The possible quantum numbers for a $3d$ electron are:

In $Li^{++}$ ion,an electron in the first Bohr orbit is excited to a higher energy level by a radiation of wavelength $\lambda$. When the ion de-excites to the ground state in all possible ways (including intermediate emissions),a total of six spectral lines are observed. What is the value of $\lambda$ in $nm$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s, 1 \, eV = 1.6 \times 10^{-19} \, J$)

If the energy in the first excited state in a hydrogen atom is $23.8 \, eV$,then the potential energy of a hydrogen atom in the ground state can be assumed to be ....... $eV$.

An electron in the ground state of the hydrogen atom has an orbital radius of $5.3 \times 10^{-11} \ m$,while that for the electron in the third excited state is $8.48 \times 10^{-10} \ m$. The ratio of the de Broglie wavelengths of the electron in the ground state to that in the third excited state is:

The ionisation potential of $H$-atom is $13.6 \, eV$. When it is excited from the ground state by monochromatic radiation of $970.6 \, \mathring{A}$,the number of emission lines will be (according to Bohr's theory):

Energy levels $A, B, C$ of a certain atom correspond to increasing values of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1, \lambda_2, \lambda_3$ are the wavelengths of radiations corresponding to the transitions $C$ to $B$,$B$ to $A$,and $C$ to $A$ respectively,which of the following statements is correct?