The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line in the same series?

If the wavelengths of the first Lyman line for Hydrogen,$He^+$,and $Li^{2+}$ ions are $\lambda_1$,$\lambda_2$,and $\lambda_3$ respectively,then the ratio of these wavelengths is:

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

The first wavelength of the Paschen series in hydrogen is $18,800 \, \mathring A$. What is the minimum wavelength of the Paschen series in $\mathring A$?

The first line of the Balmer series has a wavelength of $6563 \mathring{A}$. What will be the wavelength of the first member of the Lyman series in $\mathring{A}$?

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.