The frequencies for the series limit of the Balmer and Paschen series are $v_1$ and $v_3$ respectively. If the frequency of the first line of the Balmer series is $v_2$,then the relation between $v_1, v_2,$ and $v_3$ is:

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

The wavelength of a spectral line emitted by a hydrogen atom in the Balmer series is $\frac{16}{3 R}$ ($R$ is the Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?

The ratio of maximum to minimum wavelength in the Balmer series of a hydrogen atom is

The shortest wavelength in the Lyman series is $912 \ \text{Å}$. Then, the longest wavelength in the series must be (in $\text{Å}$)

The stopping potential for the photoelectrons emitted from a metal surface of work function $1.7 \ eV$ is $10.4 \ V$. Identify the energy levels corresponding to the transitions in a hydrogen atom which will result in the emission of a wavelength equal to that of the incident radiation for the above photoelectric effect.