If a hydrogen atom is excited from the ground state to another state with a principal quantum number $n = 4$,the number of spectral lines in the emission spectrum will be:

Which line has the maximum wavelength and frequency in each series of the hydrogen spectrum?

Ratio of the wavelengths of the first line of the Lyman series and the first line of the Balmer series is

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively:

Whenever a hydrogen atom emits a photon in the Balmer series,

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: