To determine the atomic number (Z) of a hydrogen-like ion, given that the energy difference between its second excitation state and the ground state is 108.8 eV, we use the following formula for the energy difference between two energy levels in a hydrogen-like atom:

$ \Delta E = 13.6 \, \text{eV} \times Z^2 \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right] $

Here, for the second excitation, $ n_2 = 3 $ (since excitation means moving to the second higher energy level beyond the ground state, which means $ n_1 = 1 $ for the ground state to $ n_2 = 3 $). Plug these into the energy difference formula:

$ \Delta E = 13.6 \, Z^2 \left[\frac{1}{1^2} - \frac{1}{3^2}\right] $

Simplifying the term in brackets:

$ \Delta E = 13.6 \, Z^2 \left[1 - \frac{1}{9}\right] = 13.6 \, Z^2 \times \frac{8}{9} $

Given that $\Delta E = 108.8 \, \text{eV}$, set the equations equal:

$ 13.6 \, \frac{8Z^2}{9} = 108.8 $

Solving for $ Z $, first simplify:

$ 13.6 \times \frac{8}{9} \times Z^2 = 108.8 $

$ \frac{108.8 \times 9}{13.6 \times 8} = Z^2 $

$ Z^2 = 9 $

Thus, the atomic number $ Z $ is:

$ Z = 3 $

The atomic number of the ion is 3.