In the hydrogen spectrum,the shortest wavelengths of the Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Which of the following spectral series in a hydrogen atom gives a spectral line of $4860 \mathring A$?

The wavelengths corresponding to the first four spectral lines of the Lyman series of the $H$-atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}$,and $951.4 \, \mathring{A}$. Now,consider a deuterium atom instead of a hydrogen atom. Given the mass of the hydrogen atom is $1.6725 \times 10^{-27} \, kg$,the mass of the deuterium atom is $3.3374 \times 10^{-27} \, kg$,and the mass of the electron is $9.109 \times 10^{-31} \, kg$,calculate the percentage change in the wavelength of the first spectral line of the Lyman series for the deuterium atom relative to the hydrogen atom.

The wavelength of the first line of the Balmer series of a hydrogen atom is $\lambda \ \mathring{A}$. The wavelength of the same line for a doubly ionized lithium atom $(Z = 3)$ is:

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?