It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of `sqrt2 (h)/(2pi)` corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum :
Orbital angular momentum, `L=sqrt(l(l+1))h/(2pi)` and spin angular momentum, 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 carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as `m((h)/(2pi))` the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively
The orbital angular momentum of an electron in p-orbital makes an angle of `45^@` from Z-axis. Hence Z-component of orbital angular momentum of election is :

Options

A. `h/pi`
B. `((h)/(2pi))`
C. `- h/pi`
D. `-((h)/(2pi))`

Answer

B

Explanation

`m= sqrt(l (l+ 1)cos theta), l = 1 ` for `p^-` orbital,
`m = sqrt(1(2)) cos 45^@`
Z- component of O.A.M = `h/(2pi)`