# Orbital Angular Momentum of an Electron

Recently I studied the wave equation.Then I read about quantum numbers.Now the quantum number about which I am concerned about is the Azimuthal Quantum number or the orbital angular momentum quantum $$l$$ . Now, $$l$$ can have the values

$$l$$ = 1 , 2 , 3, ........... , ($$n$$-1)

Now I further read that for $$s$$ orbital $$l$$ = 0 . Again for a single electron magnitude of orbital angular momentum is given by

$$\sqrt{l(l + 1)}$$ $$\hbar$$

Now if we put $$l$$ in the above equation we get the magnitude of the orbital angular momentum .i.e., $$l$$ = 0

Means the electron is not at all moving around the nucleus, and still the electron doesn't fall in the nucleus ??

For a while I considered the spin angular momemntum $$s$$ , but it doesn't help in any way.If the electron is fixed at a point , spinning( I know spin does not literraly mean spinning on the axis like the planets.Its an intrinsic property like charge) at a point does not help either.

Further, even spin does affect it, the two electrons in the same orbital have opposite spin which would cancel out(opposite direction).

Note by A Brilliant Member
5 years, 6 months ago

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if I am not wrong then - the value of orbital angular momentum = 0 does not describes the velocity of the electron (i.e. it is not moving). But it tells the total magnitude of the "angular momentum" of an electron in an orbital.

- 5 years, 6 months ago

I am afraid you are wrong

Firstly, angular momentum includes velocity as $$L$$ = $$r \times (mv)$$

And hence angular momentum describes the velocity too. Second, as you mentioned the total angular momentum is given by the quatum number $$j$$ which has the values

$$j$$ = $$l$$+$$s$$, $$l$$+$$s$$-1 ,...........|$$l$$-$$s$$|

And gives the magnitude of TOTAL angular momentum(orbital angular momentum + spin angular momentum) according to the relation

Total angular Momentum = $$\sqrt{l(l+1)}$$ $$\hbar$$

- 5 years, 6 months ago