The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \text{ eV}$. The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. The maximum wavelength of the emitted radiation corresponds to the transition between:

In a hydrogen sample,if the atoms are excited to states with principal quantum number $n = 20$,then the number of different wavelengths which may be observed in the spectrum is:

In a hydrogen atom,a transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

The minimum wavelength of Lyman series lines is $P$. What is the maximum wavelength of these lines?

The wavelength of the second Balmer line in the hydrogen spectrum is $600 \ nm$. The wavelength for its third line in the Lyman series is:

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.