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).

Hope I get an answer.

Note by Pranjal Rs
5 years ago

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4 votes

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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.

Kishlaya Jaiswal - 5 years ago

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

Pranjal Rs - 5 years ago

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