In Figure 25.1, we’ve drawn a few of the energy levels of a hypothetical atom. Let’s start by looking at E1. This is the lowest energy level, and it is called the ground state energy of the atom. When an electron is sitting as close to the nucleus as possible, and when it’s completely unexcited, it will have the energy of E1, -10 eV.

To increase the energy of the electron—to move the electron into a higher energy level—energy must somehow be transferred to the electron. This energy transfer is done by a photon. To jump up in energy, an electron absorbs a photon, and to drop down in energy, an electron emits a photon.

The energy diagram in Figure 25.1 tells us that to get to E2 from the ground state, an electron must absorb a photon carrying 3 eV—and only 3 eV!—of energy. If a photon with an energy of 2.9 eV comes along and knocks into the electron, nothing will happen. Similarly, if a photon with an energy of 3.1 eV comes along, nothing will happen. It’s all or nothing; either the electron gets just the right amount of energy to go from one energy level to the next, or it doesn’t.

How does the electron get from the ground state to E3? There are two ways. The electron could first absorb a photon with an energy of 3 eV, taking it from E1 to E2, and then it could absorb another photon with an energy of 1 eV, taking it from E2 to E3. Or, the electron could start in E1 and simply absorb a photon with an energy of 4 eV.

We’ve been talking about photons having certain energies, but we haven’t yet told you how to figure out the energy of a photon. Here’s the formula:

E = hf

This formula tells us that the energy of a photon is equal to Planck’s constant, h, which is 6.63 × 10-34 J·s (this value is given to you on the constants sheet), multiplied by the frequency of the photon. You should remember from Chapter 23 that the frequency of a wave is related to the wavelength by the formula

v = λf

For light, the velocity is c, or 3 . 108 m/s, so we can instead write

c = λf

This means that we can rewrite the equation for the energy of a photon to read

E = hc/λ

These formulas tell us that a photon with a high frequency, and therefore with a small wavelength, is higher in energy than a photon with a low frequency and long wavelength. So gamma rays, for example, are a lot higher energy than radio waves because gamma rays have a higher frequency.

This is a simple plug-and-chug problem. Figure 25.1 gives us the energy levels, and we just need to use our formula to find the wavelength of the absorbed photon. But what value do we use for the energy? -10 eV? -7 eV? Wrong… the value we use is the energy of the jump, not the energy of one of the states. To get from E1 to E2, an electron must absorb 3 eV of energy, so that’s the value we use in the formula:

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. This value is on the constants sheet, and it is very important to know! 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 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 energy^{2}—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.

1eV(1.6 x 10_{}^{-19}J/eV) = ½ mv^{2}

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**.

**Photoelectric Effect:** Energy in the form of light can cause an atom to eject one of its electrons, but only if the frequency of the light is above a certain value.

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.

A metal surface has a work has a work function of 10eV. What is the cutoff frequency for this metal?

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

λ = 1240eV.nm/10eV

= 124nm

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.

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