The ground state energy of a hydrogen atom is $$-13.6$$ eV. The energy of second excited state of He$$^+$$ ion in eV is:

We are dealing with hydrogen-like (single-electron) species. For any such species, the energy of an electron in the $$n^{\text{th}}$$ stationary state is given by the well-known expression

$$E_n = -13.6\;{\rm eV}\;\frac{Z^2}{n^2},$$

where $$Z$$ is the atomic number of the nucleus and $$n$$ is the principal quantum number. The factor $$-13.6\;{\rm eV}$$ is the ground-state energy of the hydrogen atom ($$Z = 1,\; n = 1$$) that has been provided in the question.

For the helium ion He$$^+$$ we have only one electron but the nuclear charge is double that of hydrogen, so

$$Z = 2.$$

Next we interpret the phrase “second excited state.” The counting of states goes as follows:

• Ground state → $$n = 1$$
• First excited state → $$n = 2$$
• Second excited state → $$n = 3$$

Hence we must use $$n = 3$$ for our calculation.

Substituting $$Z = 2$$ and $$n = 3$$ into the energy formula gives

$$E_3 = -13.6\;{\rm eV}\;\frac{(2)^2}{(3)^2}.$$

We now carry out the algebra step by step:

$$E_3 = -13.6\;{\rm eV}\;\frac{4}{9}.$$

Multiplying numerator and denominator we get

$$E_3 = -13.6 \times \frac{4}{9}\;{\rm eV}.$$

First compute the product in the numerator:

$$13.6 \times 4 = 54.4.$$

Therefore,

$$E_3 = -\frac{54.4}{9}\;{\rm eV}.$$

Finally, dividing $$54.4$$ by $$9$$ yields

$$E_3 = -6.044\ldots\;{\rm eV} \approx -6.04\;{\rm eV}.$$

Looking at the given options, this value matches Option B.

Hence, the correct answer is Option B.