In a hydrogen like ion, the energy difference between the 2nd excitation energy state and ground state is 108.8 eV. The atomic number of the ion is

For a hydrogen-like ion, the energy of the electron in the $$n^{\text{th}}$$ Bohr orbit is

$$E_n = -\,\frac{13.6\,Z^{2}}{n^{2}}\ \text{eV} \qquad -(1)$$

where $$Z$$ is the atomic number of the ion.

The ground state corresponds to $$n = 1$$.

“2nd excitation energy state’’ means the electron is in the second excited level, i.e. $$n = 3$$ (first excited is $$n = 2$$, second excited is $$n = 3$$).

Using (1):

Energy of ground state ($$n = 1$$): $$E_1 = -\,13.6\,Z^{2}\ \text{eV} \qquad -(2)$$

Energy of 2nd excited state ($$n = 3$$): $$E_3 = -\,\frac{13.6\,Z^{2}}{9}\ \text{eV} \qquad -(3)$$

The energy difference between these two states is

$$\Delta E = E_3 - E_1$$

Substituting (2) and (3):

$$\Delta E = -\,\frac{13.6\,Z^{2}}{9} - \bigl(-\,13.6\,Z^{2}\bigr)$$

$$\Delta E = 13.6\,Z^{2}\left(1 - \frac{1}{9}\right)$$

$$\Delta E = 13.6\,Z^{2}\left(\frac{8}{9}\right)$$

$$\Delta E = \frac{108.8\,Z^{2}}{9}\ \text{eV} \qquad -(4)$$

According to the question, this difference equals $$108.8\ \text{eV}$$, so from (4)

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

$$\Rightarrow \frac{Z^{2}}{9} = 1$$

$$\Rightarrow Z^{2} = 9$$

$$\Rightarrow Z = 3$$

Thus the atomic number of the ion is $$Z = 3$$, which corresponds to Option D.