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The Photoelectric Effect

Light collides with a metal surface, releasing electrons.

Distributed through Wikimedia Commons

A basic set-up of the photoelectric effect.

Andrew Zimmerman Jones

The photoelectric effect posed a significant challenge to the study of optics in the latter portion of the 1800s. It challenged the classical wave theory of light, which was the prevailing theory of the time. It was the solution to this physics dilemma that catapulted Einstein into prominence in the physics community, ultimately earning him the 1921 Nobel Prize.

What is the Photoelectric Effect?

Though originally observed in 1839, the photoelectric effect was documented by Heinrich Hertz in 1887 in a paper to the Annalen der Physik. It was originally called the Hertz effect, in fact, though this name fell out of use.

When a light source (or, more generally, electromagnetic radiation) is incident upon a metallic surface, the surface can emit electrons. Electrons emitted in this fashion are called photoelectrons (although they are still just electrons). This is depicted in the image to the right.

Setting Up the Photoelectric Effect

To observe the photoelectric effect, you create a vacuum chamber with the photoconductive metal at one end and a collector at the other. When a light shines on the metal, the electrons are released and move through the vacuum toward the collector. This creates a current in the wires connecting the two ends, which can be measured with an ammeter. (A basic example of the experiment can be seen by clicking on the image to the right, and then advancing to the second image available.)

By administering a negative voltage potential (the black box in the picture) to the collector, it takes more energy for the electrons to complete the journey and initiate the current. The point at which no electrons make it to the collector is called the stopping potential Vs, and can be used to determine the maximum kinetic energy Kmax of the electrons (which have electronic charge e) by using the following equation:

Kmax = eVs
It is significant to note that not all of the electrons will have this energy, but will be emitted with a range of energies based upon the properties of the metal being used. The above equation allows us to calculate the maximum kinetic energy or, in other words, the energy of the particles knocked free of the metal surface with the greatest speed, which will be the trait that is most useful in the rest of this analysis.

The Classical Wave Explanation

In classical wave theory, the energy of electromagnetic radiation is carried within the wave itself. As the electromagnetic wave (of intensity I) collides with the surface, the electron absorbs the energy from the wave until it exceeds the binding energy, releasing the electron from the metal. The minimum energy needed to remove the electron is the work function phi of the material. (Phi is in the range of a few electron-volts for most common photoelectric materials.)

Three main predictions come from this classical explanation:

1. The intensity of the radiation should have a proportional relationship with the resulting maximum kinetic energy.
2. The photoelectric effect should occur for any light, regardless of frequency or wavelength.
3. There should be a delay on the order of seconds between the radiation’s contact with the metal and the initial release of photoelectrons.

The Experimental Result

By 1902, the properties of the photoelectric effect were well documented. Experiment showed that:
1. The intensity of the light source had no effect on the maximum kinetic energy of the photoelectrons.
2. Below a certain frequency, the photoelectric effect does not occur at all.
3. There is no significant delay (less than 10-9 s) between the light source activation and the emission of the first photoelectrons.
As you can tell, these three results are the exact opposite of the wave theory predictions. Not only that, but they are all three completely counter-intuitive. Why would low-frequency light not trigger the photoelectric effect, since it still carries energy? How do the photoelectrons release so quickly? And, perhaps most curiously, why does adding more intensity not result in more energetic electron releases? Why does the wave theory fail so utterly in this case, when it works so well in so many other situation
Photoelectric Effect - the solution

Andrew Zimmerman Jones