Hydrogen atoms are excited from ground state to the state of principal quantum number $4$. Then,the number of spectral lines observed will be . . . . . . .

In the hydrogen spectrum,the wavelengths of light emitted in a series of spectral lines is given by the equation,$\frac{1}{\lambda}=R\left(\frac{1}{4^{2}}-\frac{1}{n^{2}}\right)$,where $n=5, 6, 7, \ldots$ and $R$ is Rydberg's constant. Identify the series and wavelength region.

In an $H$-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \, nm$ is emitted. What will be the wavelength of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

In the given figure,the energy levels of a hydrogen atom have been shown along with some transitions marked $A, B, C, D$ and $E$. The transitions $A, B$ and $C$ respectively represent:

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.

The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is: