What is the ratio of the velocity of an electron in the ground state of a hydrogen atom to the velocity of an electron in the second excited state of $He^{+}$?

The energy of an electron in the first Bohr orbit of the $H$ atom is $-13.6 \, eV$. The energy value of an electron in the first excited state of the $H$ atom is ............ $eV$.

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n=4$ to $n=2$ of $He^{+}$ spectrum?

For the Balmer series in the spectrum of $H$ atom,$\bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\}$,the correct statements among $(I)$ to $(IV)$ are:
$(I)$ As wavelength decreases,the lines in the series converge.
$(II)$ The integer $n_{1}$ is equal to $2$.
$(III)$ The lines of longest wavelength corresponds to $n_{2}=3$.
$(IV)$ The ionization energy of hydrogen can be calculated from wave number of these lines.

The energy of an electron in the first Bohr orbit of $H$-atom is $-13.6 \ eV$. The possible energy value of an electron in the first excited state of $Li^{2+}$ is (in $eV$)

An excited hydrogen atom emits light in the ultraviolet region at a frequency of $2.47 \times 10^{15} \ Hz$. Calculate the energy of a single photon. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)