The emission spectrum of a gaseous atom is line-based and not continuous. Why?

What is the frequency of the radiation emitted when an electron in a hydrogen atom undergoes a transition from $n = 3$ to the ground state?

Assertion : Angular momentum of an electron in any orbit is given by angular momentum $= \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Reason : The principal quantum number,$n$,can have any integral value.

The hydrogen spectrum consists of several spectral lines in the Lyman series ($L_1, L_2, L_3 \ldots$; $L_1$ has the lowest energy among the Lyman series). Similarly, it consists of several spectral lines in the Balmer series ($B_1, B_2, B_3 \ldots$; $B_1$ has the lowest energy among the Balmer lines). The energy of $L_1$ is $x$ times the energy of $B_1$. The value of $x$ is . . . . . . $\times 10^{-1}$ (Nearest integer).

$A$ certain transition in the hydrogen spectrum from an excited state in one or more steps gives rise to a total of $10$ lines. The number of lines lying in the visible spectrum is:

If the shortest wavelength of the Lyman series of the $H$ atom is $x$,then calculate the total energy of the $H$ atom in the first excited state.