Hydrogen atoms are excited from ground state to the state of principal quantum number $4$. Then,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

The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is

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

Given the value of Rydberg constant is $10^7 \, m^{-1}$,the wave number of the last line of the Balmer series in the hydrogen spectrum will be:

Missing lines in a continuous spectrum reveal