If a hydrogen atom is excited from the ground state to another state with a principal quantum number $n = 4$,the number of spectral lines in the emission spectrum will be:

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:

For the wavelength of visible radiation of the Hydrogen spectrum, Balmer gave an equation as $\lambda = \frac{xm^2}{m^2 - 4}$, where $m$ is an integer value. The value of $x$ in terms of Rydberg's constant $R$ is

The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are:-

If $R$ is the Rydberg constant in $cm^{-1}$, then the hydrogen atom does not emit any radiation of wavelength in the range of

The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is