In the Balmer series,the wavelength of the first line is $\lambda_1$ and in the Brackett series,the wavelength of the first line is $\lambda_2$. Then,the ratio $\frac{\lambda_1}{\lambda_2}$ is:

The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^7 \ m^{-1}$]. (in $Å$)

$A$ hydrogen atom is excited from the ground state to the energy level $n = 3$. According to Bohr's model,the number of spectral lines emitted is:

In the spectrum of a hydrogen atom,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.