The ratio of wavelengths of the second line in the Balmer series and the first line in the Lyman series of a hydrogen atom is:

What is the shortest wavelength present in the Paschen series of spectral lines (in $;nm$)?

$A$ hydrogen atom in the ground state is given an energy of $10.2 \ eV$. How many spectral lines will be emitted due to the transition of electrons?

For a hydrogen atom,the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:

The wavelengths corresponding to the first four spectral lines of the Lyman series of the $H$-atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}$,and $951.4 \, \mathring{A}$. Now,consider a deuterium atom instead of a hydrogen atom. Given the mass of the hydrogen atom is $1.6725 \times 10^{-27} \, kg$,the mass of the deuterium atom is $3.3374 \times 10^{-27} \, kg$,and the mass of the electron is $9.109 \times 10^{-31} \, kg$,calculate the percentage change in the wavelength of the first spectral line of the Lyman series for the deuterium atom relative to the hydrogen atom.

Find the maximum wavelength of the Brackett series for a hydrogen atom in $\mathring A$.