Energy Levels in an Atom for AP Physics B

At this point, the problem is simply a matter of converting units and using your calculator. However, solving it is a lot easier if you know that the quantity hc equals 1240 eV·nm. (Check the math yourself, just to be sure.) This value is on the constants sheet, and it is very important to know! Look at how, by knowing this value, we can solve for the photon's wavelength very quickly.

In other words, for an electron in Figure 25.1 to go from the ground state to E₂, it must absorb light with a wavelength of exactly 410 nm, which happens to be a lovely shade of violet.

If you look back at Figure 25.1, you might wonder what's going on in the gap between E₃ and E_∞. In fact, this gap is likely filled with lots more energy levels—E₄, E₅, E₆… However, as you go up in energy, the energy levels get squeezed closer and closer together (notice, for example, how the energy gap between E₁ and E₂ is greater than the gap between E₂ and E₃). However, we didn't draw all these energy levels, because our diagram would have become too crowded.

The presence of all these other energy levels, though, raises an interesting question. Clearly, an electron can keep moving from one energy level to the next if it absorbs the appropriate photons, getting closer and closer to E_∞. But can it ever get beyond E_∞? That is, if an electron in the ground state were to absorb a photon with an energy greater than 10 eV, what would happen?

The answer is that the electron would be ejected from the atom. It takes exactly 10 eV for our electron to escape the electric pull of the atom's nucleus—this minimum amount of energy needed for an electron to escape an atom is called the ionization energy²—so if the electron absorbed a photon with an energy of, say, 11 eV, then it would use 10 of those eV to escape the atom, and it would have 1 eV of leftover energy. That leftover energy takes the form of kinetic energy.

As we said above, it takes 10 eV for our electron to escape the atom, which leaves 1 eV to be converted to kinetic energy. The formula for kinetic energy requires that we use values in standard units, so we need to convert the energy to joules.

If we plug in the mass of an electron for m, we find that v = 5.9 × 10⁵ m/s. Is that a reasonable answer? It's certainly quite fast, but electrons generally travel very quickly. And it's several orders of magnitude slower than the speed of light, which is the fastest anything can travel. So our answer seems to make sense.

The observation that electrons, when given enough energy, can be ejected from atoms was the basis of one of the most important discoveries of twentieth century physics: the photoelectric effect.

This discovery was surprising, because physicists in the early twentieth century thought that the brightness of light, and not its frequency, was related to the light's energy. But no matter how bright the light was that they used in experiments, they couldn't make atoms eject their electrons unless the light was above a certain frequency. This frequency became known, for obvious reasons, as the cutoff frequency. The cutoff frequency is different for very type of atom.

Remembering that hc equals 1240 eV·nm, we can easily find the wavelength of a photon with an energy of 10 eV:

Using the equation c = λf, we find that f = 2.42 × 10¹⁵ Hz. So any photon with a frequency equal to or greater than 2.42 × 10¹⁵ Hz carries enough energy to eject a photon from the metal surface.

Practice problems for these concepts can be found at:

Atomic and Nuclear Physics Practice Problems for AP Physics B