In a hydrogen atom, the energy levels of the electron are quantized, meaning they can only have certain discrete values. The energy of an electron in the ground state of a hydrogen atom is given as -13.6 eV. To determine the energy of the electron in the 2nd excited state, we need to understand the energy levels and transitions in the hydrogen atom.

1. Energy Levels in a Hydrogen Atom:
The energy levels in a hydrogen atom are represented by a series of numbers called the principal quantum numbers (n). The ground state corresponds to the lowest energy level, n=1, and the energy of the electron in this state is -13.6 eV. The energy levels increase as the principal quantum number increases.

2. Excited States and Transitions:
When an electron in a hydrogen atom absorbs energy, it can move to a higher energy level, called an excited state. The energy absorbed must be equal to the difference in energy between the initial and final states. Similarly, when an excited electron emits energy, it transitions back to a lower energy level.

3. Energy Difference between States:
The energy difference between two energy levels can be calculated using the formula:

ΔE = E_final - E_initial

where ΔE is the energy difference, E_final is the energy of the final state, and E_initial is the energy of the initial state.

4. Determining the Energy in the 2nd Excited State:
To find the energy of the electron in the 2nd excited state, we need to determine the energy difference between the ground state (n=1) and the 2nd excited state (n=3).

ΔE = E_final - E_initial
ΔE = E_3rd state - E_1st state

Since the ground state energy is -13.6 eV (E_1st state = -13.6 eV), we need to find the energy of the 3rd state (E_3rd state).

The energy of each state is given by the formula:

E = -13.6 eV/n^2

Substituting n=3 into the formula, we get:

E_3rd state = -13.6 eV/3^2 = -13.6 eV/9 = -1.51 eV

Therefore, the energy in the 2nd excited state (or the 3rd energy level) is -1.51 eV.