The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is expressed as $E_n = \frac{-13.6}{n^2} \, eV$. The shortest and longest wavelengths of the Lyman series will be:

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.

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

The ratio of the longest to shortest wavelengths in the Brackett series of hydrogen spectra is

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

The wavelength of radiation emitted is ${\lambda _0}$ when an electron jumps from the third to the second orbit of a hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom,the wavelength of radiation emitted will be