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:

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:

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

In a hydrogen spectrum,let $\lambda$ be the wavelength of the first transition line of the Lyman series. The wavelength difference between the $3^{\text{rd}}$ transition line of the Paschen series and the $2^{\text{nd}}$ transition line of the Balmer series is $a\lambda$,where $a = ........$

If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

In the hydrogen atom spectrum, ($R$ is Rydberg's constant):
$A$. The maximum wavelength of the radiation of the Lyman series is $\frac{4}{3R}$.
$B$. The Balmer series lies in the visible region of the spectrum.
$C$. The minimum wavelength of the radiation of the Paschen series is $\frac{9}{R}$.
$D$. The minimum wavelength of the Lyman series is $\frac{5}{4R}$.
Choose the correct answer from the options given below: