If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

Which series of the hydrogen spectrum was observed first?

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first member of the Balmer and Paschen series in a hydrogen atom,respectively,then the ratio of the respective frequencies,$f_1 / f_2$,is:

In the hydrogen spectrum,the shortest and longest wavelengths of the Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant $R$ of hydrogen is:

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \text{ eV}$. The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. The maximum wavelength of the emitted radiation corresponds to the transition between: