Line spectra are due to

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:

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

The ratio of the longest to the shortest wavelength emitted in the Paschen series of a hydrogen atom is:

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

$A$ hydrogen atom is excited from the ground state to another state with a principal quantum number equal to $4$. The number of spectral lines in the emission spectra will be: