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).

Hope I get an answer.

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## Comments

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TopNewestif 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.

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

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