Determine the initial energy level of the hydrogen atom.

Hydrogen atoms are in the ground state, which corresponds to the principal quantum number $n_i = 1$.
The energy of an electron in the $n$-th orbit of a hydrogen atom is given by:

$$E_n = -13.6 \frac{1}{n^2} \text{ eV}$$

For the ground state ($n_i = 1$):

$$E_1 = -13.6 \frac{1}{1^2} = -13.6 \text{ eV}$$

The atoms are excited by monochromatic radiation with a photon energy of 12.1 eV.
The final energy $E_f$ of the electron will be the initial energy plus the absorbed photon energy:

$$E_f = E_1 + \text{Photon Energy}$$ $$E_f = -13.6 \text{ eV} + 12.1 \text{ eV}$$ $$E_f = -1.5 \text{ eV}$$

Using the energy formula for $E_f$:

$$E_f = -13.6 \frac{1}{n_f^2}$$ $$-1.5 = -13.6 \frac{1}{n_f^2}$$ $$n_f^2 = \frac{-13.6}{-1.5}$$ $$n_f^2 = \frac{13.6}{1.5}$$ $$n_f^2 \approx 9.066...$$ $$n_f = \sqrt{9.066...} \approx 3.01$$

Since the principal quantum number must be an integer, the electron is excited to the $n_f = 3$ energy level.

When an electron de-excites from an excited state $n$ to lower energy levels, the total number of possible spectral lines $N$ is given by the formula:

$$N = \frac{n(n-1)}{2}$$

In this case, the electron is excited to $n=3$.

$$N = \frac{3(3-1)}{2}$$ $$N = \frac{3 \times 2}{2}$$ $$N = \frac{6}{2}$$ $$N = 3$$

The possible transitions are $3 \to 2$, $3 \to 1$, and $2 \to 1$.

The final answer is $\text{3}$.