Quantum Mechanical Model of the Atom

The number of orbitals possible for the following quantum numbers is/are

$\mathrm{n}=3,\hspace{0.17em}\hspace{0.17em}\mathrm{l}=2, \mathrm{m}=+2$

In hydrogen atom, the degeneracy of the level with energy $\frac{-{\mathrm{R}}_{\mathrm{H}}}{9}$ is

The orbital angular momentum of a $\mathrm{p}-$electron is given as:

Which of the following elements contain non-spherical electron density?

The number of nodal planes in $5\mathrm{d}$ orbital is _____.

The electrons occupying the same orbital are distinguished by_____ quantum number. (azimuthal/spin)

The number of nodal planes in $5\mathrm{d}$ orbital is

What is the maximum number of electrons that can be associated with the following set of quantum numbers?
$\mathrm{n}=3, \mathrm{l}=1 \mathrm{and} \mathrm{m}=-1$

The correct set of four quantum numbers for the valence electron of rubidium ($\mathrm{Z}=37$) is

What is the maximum number of orbitals that can be identified with the following quantum numbers? $\mathrm{n}=3; \mathrm{l}=\mathrm{l}; {\mathrm{m}}_{\mathrm{l}}=0;$

For Principal quantum number $\mathrm{n}=4$, the total number of orbitals having $\mathrm{l}=3$ is

Maximum number of electrons in a subshell of an atom is determined by the following

Principal, Azimuthal and magnetic quantum numbers are respectively related to

What is the maximum number of electrons that can be associated with the following set of quantum numbers?
$\mathrm{n}=3, \mathrm{l}=1 \mathrm{and} \mathrm{m}=-1$

The correct set of quantum numbers for the unpaired electron of $\mathrm{Cu}$ atom is :

Among the following, the incorrect statement is

The number of nodes in the $4\mathrm{f}$ orbital is _____

The maximum orbital angular momentum of an electron with n = 5 is

It is tempting to think that all possible transitions 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 has an intrinsic spin angular momentum of $\frac{\sqrt{2}\mathrm{ }\mathrm{h}}{2\mathrm{\pi }}$ corresponding to $\mathrm{s}=1$, 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, $\text{L}=\left[\sqrt{l\left(l+1\right)}\right]\text{h}/2\text{π}$ and spin angular momentum, ${\text{L}}_{\text{s}}=\sqrt{\text{s}\left(\text{s}+1\right)}\text{h}/2\text{π}$ arising from orbital motion and spin motion of electron, respectively. The change in angular momentum of the electron during any electronic transition must compensates for the angular momentum carried away by the photon. To satisfy this condition, the difference between azimuthal quantum number of the orbitals within which the transition takes place must differ by $1$. Thus, an electron in a $\mathrm{d}$-orbital $\left(\mathit{\text{l}}=2\right)$ cannot make a transition into an $\mathrm{s}$-orbital $\left(\mathit{\text{l}}=0\right)$ because the photon cannot carry away enough angular momentum.

The maximum orbital angular momentum of an electron with $\mathrm{n}=5$ is