If the first line in the Lyman series has wavelength $\lambda$,then the first line in the Balmer series has the wavelength

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:

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 $\text{ eV}$)

The shortest wavelength of the Lyman series of a hydrogen atom is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is equal to:

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

The first member of the Paschen series in the hydrogen spectrum has a wavelength of $18,800 \,\mathring{A}$. The short wavelength limit of the Paschen series is ...... $\mathring{A}$.