B

Step 1: Determine the energy levels of the hydrogen atom. The energy levels are given by the formula:
$$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $$
where $n$ is the principal quantum number.
Step 2: To find the possible energy states that can be reached by exciting the hydrogen atom with 12.1 eV, calculate the maximum principal quantum number $n$ that can be reached:
1. Initial energy in ground state ($n=1$):
$$ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} $$
2. Final energy after absorption of 12.1 eV:
$$ E_{final} = -13.6 + 12.1 = -1.5 \, \text{eV} $$
3. Setting $E_n = -1.5 \, \text{eV}$, we find the principal quantum number:
$$ -\frac{13.6 \, \text{eV}}{n^2} = -1.5 \implies n^2 = \frac{13.6}{1.5} \approx 9.067 $$
so $n \approx 3$.
Step 3: Possible energy levels after excitation are $n=1$, $n=2$, and $n=3$.
Step 4: Calculate the number of spectral lines emitted: The number of lines emitted when an electron transitions between energy levels is given by the formula:
$$ N = \frac{n(n-1)}{2} $$
Applying for $n=3$:
$$ N = \frac{3(3-1)}{2} = \frac{3 \cdot 2}{2} = 3 $$
Therefore, the number of spectral lines emitted by hydrogen atoms will be 3.
Final Answer: Based on the options available, select option B.