It is tempting to think that all possible transition are permissible and that an atomic spectrum arises from the transition of an electron from any initial orbital to any other orbital .However this is not so because a photon a photon has as intrinsic spin angular momentum of `sqrt(2) h// 2pi` corresponding to `S = 1` although it has no charge and no rest mass
On the other hand , an electron has got two type of angular momentum: orbital angular momentum
`L = [sqrt(l(l+1))] h//2pi`,and spin angular momentum `L_(1) = sqrt(s(s + 1)) h//2pi` arising from orbital motion and spin motion of the electron during any electronic transition must compensate for the angular momentum carried away by the photon .To satisfy this condition the different between the azimuthal quantum number of the orbital within which the transition take place must differ by 1.thus, an electron in a d-orbital `(l = 2)` cannot make a transition into as s-orbital `(l = 0)`because the photon cannot carry away enough angular momentum
The maximum orbital angular momentum of an electron with `n = 5` is

Options

A. There will be no change in the orbital angular momentum of electron athough the emitted photon has angular momentum
B. There will be change in the orbital angular momentum whereas the emitte photon has to momentum
C. `Delta m_(1) `valuee between `4s`1 and `3s` is not zero , which is an important selection slection rule for allowed transition
D. In `4s` and `3s` orbitals the wavelength of the electeron wave `n = 5` is

Answer

A

Explanation

a. It does not involve any change in the value of l value of l should change by `+- 1`
b. There will be no change in angular momentum because photon have angular momentum
c. `Delta m` for `4s` and `3s` is zero
d. Wavelength of electron wave are not the same because n value is difference