The Photoelectric Effect

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Introduction

Previously, we have discussed light as a wave, with properties such as wavelength and frequency. Here, we will think of light as individual particles, known as photons. The truth is, light has the unique property of acting as either a wave or a particle! When we analyze the interactions between light and matter, it is convenient to think of light as a particle.

"The photoelectric effect" is the name for the process whereby light can induce a charge on a metal plate. It was first observed in 1887 by Heinrich Hertz when light was beginning to be thought of as particles. Einstein's explanation of this effect, written in 1905, earned him the Nobel Prize in 1921. It occurs when a photon collides with an electron, disappearing and imparting its energy into the electron. If the photon had enough energy, then the electron is expelled from the atom. Einstein's motivation for writing this paper was to explain how a weak beam of violet light would cause this effect while a very bright source of red light would not. This will be explained later in this page.

The Work Function

Electrons in the metal plate are held to the atoms in the plate with a certain energy. For the purposes of describing the photoelectric effect, this energy is called the "work function." It is defined more rigorously as the energy that would be needed to push the electrons out an infinite distance from the atomic nucleus. We can calculate this fairly easily for simple atoms by utilizing the equation for electric potential energy, V = 1/(4 \pi \epsilon_0)*(q_1)(q_2)/r For anything beyond a hydrogen atom, the work function becomes too complicated to calculate, so we look it up. The energy depends on the charge of the nucleus, the charge of the electron, and the distance between the nucleus and the electron.

Quanta of Energy

The particles of light in modern theories about electricity and magnetism are called "photons." This theory, quantum electrodynamics, tells us that the energy of each of these photons is E = h \NU, where h is a fundamental constant with a value of apprx. 1.05457148*10-34 m2kg/s, and where \NU, "nu," is the frequency of the light.

Because the energies involved here are many orders of magnitude smaller than the energies we usually deal with, we do not use Joules. Instead, we define a new unit, called an electron-volt (eV). One eV is equal to 1.602*10-19 Joules. This number is not arbitrary; it is the difference in energy of an electron in two places whose potentials differ by one volt. The formula for the energy of a photon, E = h \NU, gives a result in Joules, so for many applications it must be converted into eV.

Calculating Frequencies for Incident Light

Light Energy

The frequency of light is inversely proportional to the wavelength of the light. Specifically, \lambda = c/\nu , where the constant c is the speed of light, lambda is wavelength, and nu is frequency. Therefore, going back to our previous equations, a photon's energy is also inversely proportional to its wavelength. Red light has the longest wavelength (and therefore lowest energy) of all visible light, and blue light has the shortest wavelength (highest energy).

high-energy photon-electron interaction

In the image above, a blue photon (wavelength = 475 nm) hits a metal plate and knocks off an electron. Blue light has more energy than most visible light; the spectrum of visible light ranges from 450(violet)-700(red)nm.

medium-energy photon-electron interaction

In this image, a green photon (wavelength = 510 nm) hits the same metal plate. Since green light has less energy (per photon) than blue light, the electron is knocked off with less energy (thus less velocity).

low-energy photon-electron interaction

In this image, a red photon (wavelength = 650 nm) hits the metal plate. Red light has the highest wavelengt (lowest frequency), thus the least energy of all visible light. In this case, the photon's energy was less than the "work function." Because of this, the electron absorbed the photon without gaining enough energy to escape the metal plate.

Interestingly, and contrary to conventional thinking, the energy of individual photons does not depend on the intensity of the light. The intensity simply measures how many photons are moving, while frequency relates to the energy of the individual photons. An increase in intensity does, though, increase the number of electrons knocked off. There is a one-to-one interaction; each photon can only hit one electron, no matter how much energy it has. Because intensity measures the number of photons, higher intensity does mean higher net energy of the light, but it does not mean higher energy per photon.

Sample Light Energy Problem

Find a)the wavelength and b)the energy-per-photon (in eV) of light with a frequency of 5.26*1014 Hz. (Hint: the speed of light is approximately 300000000 m/s!)

Answer:

For part (a), we know that \lambda = c/\nu , and the problem tells us that the frequency (\nu) is 5.26*1014. Thus the wavelength is 3*108/5.26*1014 = 5.7*10-7 = 570*10-9 = 570 nm. The color of 570nm light is yellow.
For part (b), we know that E = h \NU, where h is about 6.62*10-34. The frequency (5.26*1014) is given to us in the problem. Thus, E= (6.62*10-34)(5.26*1014) = 3.48*10-19 BUT WAIT! We aren't done because the answer here is in Joules. We must convert this to electron volts (eV) by multiplying by 6.24*1018, to get 2.17 eV.

Electron Energy

A photon is like a tiny blob of pure energy. In the photoelectric effect, an electron is hit by a wandering blob of energy and is so excited that it breaks its bond with the atom to which it is held. If we want to find the energy with which it leaves, we must know two things: the energy it is given by the photon, and the energy that it takes to break the bond with its atom. From conservation of energy, we know that for any interaction, energy in = energy out. In this case, energy in takes the form of the incident photon's energy (which we know how to calculate from above). The energy out is split into two bits; part of it goes into breaking the electron's bond with the atom, and part of it becomes kinetic energy of the electron. From this information, we can form an equality as follows: photon energy = work function + electon kinetic energy. We can rephrase this to read: electron kinetic energy = photon energy - work function. (At this point, it is obvious that the photon must have more energy than the work fuction in order to move the electron: you cannot have negative kinetic energy, as it would imply imaginary velocity).

We can now calculate the energy of ejected electrons, if we know a material's work function (which is possible to calculate, but which we will always look up). Rephrasing the above logic mathematically, we can say that KE_e = h \nu - w, where KEe is the energy of the electron and w is the work function fo the material.

Going back to Image 1 above, we will find the energy of the ejected electron, assuming the metal plate is made of Cesium, which has a work function of 2.1eV. The incident photon is blue, with a wavelength of 475nm. This means that its frequency is 3*108/475*10-9 = 6.31*1014 Hz. Its energy is that times planck's constant, (6.31*1014)x(6.62*10-34) = 4.18*10-19J. Expressed in eV, it is 2.61eV. The energy of the photon (2.61 eV) is greater than the work function of the material, 2.1eV, so we know that electrons will be knocked off. According to conservation of energy, KE_e = h \nu - w. The kinetic energy of the ejected electron is the difference between the energy of the photon and the work function of the material, which is .51 eV.

Additional Information

  • For a list of wavelengths that correspond to certain colors, visit this site.
  • If you want to calculate the velocity of an ejected electron, use the formula KE_e = 1/2 m_e v^2.
  • For a good explanation on how to calculate the work function, click here .

Practice Problem

Monochromatic light of wavelength 400 nm strikes a plate of Cesium. Cesium has a work function of 2.14 electron-volts.
a) Does the light knock off any electrons? If so, with what energy are the electrons ejected?
b) Light from a new light source of wavelength 750 nm is shone onto the plate. Does this light knock off any electrons? If so, with what energy are the electrons ejected?

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