# Angular momentum of an electron in an atom

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?

• 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. Commented Jan 16, 2019 at 20:16
• 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.
– Greg
Commented Jan 17, 2019 at 3:28
• 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.) Commented Jan 17, 2019 at 8:50
• @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. Commented Jan 17, 2019 at 10:12

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!
• 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. Commented Jan 17, 2019 at 10:14