It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of an electron from nay initial orbital to any other orbital. However, this is no so, because a photon has an intrinsic spin angular momentum of `sqrt(2)(h)/(2pi)` corresponding to `S = 1` although it has no charge and no rest mass. on the otherhand, an electron has got two types of angular momentum: Orbital angular momentum, `L = = sqrt(l(l+1))(h)/(2pi)` and spin angular momentum, `L_(s) (=sqrt(s(s+1))(h)/(2pi))` arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carried away by the photon. To satisfy this condition the difference between the azimuthal quantum numbers of the orbitals within which transition takes place must differ by one. Thus, an electron in a d-orbital `(l=2)` cannot make a transition into an s-orbital `(l=0)` because the photon cannot carry away enough angular momentum. An electron, possess four quantum numbers, n l, m and s. Out of these four l determines the magnitude of orbital angular momentum (mentioned above) while m determines its Z-component as `m((h)/(2pi))`. The permissible values of only integers right from `-l` to `+l`. While those for l are also integers starting from 0 to `(n-1)`. The values of l denotes the sub-shell. For `l = 0,1,2,3,4...` the sub-shells are denoted by the symbols s,p,d,f,g....respectively.
The spin-only magnetic moment of a free ion is `sqrt(8)B.M`. The spin angular momentum of electron will be

Options

A. `sqrt(2)(h)/(2pi)`
B. `sqrt(8)(h)/(2pi)`
C. `sqrt(6)(h)/(2pi)`
D. `sqrt((3)/(4))(h)/(2pi)`

Answer

A

Explanation

`mu_(s) = sqrt(n(n+2)) BM`. Where `n = 2`
spin angular momentum: `sqrt(s(s+1)).(h)/(2pi)`