When the electron in a hydrogen atom jumps from the fourth Bohr orbit to the second Bohr orbit,one gets the:

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

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 $a = ........$

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:

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.
Reason : $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is