What is the ratio of the total energy of an electron in a hydrogen atom in the first excited state to that in the third excited state?

The following diagram indicates the energy levels of a certain atom. When the system moves from the $4E$ level to the $E$ level,a photon of wavelength ${\lambda _1}$ is emitted. The wavelength of the photon produced during its transition from the $\frac{7}{3}E$ level to the $E$ level is ${\lambda _2}$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be:

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

In a hydrogen atom,if the energy difference between the $n = 2$ and $n = 3$ orbits is $E$,how much energy (in terms of $E$) is required to remove an electron from the ground state?

In the hydrogen atom spectrum,let $E_1$ and $E_2$ be the energies for the transitions $n=2 \rightarrow n=1$ and $n=3 \rightarrow n=2$ respectively. The ratio $E_2 / E_1$ is

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths,$\lambda_1/\lambda_2$,of the photons emitted in this process is