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:

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

In the Lyman series,the series limit of wavelength is $\lambda_1$. The wavelength of the first line of the Lyman series is $\lambda_2$ and in the Balmer series,the series limit of wavelength is $\lambda_3$. Then the relation between $\lambda_1$,$\lambda_2$,and $\lambda_3$ is

Missing lines in a continuous spectrum reveal

An electron jumps from the $4^{\text{th}}$ orbit to the $2^{\text{nd}}$ orbit of a hydrogen atom. Given Rydberg's constant $R_{H}=10^7 \ m^{-1}$,calculate the frequency in $Hz$ of the emitted radiation. (Take $c=3 \times 10^8 \ m/s$)

In terms of Rydberg constant $R,$ the shortest wavelength in the Balmer series of the Hydrogen atom spectrum will be: