Determine the energy of a hydrogen atom in the first excited state...

The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

The potential energy is twice the total energy, so $U_n = 2E_n = -\dfrac{27.2}{n^2} \text{ eV}$.

In the ground state ($n=1$), the standard potential energy is $U_1 = -27.2 \text{ eV}$.

If the potential energy in the ground state is taken to be zero, the reference level of potential energy is shifted by $+27.2 \text{ eV}$. Since kinetic energy remains unaffected by the choice of reference, the total energy of every state also increases by $+27.2 \text{ eV}$.

The standard total energy in the first excited state ($n=2$) is:
$E_2 = -\dfrac{13.6}{2^2} = -3.4 \text{ eV}$

The new total energy in the first excited state will be:
$E'_2 = E_2 + 27.2 \text{ eV} = -3.4 \text{ eV} + 27.2 \text{ eV} = 23.8 \text{ eV}$