A beam of electrons, each with a kinetic energy of 12.9 eV, passes through hydrogen gas whose atoms…

Step 1. The levels. The energy to reach $n = 3$ is $13.6\left(1 - \dfrac{1}{9}\right) = 12.09\ \text{eV}$. To reach $n = 4$ it is $13.6\left(1 - \dfrac{1}{16}\right) = 12.75\ \text{eV}$. To reach $n = 5$ it is $13.06\ \text{eV}$, more than the $12.9\ \text{eV}$ available. The highest level is $n = 4$.

Step 2. The lines. From $n = 4$ the atoms can come down in $\dfrac{4 \times 3}{2} = 6$ ways: 4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1 and 2 to 1.

Step 3. The longest wavelength. It belongs to the smallest gap, $4 \to 3$: $E = 13.6\left(\dfrac{1}{9} - \dfrac{1}{16}\right) = 0.661\ \text{eV}$. So $\lambda = \dfrac{1240\ \text{eV nm}}{0.661\ \text{eV}} = 1875\ \text{nm}$, in the infrared.

Step 4. Visible lines. Only jumps that end on $n = 2$ are visible. Here they are $4 \to 2$ at $486\ \text{nm}$ and $3 \to 2$ at $656\ \text{nm}$. So there are 2.