The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light. Find the maximum work function that a metal can have so that $H_{\alpha}$-light can emit photoelectrons from it. (in $\text{ eV}$)

For the wavelength of visible radiation of the Hydrogen spectrum, Balmer gave an equation as $\lambda = \frac{xm^2}{m^2 - 4}$, where $m$ is an integer value. The value of $x$ in terms of Rydberg's constant $R$ is

The shortest wavelength in the Paschen series of the Hydrogen spectrum is (Rydberg constant of hydrogen $R_H = 1.097 \times 10^7 \ m^{-1}$). (in $nm$)

$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

The ratio of longest wavelengths of the spectral lines in the Lyman and Balmer series of hydrogen spectrum 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: