If the value of azimuthal quantum number is $3$,the possible values of magnetic quantum number would be

In an atom having $2$ $K$,$8$ $L$,$8$ $M$ and $2$ $N$ electrons,the number of electrons with $m=0$ and $s=+\frac{1}{2}$ is $:-$

In a multi-electron system,the energy of an electron of $2p_y$ orbital is :-

The wave function $(\Psi)$ of $2s$ is given by
$\Psi_{2s} = \frac{1}{2\sqrt{2\pi}} \left(\frac{1}{a_0}\right)^{3/2} \left(2 - \frac{r}{a_0}\right) e^{-r/2a_0}$
At $r = r_0$,a radial node is formed. Thus,$r_0$ in terms of $a_0$ is:

For principal quantum number $n = 4$,the total number of orbitals having $l = 3$ is:

If $Br^{-}$ configuration is $[Ar]\, 3d^{10}\,4s^2\,4p^6$,then $Br^{+2}$ configuration will be identical to which element?