293. The Photoelectric EffectLearning Objectives
When light strikes materials, it can eject electrons from them. This is called the This effect has been known for more than a century and can be studied using a device such as that shown in Figure 1. This figure shows an evacuated tube with a metal plate and a collector wire that are connected by a variable voltage source, with the collector more negative than the plate. When light (or other EM radiation) strikes the plate in the evacuated tube, it may eject electrons. If the electrons have energy in electron volts (eV) greater than the potential difference between the plate and the wire in volts, some electrons will be collected on the wire. Since the electron energy in eV is $\text{qV}$, where $q$ is the electron charge and $V$ is the potential difference, the electron energy can be measured by adjusting the retarding voltage between the wire and the plate. The voltage that stops the electrons from reaching the wire equals the energy in eV. For example, if $\mathrm{\u20133.00\; V}$ barely stops the electrons, their energy is 3.00 eV. The number of electrons ejected can be determined by measuring the current between the wire and plate. The more light, the more electrons; a little circuitry allows this device to be used as a light meter. What is really important about the photoelectric effect is what Albert Einstein deduced from it. Einstein realized that there were several characteristics of the photoelectric effect that could be explained only if EM radiation is itself quantized: the apparently continuous stream of energy in an EM wave is actually composed of energy quanta called photons. In his explanation of the photoelectric effect, Einstein defined a quantized unit or quantum of EM energy, which we now call a where $E$ is the energy of a photon of frequency $f$ and $h$ is Planck’s constant. This revolutionary idea looks similar to Planck’s quantization of energy states in blackbody oscillators, but it is quite different. It is the quantization of EM radiation itself. EM waves are composed of photons and are not continuous smooth waves as described in previous chapters on optics. Their energy is absorbed and emitted in lumps, not continuously. This is exactly consistent with Planck’s quantization of energy levels in blackbody oscillators, since these oscillators increase and decrease their energy in steps of $\text{hf}$ by absorbing and emitting photons having $E=\text{hf}$. We do not observe this with our eyes, because there are so many photons in common light sources that individual photons go unnoticed. (See Figure 2.) The next section of the text (Photon Energies and the Electromagnetic Spectrum) is devoted to a discussion of photons and some of their characteristics and implications. For now, we will use the photon concept to explain the photoelectric effect, much as Einstein did. The photoelectric effect has the properties discussed below. All these properties are consistent with the idea that individual photons of EM radiation are absorbed by individual electrons in a material, with the electron gaining the photon’s energy. Some of these properties are inconsistent with the idea that EM radiation is a simple wave. For simplicity, let us consider what happens with monochromatic EM radiation in which all photons have the same energy $\text{hf}$.
Einstein’s idea that EM radiation is quantized was crucial to the beginnings of quantum mechanics. It is a far more general concept than its explanation of the photoelectric effect might imply. All EM radiation can also be modeled in the form of photons, and the characteristics of EM radiation are entirely consistent with this fact. (As we will see in the next section, many aspects of EM radiation, such as the hazards of ultraviolet (UV) radiation, can be explained only by photon properties.) More famous for modern relativity, Einstein planted an important seed for quantum mechanics in 1905, the same year he published his first paper on special relativity. His explanation of the photoelectric effect was the basis for the Nobel Prize awarded to him in 1921. Although his other contributions to theoretical physics were also noted in that award, special and general relativity were not fully recognized in spite of having been partially verified by experiment by 1921. Although heroworshipped, this great man never received Nobel recognition for his most famous work—relativity. Example 1: Calculating Photon Energy and the Photoelectric Effect: A Violet Light(a) What is the energy in joules and electron volts of a photon of 420nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420nm violet light, given that the binding energy (or work function) of electrons for calcium metal is 2.71 eV? Strategy To solve part (a), note that the energy of a photon is given by $E=\text{hf}$. For part (b), once the energy of the photon is calculated, it is a straightforward application of ${\text{KE}}_{e}=\text{hf}\text{\u2013BE}$ to find the ejected electron’s maximum kinetic energy, since BE is given. Solution for (a) Photon energy is given by $$E=\text{hf}$$Since we are given the wavelength rather than the frequency, we solve the familiar relationship $c=\mathrm{f\lambda}$ for the frequency, yielding $$f=\frac{c}{\lambda}.$$Combining these two equations gives the useful relationship $$E=\frac{\text{hc}}{\lambda}.$$Now substituting known values yields $$E=\frac{\left(6\text{.}\text{63}\times {\text{10}}^{\text{\u201334}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}\right)\left(3.00\times {\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)}{\text{420}\times {\text{10}}^{\text{\u20139}}\phantom{\rule{0.25em}{0ex}}\text{m}}=\text{4.74}\times {\text{10}}^{\text{\u201319}}\phantom{\rule{0.25em}{0ex}}\text{J}.$$Converting to eV, the energy of the photon is $$E=\left(4\text{.}\text{74}\times {\text{10}}^{\text{\u201319}}\phantom{\rule{0.25em}{0ex}}\text{J}\right)\frac{\mathrm{1\; eV}}{1.6\times {\text{10}}^{\text{\u201319}}\phantom{\rule{0.25em}{0ex}}\mathrm{J}}=\text{2.96 eV}.$$Solution for (b) Finding the kinetic energy of the ejected electron is now a simple application of the equation ${\text{KE}}_{e}=\text{hf}\text{\u2013BE}$. Substituting the photon energy and binding energy yields $${\text{KE}}_{e}=\text{hf}\text{\u2013 BE}=\text{2}\text{.}\text{96 eV \u2013 2}\text{.}\text{71 eV}=\text{0}\text{.}\text{246 eV}.$$Discussion The energy of this 420nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the UV photon in this example could have biological effects. The ejected electron (called a photoelectron) has a rather low energy, and it would not travel far, except in a vacuum. The electron would be stopped by a retarding potential of but 0.26 eV. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420nm photons with their 2.96eV energy are not much above the frequency threshold. You can show for yourself that the threshold wavelength is 459 nm (blue light). This means that if calcium metal is used in a light meter, the meter will be insensitive to wavelengths longer than those of blue light. Such a light meter would be completely insensitive to red light, for example. PhET Explorations: Photoelectric EffectSee how light knocks electrons off a metal target, and recreate the experiment that spawned the field of quantum mechanics. Section Summary
Conceptual QuestionsExercise 1Is visible light the only type of EM radiation that can cause the photoelectric effect? Exercise 2Which aspects of the photoelectric effect cannot be explained without photons? Which can be explained without photons? Are the latter inconsistent with the existence of photons? Exercise 3Is the photoelectric effect a direct consequence of the wave character of EM radiation or of the particle character of EM radiation? Explain briefly. Exercise 4Insulators (nonmetals) have a higher BE than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors. Exercise 5If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the object as compared with heating it. Problems & ExercisesExercise 1What is the longestwavelength EM radiation that can eject a photoelectron from silver, given that the binding energy is 4.73 eV? Is this in the visible range? Show/Hide Solution Solution263 nm Exercise 2Find the longestwavelength photon that can eject an electron from potassium, given that the binding energy is 2.24 eV. Is this visible EM radiation? Exercise 3What is the binding energy in eV of electrons in magnesium, if the longestwavelength photon that can eject electrons is 337 nm? Show/Hide Solution Solution3.69 eV Exercise 4Calculate the binding energy in eV of electrons in aluminum, if the longestwavelength photon that can eject them is 304 nm. Exercise 5What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450nm EM radiation, given that the binding energy is 2.28 eV? Show/Hide Solution Solution0.483 eV Exercise 6UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kinetic energy of the ejected photoelectrons? Exercise 7Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the binding energy of electrons to sodium metal? Show/Hide Solution Solution2.25 eV Exercise 8UV radiation having a 300nm wavelength falls on uranium metal, ejecting 0.500eV electrons. What is the binding energy of electrons to uranium metal? Exercise 9What is the wavelength of EM radiation that ejects 2.00eV electrons from calcium metal, given that the binding energy is 2.71 eV? What type of EM radiation is this? Show/Hide Solution Solution(a) 264 nm (b) Ultraviolet Exercise 10Find the wavelength of photons that eject 0.100eV electrons from potassium, given that the binding energy is 2.24 eV. Are these photons visible? Exercise 11What is the maximum velocity of electrons ejected from a material by 80nm photons, if they are bound to the material by 4.73 eV? Show/Hide Solution Solution$1.95\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ Exercise 12Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device? Exercise 13A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.71 eV? Show/Hide Solution Solution(a) $4.02\times {\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\text{/s}$ (b) 0.256 mW Exercise 14(a) Calculate the number of photoelectrons per second ejected from a 1.00mm ^{2} area of sodium metal by 500nm EM radiation having an intensity of $1\text{.}{\text{30 kW/m}}^{2}$ (the intensity of sunlight above the Earth’s atmosphere). (b) Given that the binding energy is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons. Where does the other power go? How can it be recovered? Exercise 15Unreasonable Results Red light having a wavelength of 700 nm is projected onto magnesium metal to which electrons are bound by 3.68 eV. (a) Use ${\text{KE}}_{e}=\text{hf}\u2013\text{BE}$ to calculate the kinetic energy of the ejected electrons. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? Show/Hide Solution Solution(a) $\mathrm{\u20131.90\; eV}$ (b) Negative kinetic energy (c) That the electrons would be knocked free. Exercise 16Unreasonable Results (a) What is the binding energy of electrons to a material from which 4.00eV electrons are ejected by 400nm EM radiation? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
