1. How much energy would a photon of red light have if it is produced by the transition of an electron from the n = 3 to the n = 2 energy level in the hydrogen atom?

First we must realize that the energy of any specific energy level in the hydrogen atom is related to the ground state energy, -13.60 eV, in the following way:

E_n = -13.60 eV / n²

The two specific energies that we need here are:

E₂ = -13.60 eV / 2² = -3.40 eV, and E₃ = -13.60 eV / 3² = -1.51 eV

The difference between these two energy levels will then give us the actual energy of the red-light photon that is emitted when an electron drops from the n = 3 level to the n = 2 level.

E_photon = E_f - E_i = E₂ - E₃

where E₂ is the final energy level and E₃ is the initial energy level. Thus,

E_photon = -3.40 eV - (-1.51 eV) = -1.89 eV

Notice that when we subtract a smaller negative initial energy from a larger final negative energy, the double negative in the subtracted term makes it positive, and we are still left with a negative energy for the photon. This means that the photon is given off, or emitted, when the electron jumps down from a higher energy level to a lower one.

In atomic physics, energy is often expressed in electron volts. One eV is equal to 1.6 x 10^-19 joules, so 1.89 eV will be equal to 3.03 x 10^-19 joules. This is only a tiny amount of energy, but the idea is still very important in explaining the emission spectrum of excited chemical elements.

If we look at Figure 9.13 in the textbook, we will find that this transition is shown as the fifth descending arrow from the left, and it is easy to see that this energy is exactly the difference between two of the energy levels in the hydrogen atom. To be specific, this color of photon is emitted when an electron drops from the n = 3 level to the n = 2 level. This is the red line in the emission spectrum series discovered by Balmer. Let's look at this transition in another way.


2. How much energy is required to raise an electron in the hydrogen atom from the first excited state to the second excited state?

The first excited energy state of the hydrogen atom is characterized by the principal quantum number n = 2, whereas the second excited state has n = 3. Remember that n = 1 for the ground state of hydrogen and both of these levels are above that in energy. The energy needed to cause this transition is again simply the difference between the energy of each of these excited states.

E_photon = E_f - E_i = E₃ - E₂ = -13.60 eV / 3² - (-13.60 eV / 2²)

E_photon = -1.51 eV -(-3.40 eV)

E_photon = 1.89 eV

where E₃ is now the final energy level and E₂ in the initial one.

Notice that we are now subtracting a larger negative energy from a smaller negative energy, and the double negative term makes the final energy difference positive. This happens when the photon is absorbed, which causes the electron to be kicked into a higher energy level. The energy difference is, in fact, the same magnitude as the one we calculated in problem 1, but in this case the sign is positive, indicating that energy is absorbed instead of emitted as it was in the first example.


3. Determine how tightly an electron in the ground state is bound to the proton in a hydrogen atom.

The binding energy of the electron to the nucleus can be found for the hydrogen atom in much the same way that we calculated the energy difference in problem 2. In this case the final potential energy of the electron is zero because it is now totally free and no longer bound to the nucleus.

Binding Energy = E_f - E_i = 0 - (-13.60 eV / n²)

Because the electron is starting in the ground state, n = 1,

Binding Energy = 0 -(-13.60 eV / 1²)

Binding Energy = 13.60 eV

Thus the binding energy for an electron in the ground state of the hydrogen atom has the same magnitude as the ground state energy of the electron, but it has a positive value. When an electron is removed from the atom, the process is called ionization, and the ionization energy is less for electrons in higher energy levels.

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