The electron in a hydrogen atom is initially in the third excited state. When it finally moves to the ground state,the maximum number of spectral lines emitted is:

Hydrogen atoms are excited from the ground state to the principal quantum number $n = 5$. The number of spectral lines observed will be:

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of the Lyman series and the first member of the Paschen series respectively,then the value of $\lambda_{1} : \lambda_{2}$ is

If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

In a hydrogen atom spectrum,when an electron jumps from the second excited state to the first excited state,the wavelength of radiation emitted is $\lambda$. If the electron jumps from the third excited state to the second orbit,the wavelength of radiation emitted will be $20 \lambda / x$. The value of $x$ 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.