Find the maximum wavelength of the Brackett series for a hydrogen atom in $\mathring A$.

Match List $I$ with List $II$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

Choose the correct answer from the options given below:

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first line of the Lyman and Paschen series respectively,then $\lambda_2 : \lambda_1$ is

The difference between the frequencies of the second and first Paschen lines of the hydrogen atom is (where $R$ is the Rydberg constant and $c$ is the speed of light in vacuum).

How many spectral lines are obtained when an electron in the ground state of hydrogen is excited to the principal quantum number $n = 3$?

Every series of the hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to $18752 \mathring{A}$ is:
(Rydberg constant $R = 1.097 \times 10^7 \ m^{-1}$)