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?

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

The maximum number of emission lines for atomic hydrogen that you would expect to see with the naked eye if the only electronic levels involved are those shown in the figure,is

The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light. Find the maximum work function that a metal can have so that $H_{\alpha}$-light can emit photoelectrons from it. (in $\text{ eV}$)

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.

$A$ hydrogen atom transitions from an excited state to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant,the principal quantum number $n$ of the excited state is: