If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li$$^{++}$$ is:

The binding energy of an electron in a hydrogen atom is given as 13.6 eV, which corresponds to the energy required to remove the electron from the ground state (n=1) of hydrogen. For hydrogen-like atoms, the energy of an electron in a particular state is given by the formula:

where $$Z$$ is the atomic number and $$n$$ is the principal quantum number.

The question asks for the energy required to remove the electron from the first excited state of Li⁺⁺. Lithium has an atomic number $$Z = 3$$, so Li⁺⁺ is a hydrogen-like ion with one electron. The first excited state corresponds to $$n = 2$$ because the ground state is $$n = 1$$, and the next state is $$n = 2$$.

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

First, compute $$3^2 = 9$$ and $$2^2 = 4$$:

Now, multiply 13.6 by 9:

So the expression becomes:

Divide 122.4 by 4:

Therefore:

The energy required to remove the electron (binding energy) is the magnitude of this energy, since it represents the energy needed to take the electron from its bound state to infinity (where energy is zero). Thus, the binding energy is:

Comparing with the options:

A. 13.6 eV

B. 3.4 eV

C. 122.4 eV

D. 30.6 eV

Hence, the correct answer is Option D.