As an electron moves away from the nucleus,its potential energy

The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of $H = 2.18 \times 10^{-18} \ J \ atom^{-1}$ and $h = 6.625 \times 10^{-34} \ Js$)

Find the wave number of the shortest wavelength in the Balmer series.

The energy of a photon of wavelength $(\lambda) = 1 \, m$ is (Planck's constant $= 6.626 \times 10^{-34} \, J \, s$,speed of light $= 3 \times 10^8 \, ms^{-1}$)

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

The ratio of minimum wavelengths of Lyman and Balmer series will be $:-$