# 1.3: Photoelectric Effect Explained with Quantum Hypothesis

Overview

Although Hertz discover the photoelectron in 1887, it was not until 1905 that a theory was proposed that explained the effect completely. The theory was proposed by Einstein and it made the claim that electromagnetic radiation had to be thought of as a series of particles, called photons, which collide with the electrons on the surface and emit them. This theory ran contrary to the belief that electromagnetic radiation was a wave and thus it was not recognized as correct until 1916 when Robert Millikan experimentally confirmed the theory

The photoelectric effect was first documented in 1887 by the German physicist Heinrich Hertz and is therefore sometimes referred to as the Hertz effect. While working with a spark-gap transmitter (a primitive radio-broadcasting device), Hertz discovered that upon absorption of certain frequencies of light, substances would give off a visible spark. In 1899, this spark was identified as light-excited electrons (also called photoelectrons) leaving the metal's surface by J.J. Thomson (Figure $$\PageIndex{1}$$).

Figure $$\PageIndex{1}$$: Simplified picture of the photoelectron effect. Image used with permission from Wikipedia.

The classical picture underlying the photoelectron effect was that the atoms in the metal contained electrons, that were shaken and caused to vibrate by the oscillating electric field of the incident radiation. Eventually some of them would be shaken loose, and would be ejected from the cathode. It is worthwhile considering carefully how the number and speed of electrons emitted would be expected to vary with the intensity and color of the incident radiation.

• Increasing the intensity of radiation would shake the electrons more violently, so one would expect more to be emitted, and they would shoot out at greater speed, on average.
• Increasing the frequency of the radiation would shake the electrons faster, so it might cause the electrons to come out faster. For very dim light, it would take some time for an electron to work up to a sufficient amplitude of vibration to shake loose.

### Lenard's Experimental Results

In 1902, Hertz's student, Philipp Lenard, studied how the energy of the emitted photoelectrons varied with the intensity of the light. He used a carbon arc light, and could increase the intensity a thousand-fold. The ejected electrons hit another metal plate, the collector, which was connected to the cathode by a wire with a sensitive ammeter, to measure the current produced by the illumination. To measure the energy of the ejected electrons, Lenard charged the collector plate negatively, to repel the electrons coming towards it. Thus, only electrons ejected with enough kinetic energy to get up this potential hill would contribute to the current.

Lenard discovered that there was a well defined minimum voltage that stopped any electrons getting through ($$V_{stop}$$). To Lenard's surprise, he found that $$V_{stop}$$ did not depend at all on the intensity of the light! Doubling the light intensity doubled the number of electrons emitted, but did not affect the kinetic energies of the emitted electrons. The more powerful oscillating field ejected more electrons, but the maximum individual energy of the ejected electrons was the same as for the weaker field (Figure $$\PageIndex{2}$$).

Figure $$\PageIndex{2}$$: Current Intensity and Threshold Frequency

However, Lenard did something else. With his very powerful arc lamp, there was sufficient intensity to separate out the colors and check the photoelectric effect using light of different colors. He found that the maximum energy of the ejected electrons did depend on the color - the shorter wavelength, higher frequency light caused electrons to be ejected with more energy (Figures $$\PageIndex{3}$$ and $$\PageIndex{4}$$).

Figure $$\PageIndex{3}$$: Lenard's photoelectric experiment with high-energy blue light. The battery represents the potential Lenard used to charge the collector plate negatively, which would actually be a variable voltage source. Since the electrons ejected by the blue light are getting to the collector plate, the potential supplied by the battery is less than $$V_{stop}$$, for blue light.

Figure $$\PageIndex{4}$$: Lenard's photoelectric experiment with low-energy red light. Since the electrons ejected by the blue light are not getting to the collector plate, the potential supplied by the battery exceeds $$V_{stop}$$ for red light.

This was, however, a fairly qualitative conclusion -- the energy measurements were not very reproducible, because they were extremely sensitive to the condition of the surface, in particular its state of partial oxidation. In the best vacuum available at that time, significant oxidation of a fresh surface took place in tens of minutes. The details of the surface are crucial because the fastest electrons emitted are those from right at the surface, and their binding to the solid depends strongly on the nature of the surface -- is it pure metal or a mixture of metal and oxygen atoms?

Classical theory predicts that energy carried by light is proportional to its amplitude independent of its frequency, and this fails to correctly explain the experiment.

As shown in Figure $$\PageIndex{4}$$, just the opposite behavior from classical is observed from Lenard's experiments. The intensity affects the number of electrons, and the frequency affects the kinetic energy of the emitted electrons. From these sketches, we see that

• the kinetic energy of the electrons is linearly proportional to the frequency of the incident radiation above a threshold value of $$ν_0$$ (no current is observed below $$ν_0$$), and the kinetic energy is independent of the intensity of the radiation, and
• the number of electrons (i.e. the electric current) is proportional to the intensity and independent of the frequency of the incident radiation above the threshold value of $$ν_0$$ (no current is observed below $$ν_0$$).

Figure $$\PageIndex{5}$$: Schematic drawings showing the characteristics of the photoelectric effect. (a) The kinetic energy of any single emitted electron increases linearly with frequency above some threshold value and is independent of the light intensity. (b) The number of electrons emitted per second (i.e. the electric current) is independent of frequency and increases linearly with the light intensity.

### The Quantum Picture Behind Photoelectrons

In 1905 Einstein gave a very simple interpretation of Lenard's results. He borrowed Planck's hypothesis about the quantized energy from his blackbody research and assumed that the incoming radiation should be thought of as quanta of frequency $$h\nu$$, with $$\nu$$ the frequency. In photoemission, one such quantum is absorbed by one electron. If the electron is some distance into the material of the cathode, some energy will be lost as it moves towards the surface. There will always be some electrostatic cost as the electron leaves the surface, this is usually called the work function, $$\Phi$$. The most energetic electrons emitted will be those very close to the surface, and they will leave the cathode with kinetic energy

$E = h\nu - \Phi \label{Eq1}$

On cranking up the negative voltage on the collector plate until the current just stops, that is, to $$V_{stop}$$, the highest kinetic energy electrons ($$KE_e$$) must have had energy $$eV_{stop}$$ upon leaving the cathode. Thus,

$eV_{stop} = h\nu - \Phi \label{Eq2}$

Thus, Einstein's theory makes a very definite quantitative prediction: if the frequency of the incident light is varied, and $$V_{stop}$$ plotted as a function of frequency, the slope of the line should be $$\frac{h}{e}$$ (Figure $$\PageIndex{5}$$a). It is also clear that there is a minimum light frequency for a given metal $$\nu_o$$, that for which the quantum of energy is equal to the work function (Figure $$\PageIndex{2}$$). Light below that frequency, no matter how bright, will not eject electrons.

Since, according to both Planck and Einstein, the energy of light is proportional to its frequency rather than its amplitude, there will be a minimum frequency $$\nu_0$$ needed to eject an electron with no residual energy.

Since every photon of sufficient energy excites only one electron, increasing the light's intensity (i.e. the number of photons/sec) only increases the number of released electrons and not their kinetic energy. In addition, no time is necessary for the atom to be heated to a critical temperature and therefore the release of the electron is nearly instantaneous upon absorption of the light. Finally, because the photons must be above a certain energy to satisfy the work function, a threshold frequency exists below which no photoelectrons are observed. This frequency is measured in Hertz (1/second) in honor of the discoverer of the photoelectric effect.

The kinetic energy of an ejected electron equals the photon energy minus the work function of the electron in the specific material. An individual photon can give all of its energy to an electron. The photon’s energy is partly used to break the electron away from the material. The remainder goes into the ejected electron’s kinetic energy. In equation form, this is given by

$KE_e=h\nu−\Phi \label{Eq3}$

where

• $$KE_e$$ is the maximum kinetic energy of the ejected electron,
• $$h\nu$$ is the photon’s energy, and
• $$\Phi$$ is the work function (sometimes call the binding energy of the electron) to the particular material.

Equation $$\ref{Eq3}$$, due to Einstein in 1905, explains the properties of the photoelectric effect quantitatively. A strange implication of this experiment is that light can behave as a kind of massless "particle" now known as a photon whose energy $$E=h\nu$$ can be transferred to an actual particle (an electron), imparting kinetic energy to it, just as in an elastic collision between to massive particles such as billiard balls.

Example $$\PageIndex{1}$$

1. What is the energy in joules and electron volts of a photon of 420-nm violet light?
2. What is the maximum kinetic energy of electrons ejected from calcium by 420-nm violet light, given that the work function (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 $$E=h\nu$$. For part (b), once the energy of the photon is calculated, it is a straightforward application of $$KE=h\nu–\Phi$$ to find the ejected electron’s maximum kinetic energy, since $$\Phi$$ is given.

Solution for (a)

Photon energy is given by

$E = h\nu$

Since we are given the wavelength rather than the frequency, we solve the familiar relationship $$c=\nu\lambda$$ for the frequency, yielding

$\nu=\dfrac{c}{\lambda}$

Combining these two equations gives the useful relationship

$E=\dfrac{hc}{\lambda}$

Now substituting known values yields

$E=\dfrac{(6.63 \times 10^{-34}\; J \cdot s)(3.00 \times 10^8 m/s)}{420 \times 10^{-9}\; m} = 4.74 \times 10^{-19}\; J$

Converting to eV, the energy of the photon is

$E=(4.74 \times 10^{-19}\; J) \dfrac{1 \;eV}{1.6 \times 10^{-19}\;J} = 2.96\; eV.$

Solution for (b)

Finding the kinetic energy of the ejected electron is now a simple application of the equation $$KE_e=h\nu–\Phi$$. Substituting the photon energy and binding energy yields

$KE_e=h\nu – \Phi = 2.96 \;eV – 2.71 \;eV = 0.246\; eV.$

Discussion

The energy of this 420-nm 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 420-nm photons with their 2.96-eV 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.

Exercise $$\PageIndex{1}$$

What is the longest-wavelength electromagnetic radiation that can eject a photoelectron from silver, given that the work function is 4.73 eV? Is this in the visible range?

263 nm. This is ultraviolet and not in the visible range.

### The Work Function

The required energy to free an electron from an atom is called the work function and is designated by the symbol $$\Phi$$. The work function is an intrinsic property of the metal. The following table summarizes the work functions for several elements:

Element Work function $$\Phi$$ (eV) Ionization Energy (eV)
Table $$\PageIndex{1}$$: Work functions of Select Elements
Silver (Ag) 4.64 7.57
Aluminum (Al) 4.20 5.98
Gold (Au) 5.17 9.22
Boron (B) 4.45 8.298
Beryllium (Be) 4.98 9.32
Bismuth (Bi) 4.34 7.29
Carbon (C) 5.0 11.26
Cesium (Ce) 1.95 3.89
Iron (Fe) 4.67 7.87
Gallium (Ga) 4.32 5.99
(Hg) liquid 4.47 10.43
Sodium (Na) 2.36 5.13
Lithium (Li) 2.93 5.39
Potassium 2.3 4.34
Selenium (Se) 5.9 9.75
Silicon (Si) 4.85 8.15
Tin (Sn) 4.42 7.34
Germanium (Ge) 5.0 7.89
Arsenic (As) 3.75 9.81

While the work functions and ionization energies appear as similar concepts, they are independent concepts. The work function of a metal (in the context of the photoelectric effect) is defined as:

The minimum amount of energy necessary to remove an electron from the surface of the metal.

This is qualitatively similar to ionization energy:

The amount of energy required to remove an electron from an atom or molecule in the gaseous state.

These two energies are generally different (Table $$\PageIndex{1}$$). For instance, copper has a work function of about 4.7 eV, but has a higher ionization energy of about 746 kJ mol-1 or 7.7 eV. Generally, the ionization energies for metals are greater than the corresponding work functions.

Work function deals with "free" electrons in the metallic bonding of the metal, while ionization energy addressed the valence electrons still bound within the atom.

Exercise $$\PageIndex{2}$$

Since both the work function and the ionization energy refers to the energy required to remove an electron, why do they differ? Why is the work function generally lower than the ionization energy?

Millikan's Failed effort to Disprove Quantum Theory

Einstein's theory of the photoelectron presented a completely different way to measure Planck's constant. The American experimental physicist Robert Millikan did not accept Einstein's theory, which he saw as an attack on the wave theory of light, and worked for ten years until 1916, on the photoelectric effect. He even devised techniques for scraping clean the metal surfaces inside the vacuum tube. For all his efforts he found disappointing results: he confirmed Einstein's theory, measuring Planck's constant to within 0.5% by this method. One consolation was that he did get a Nobel prize for this series of experiments.

### Summary

• The photoelectric effect is the process in which electromagnetic radiation ejects electrons from a material.
• Einstein proposed photons to be quanta of electromagnetic radiation having energy $$E=h\nu$$ is the frequency of the radiation.
• All electromagnetic radiation is composed of photons. As Einstein explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons.
• The maximum kinetic energy $$KE_e$$ of ejected electrons (photoelectrons) is given by $$KE_e=h\nu – \Phi$$, where $$h\nu$$ is the photon energy and $$\Phi$$ is the work function (or binding energy) of the electron to the particular material.

Einstein's simple explanation completely accounted for the observed phenomenon in Lenard's experiment and began an investigation into the field we now call quantum mechanics. This new field seeks to provide a quantum explanation for classical mechanics and create a more unified theory of physics and thermodynamics. The study of the photoelectric effect has also lead to the creation of new photoelectron spectroscopy theory and applications.

### Conceptual Questions

1. Is visible light the only type of electromagnetic radiation that can cause the photoelectric effect?
2. Which 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?
3. Is the photoelectric effect a direct consequence of the wave character of electromagnetic radiation or of the particle character of electromagnetic radiation? Explain briefly.
4. Insulators (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.
5. If 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.