – Greg
    Commented Jan 17, 2019 at 3:28
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    $\begingroup$ You should start by looking for LS coupling (Russell-Saunders coupling) in a undergraduate level spectroscopy or phys. chem. textbook. The angular momenta are added in a vectorial like manner (vector model of angular momentum). There is a standard method by which to find the total angular momentum quantum number, usually called J. (The Hyperphysics site has some good pics.) $\endgroup$
    – porphyrin
    Commented Jan 17, 2019 at 8:50
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    $\begingroup$ @porphyrin, Thanks for the reference. At present, this seems to be somewhat out of my scope, since I'm only in Grade 12, but I'll try my best to understand. $\endgroup$
    – Ankit Kumar Misra
    Commented Jan 17, 2019 at 10:12

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The question does not state what kind of atom, but I assume this is about the Hydrogen atom. In this case you can combine orbital and spin angular momentum according to Russell–Saunders coupling.

Since we are talking about the Hydrogen atom, the ground state is essentially just the $1s$ orbital. In general however, for many-electron systems, electronic states and orbitals are different things!

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  • $\begingroup$ So what would be meant by 'ground state' in any general atom, in view of the quantum mechanical model of an atom? Isn't the 1s orbital the one with the lowest energy even in a multi electronic atom? $\endgroup$
    – Ankit Kumar Misra
    Commented Jan 17, 2019 at 8:11
  • $\begingroup$ Yes $1s$ is the lowest orbital, but it is not a state! Within Hartree-Fock, a state is a Slater-Determinant, which is build up from many orbitals and therefore corresponds to an electron configuration. But this is really oversimplifying things. $\endgroup$
    – Feodoran
    Commented Jan 17, 2019 at 10:14
  • $\begingroup$ Maybe better put it this way: An orbital is a one-electron wave function, two electrons if you are talking spatial orbitals, but never more! An electronic state on the other hand is a many-electron wave function, meaning it can be more than just one or two electrons. The state is what you can physically measure, an orbital is just a mathematical tool. $\endgroup$
    – Feodoran
    Commented Jan 17, 2019 at 10:36
  • $\begingroup$ For completeness, I would suggest to mention the alternative for heavier atoms, j-j-coupling. $\endgroup$
    – TAR86
    Commented Jan 25, 2020 at 19:19
  • $\begingroup$ In case of a single electron, LS and jj coupling are equivalent. $\endgroup$
    – Paul
    Commented Sep 16, 2021 at 22:14

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