The energy required to ionize an unknown hydrogen-like species from...

The ionization energy of a hydrogen-like species from a state with principal quantum number $n$ is given by $E = 13.6 \dfrac{Z^2}{n^2} \mathrm{eV}$, where $Z$ is the atomic number.

The first excited state corresponds to $n = 2$.

Given that the ionization energy from this state is $13.6 \mathrm{eV}$, we can write:
$13.6 = 13.6 \times \dfrac{Z^2}{2^2}$

$\Rightarrow \dfrac{Z^2}{4} = 1$

$\Rightarrow Z^2 = 4$

$\Rightarrow Z = 2$

The species with atomic number $Z = 2$ is the helium ion ($\mathrm{He}^{+}$).

Answer: $\mathrm{He}^{+}$ ion