The difference between the frequencies of the first and second Lyman lines of the hydrogen atom is (where $R$ is the Rydberg constant and $c$ is the speed of light in vacuum).

In the hydrogen emission spectrum,for any series,the principal quantum number of the higher energy level is $n+1$ and the lower energy level is $n$. The corresponding maximum wavelength $\lambda$ is ($R=$ Rydberg's constant).

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

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

The first three spectral lines of the $H$-atom in the Balmer series are given as $\lambda_{1}, \lambda_{2}, \lambda_{3}$. Considering the Bohr atomic model,the wavelengths of the first and third spectral lines $\left(\frac{\lambda_{1}}{\lambda_{3}}\right)$ are related by a factor of approximately '$x$' $\times 10^{-1}$. The value of $x$,to the nearest integer,is:

The ratio of minimum wavelength of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is