A hydrogen atom initially in the n = 6 state undergoes two...

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

For the initial state $n = 6$, the energy is:
$E_6 = -\dfrac{13.6}{6^2} = -0.378 \text{ eV}$

During the first transition, a photon of energy $1.13 \text{ eV}$ is emitted. The energy of the intermediate state $n'$ is:
$E_{n'} = E_6 - 1.13 = -0.378 - 1.13 = -1.508 \text{ eV}$

Comparing this with the standard energy levels of hydrogen, we have $E_3 = -\dfrac{13.6}{3^2} = -1.51 \text{ eV}$. Thus, the intermediate state is $n' = 3$.

The second transition occurs from $n' = 3$ to the ground state $n = 1$. The energy of the ground state is $E_1 = -13.6 \text{ eV}$.

The energy of the photon emitted in the second transition is:
$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \approx 12.1 \text{ eV}$