Question

The orbital momentum and angular momentum (classical analogue) for the electron of 4s-orbital are respectively, equal to:
(A) $$\sqrt{12}{\displaystyle \frac{h}{2\pi }}$$ and $${\displaystyle \frac{h}{2\pi }}$$
(B) Zero and $${\displaystyle \frac{2h}{\pi }}$$
(C) $$\sqrt{6}h$$and $${\displaystyle \frac{2h}{\pi }}$$
(D) $$\sqrt{2}{\displaystyle \frac{h}{2\pi }}$$and$${\displaystyle \frac{3h}{2\pi }}$$

Answer 1

Hint: For finding the orbital angular momentum and angular momentum we have to write their formulas in terms of ‘n’ and ‘l’. By using direct formulas we can find these values for 4s- orbital. Use the hint that l value for s orbital is zero.

Step by step solution:
Orbital Angular Momentum: The angular momentum of a microparticle moving in a strong field that has spherical symmetry. And the term “orbital angular momentum” is due to the graphic representation of the motion of an atomic electron in a spherically symmetric field of the nucleus in a definite closed orbit.
The angular momentum of an electron by Bohr is given by mvr or $${\displaystyle \frac{nh}{2\pi }}$$ (where v is the velocity, n is the orbit in which electron is, m is mass of the electron, and r is the radius of the nth orbit).
 Orbital angular momentum:
Orbital angular momentum is the component of angular momentum of an electron in an atom arising from its orbital motion rather than from its spin.
The orbital angular momentum is given by the formula:
Orbital angular momentum $$(O.A.M.)=\sqrt{l(l+1)}{\displaystyle \frac{h}{2\pi }}$$
Where h = planck constant
‘l’ = Azimuthal quantum number
And we know that for ‘s’ orbital value of l = 0
Hence,
Orbital angular momentum for 4s = $$\sqrt{0(0+1)}{\displaystyle \frac{h}{2\pi }}$$
Orbital angular momentum = 0
And we know that Angular momentum $$A.M.=mvr={\displaystyle \frac{nh}{2\pi }}$$
For ‘4s‘ the value of n = 4
Now we will put this value of n in above equation:
Angular momentum $$A.M.={\displaystyle \frac{4h}{2\pi }}$$
Angular momentum A.M. = $${\displaystyle \frac{2h}{\pi }}$$
So, from the above derivation or calculation we can say that the answer is option “B”.

Note: In the orbital angular momentum there is no role of orbital number and in angular momentum there is no role of Azimuthal quantum number. And here you should remember the Azimuthal quantum number for every orbital, which is 0 for s, 1 for p, and 2 for d orbital.