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.
Comments
Sort by:
Top Newestif 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 · 4 years, 1 month ago
Log in to reply
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 · 4 years, 1 month ago
Log in to reply