What is the energy of the electron revolving in the third orbit expressed in $ eV $ (in $eV$)?

To excite the spectral line of wavelength $4960 \mathring A$ of an atom,an excitation energy of $7.7 \text{ eV}$ is required. The ground state energy of the atom is $10.5 \text{ eV}$. The energies of the two levels involved in the emission of the $4960 \mathring A$ line are (Assume $hc = 1240 \text{ eV nm}$,where $h$ is Planck's constant and $c$ is the speed of light).

The four lowest energy levels of an $H$-atom are shown in the figure. The number of possible emission lines would be

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be ......$V$ (Take $hc = 12420 \text{ eV-} \mathring{A}$).

$A$ sample of hydrogen atoms is in an excited state (all the atoms). The photons emitted from this sample are made to pass through a filter through which light having a wavelength greater than $800 \ nm$ can only pass. Only one type of photon is found to pass through the filter. The sample's initial excited state is: [Take $hc = 1240 \ eV \cdot nm$,ground state energy of hydrogen atom = $-13.6 \ eV$.]

$A$ $12.5 \text{ eV}$ electron beam is used to bombard gaseous hydrogen at ground state. The energy level up to which the hydrogen atoms would be excited is: