1. Find which excited states of hydrogen can be reached

For hydrogen, the energy of the $n$th level is $E_n=-\frac{13.6}{n^2}\ \text{eV}$

If the atom starts in the ground state $(n=1)$, then the excitation energy to level $n$ is $\Delta E = E_n-E_1 = 13.6\left(1-\frac{1}{n^2}\right) \text{ eV}$

The incident electrons have energy $12.5\ \text{eV}$. So only those excitations are possible for which $13.6\left(1-\frac{1}{n^2}\right) \le 12.5$

Now check successive levels:

• For $n=2$: $\Delta E = 13.6\left(1-\frac14\right)=13.6\cdot \frac34=10.2\ \text{eV}$ Possible.

• For $n=3$: $\Delta E = 13.6\left(1-\frac19\right)=13.6\cdot \frac89\approx 12.09\ \text{eV}$ Possible.

• For $n=4$: $\Delta E = 13.6\left(1-\frac1{16}\right)=13.6\cdot \frac{15}{16}=12.75\ \text{eV}$ Not possible, since $12.75>12.5$.

So the highest excited state accessible is $n=3$.

2. List all possible emitted spectral lines

Once atoms are excited up to $n=3$, de-excitation can produce all possible downward transitions among levels $3,2,1$:

• $3 \to 2$
• $3 \to 1$
• $2 \to 1$

Thus total number of distinct spectral lines is $\frac{n(n-1)}{2}=\frac{3\cdot 2}{2}=3$

3. Match with options

The number of spectral lines emitted is $\boxed{3}$

So the correct option is C.