For a hydrogen atom, determine the ratio of the ionization energy for the n = 3 excited state to the ionization energy for the ground state.

To determine the ionization energy ratio for a hydrogen atom in the n=3 excited state to the ground state, we need to understand the energy levels of hydrogen and the concept of ionization energy.

Ionization energy is the energy required to remove an electron from an atom or ion in its ground state. In the case of a hydrogen atom, the ionization energy corresponds to the transition of the electron from the lowest energy level (n=1, ground state) to an energy level where it is completely removed from the atom.

The energy levels in the hydrogen atom are given by the formula: En = -13.6 eV/n^2, where En is the energy level, and n is the principal quantum number representing the specific energy level.

For the ground state of hydrogen (n=1), the energy is calculated as: E1 = -13.6 eV/1^2 = -13.6 eV.

For the n=3 excited state, the energy is calculated as: E3 = -13.6 eV/3^2 = -13.6 eV/9 = -1.51 eV.

Now, to find the ratio of ionization energies, we need to compare the energy required to remove an electron in the n=3 state to that in the ground state (n=1).

The ionization energy for the n=3 excited state is the difference between the energy at that state and the energy at the ground state: ΔE = E3 - E1 = -1.51 eV - (-13.6 eV) = 12.09 eV.

The ionization energy for the ground state (n=1) is simply the energy at that state: E1 = -13.6 eV.

Therefore, the ratio of the ionization energy for the n=3 excited state to the ionization energy for the ground state is given by: ΔE / E1 = 12.09 eV / -13.6 eV ≈ -0.890.

Note: Since ionization energy is a positive quantity, the negative sign in the ratio indicates that the energy is released when the electron is removed from the n=3 excited state to the ground state.