Lesson Explainer: The Kinetic Energy of Photoelectrons | Nagwa Lesson Explainer: The Kinetic Energy of Photoelectrons | Nagwa

Lesson Explainer: The Kinetic Energy of Photoelectrons Physics • Third Year of Secondary School

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In this explainer, we will learn how to calculate the maximum possible kinetic energy of electrons that are ejected from the surface of a metal due to the photoelectric effect.

The photoelectric effect is the process of electrons leaving the surface of a metal after absorbing electromagnetic radiation. An experimental apparatus used for observing the photoelectric effect is shown in the diagram below.

Two separate metal plates are attached to a circuit, which has an ammeter connected in series. The metal plates are enclosed in a vacuum chamber so that air does not affect the experiment. Light is directed at one of the metal plates. If the incident light has great enough energy, electrons are ejected from the metal surface. These ejected electrons are known as โ€œphotoelectrons.โ€ The ammeter detects a current as photoelectrons reach the adjacent plate.

Recall that light can be modeled as a particle. Particles of light are known as photons. Each photon has a discrete amount of energy, ๐ธ, described by the formula ๐ธ=โ„Ž๐‘“, where โ„Ž represents the Planck constant and ๐‘“ represents the frequency of the photon.

Each single incident photon transfers energy to a single electron on the metal surface. The electron will leave the surface if the photon has great enough energy. Since photon energy is determined by frequency, it does not matter what the amplitude of the light wave isโ€”the photoelectric effect is induced so long as the light has high-enough frequency. The relationship between energy and frequency, and the independence of these values from amplitude, is shown in the table below.

Now that we have established the basics of the photoelectric effect, let us get a closer look at the energy transfer between photons and electrons.

Recall that atomic nuclei have electrons in discrete energy levels. At each level, the electrons have different amounts of energy that keep them bound to the atomic system; this amount of energy is called a โ€œwork function.โ€ We can consider the work function, denoted by ๐‘Š, as a barrier that keeps an electron bound to a material. If an amount of energy greater than the work function is transferred to an electron, the barrier is overcome and the electron is freed of its bond.

Conductive materials like metals have relatively low work functions. Thus, outermost electrons on a metal surface can somewhat readily leave the material altogether if they gain enough energy. This is what occurs in the photoelectric effect.

If an electron receives an amount of energy greater than the work function, the remaining energy becomes kinetic energy of the electron. This can be observed as photoelectrons often leave the metal surface at significant speeds.

We are able to determine the maximum kinetic energy of a photoelectron so long as we know the energy supplied by the photon and the work function for the metal surface. The amount of kinetic energy that a resulting photoelectron has is equal to the energy that a photon transferred to it minus the work function that had to be overcome.

Let us formally define this relationship.

Definition: The Maximum Kinetic Energy of a Photoelectron given Frequency

The maximum kinetic energy of a photoelectron is given by ๐ธ=โ„Ž๐‘“โˆ’๐‘Š,max where โ„Ž is the Planck constant, ๐‘“ is the frequency of the incident photon, and ๐‘Š is the work function of the metal surface.

We will practice using this equation in the following example.

Example 1: Calculating the Maximum Kinetic Energy of Photoelectrons

A polished metal surface in a vacuum is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal. The light has a frequency of 2.00ร—10๏Šง๏Šซ Hz. The work function of the metal is 1.40 eV. What is the maximum kinetic energy that the electrons can have? Use a value of 4.14ร—10๏Šฑ๏Šง๏Šซ eVโ‹…s for the Planck constant. Give your answer in electron volts.

Answer

Let us begin by recalling the equation for the maximum kinetic energy of a photoelectron, ๐ธ=โ„Ž๐‘“โˆ’๐‘Š.max

We have been given values for โ„Ž, ๐‘“, and ๐‘Š; substituting them in, we have ๐ธ=๏€น4.14ร—10โ‹…๏…๏€น2.00ร—10๏…โˆ’1.40=6.88.max๏Šฑ๏Šง๏Šซ๏Šง๏ŠซeVsHzeVeV

Thus, we have found that the maximum kinetic energy the electrons can have is 6.88 eV.

It is often useful to graph the equation for the maximum kinetic energy of a photoelectron. A plot of photoelectron kinetic energy against incident photon frequency is shown below.

Recall that for an electron to be ejected, an incident photon must have high enough frequency (and therefore energy) to overcome the work function. For this reason, we record zero photoelectron energy for low-frequency light, as illustrated by the horizontal portion of the graph. This shows where the incident light is of too low an energy to remove electrons, so we detect no photoelectrons and no kinetic energy.

However, at a high-enough photon frequency, the work function is overcome. Recall that the work function of a material is a constant value, so once it is overcome, the kinetic energy of the photoelectrons increases as the incident photon frequency increases. Thus, ๐ธmax is directly proportional to ๐‘“, and the relationship is linear, as illustrated by the sloped, increasing portion of the graph.

We can determine certain properties of an apparatus by analyzing its graph of ๐ธmax versus ๐‘“. Specifically, we are interested in the point where the graph bends from the horizontal axis, as highlighted in the figure below. This point occurs at a threshold frequency value that we will call ๐‘“๏Šฆ.

This defines a turning point in the experiment where photons transfer just enough energy for the electrons to be ejected. Here, the photoelectronsโ€™ โ€œleftoverโ€ kinetic energy is equal to zero, since the energy of the photon is barely enough to overcome the work function.

We can use this information to experimentally determine the work function of a material. To begin, let us rearrange the maximum kinetic energy formula to solve for ๐‘Š: ๐‘Š=โ„Ž๐‘“โˆ’๐ธ.max

Recall that ๐ธ=0max at the threshold frequency, ๐‘“๏Šฆ. Substituting these values in, we have ๐‘Š=โ„Ž๐‘“.๏Šฆ

Thus, at the threshold frequency, the work function is equal to the incident photon energy. We will practice this method of determining work function in the next couple of examples.

Example 2: Determining Work Function Using a Graph of Electron Energy versus Photon Energy

A tunable laser is used to illuminate the surface of a metal with different frequencies of light. Above a certain frequency of light, electrons are emitted from the surface of the metal. The graph shows the maximum kinetic energy of the electrons emitted against the energy of the photons. What is the work function of the metal?

Answer

This graph illustrates the relationship between incident photon energy and the maximum kinetic energy of a photoelectron leaving the metal surface. Recall the equation relating these values, ๐ธ=โ„Ž๐‘“โˆ’๐‘Š,max where โ„Ž๐‘“ describes the energy of an incident photon given its frequency, ๐‘“, and the Planck constant, โ„Ž. We want to find the work function for this metal surface, so we will rearrange this equation to solve for ๐‘Š: ๐‘Š=โ„Ž๐‘“โˆ’๐ธ.max

We can use coordinate values from any point on the graph to substitute into this equation. Generally, the simplest point to work with is at the โ€œthreshold frequencyโ€ ๐‘“๏Šฆ, or the graphโ€™s horizontal intercept, because ๐ธ=0max at this point. Thus, we can eliminate the ๐ธmax term in the equation, and we are left with ๐‘Š=โ„Ž๐‘“.๏Šฆ

Therefore, the photon energy at this point is equal to the work function of the material.

The graph intersects the horizontal axis at 2.6 eV, and thus we have found that the work function of the metal is 2.6 eV.

Example 3: Determining Work Function Using a Graph of Electron Energy versus Photon Energy

The graph shows the maximum kinetic energy of photoelectrons when different metals are illuminated with light of different frequencies.

  1. Which metal has the lowest work function?
  2. Which metal has the highest work function?

Answer

Part 1

Recall the formula for the maximum kinetic energy of a photoelectron, ๐ธ=โ„Ž๐‘“โˆ’๐‘Š,max where ๐‘Š is the work function and โ„Ž๐‘“ is the photon energy value, which depends on photon frequency, ๐‘“, and the Planck constant, โ„Ž.

This graph illustrates the properties of five different elements. All five lines on the graph have the same slope and are only made distinct by their horizontal axis intercepts.

We can learn about the elements from where their graphs intersect the horizontal axis because this value describes where incident photons have just enough energy to overcome the work function. Thus, ๐ธ=0max, but photoelectrons are still being created. We can substitute this value in to define a relationship between the work function and photon energy: 0=โ„Ž๐‘“โˆ’๐‘Š, or โ„Ž๐‘“=๐‘Š.

Therefore, the photon energy at this point is equal to the work function of the material.

A smaller horizontal axis intercept means that a lower photon energy value is required to overcome the work function. Thus, we can compare the magnitudes of the materialsโ€™ work functions by comparing their threshold photon energy values. Cesiumโ€™s line has the smallest horizontal intercept.

Thus, we have found that cesium has the lowest work function.

Part 2

Again inspecting the graph, we can see that platinum is the element with the greatest photon energy at the threshold where ๐ธ=0max.

Therefore, platinum has the highest work function.

We have explored how to determine the work function of a material from a graph of its electron kinetic energy against incident photon frequency. Now suppose we want to know how this relates to incident light wavelength, rather than frequency. To do this, we must devise a relationship between the frequency and the wavelength of light so that we can substitute ๐‘“ out of our equation and substitute ๐œ† in.

We can relate frequency and wavelength using the wave speed equation for an electromagnetic wave, ๐‘=๐œ†๐‘“, where ๐‘ is the speed of light. Solving this formula for frequency, we have ๐‘“=๐‘๐œ†.

Now recall the electron kinetic energy equation, ๐‘Š=โ„Ž๐‘“โˆ’๐ธ.max

Finally, we can make the substitution for frequency: ๐‘Š=โ„Ž๐‘๐œ†โˆ’๐ธ.max

This equation allows us to relate work function and maximum photoelectron kinetic energy to the wavelength of incident light.

We can rearrange this formula to define the maximum kinetic energy of a photoelectron, given incident photon wavelength, as stated below.

Definition: The Maximum Kinetic Energy of a Photoelectron given Wavelength

The maximum kinetic energy of a photoelectron is given by ๐ธ=โ„Ž๐‘๐œ†โˆ’๐‘Š,max where โ„Ž is the Planck constant, ๐‘ is the speed of light, ๐œ† is the wavelength of the incident photon, and ๐‘Š is the work function of the metal surface.

Notice that, in the frequency form of the equation, ๐‘“ appears in the numerator, allowing for a linear relationship between ๐‘“ and ๐ธmax. By contrast, in the wavelength form of the equation, ๐œ† appears in the denominator, meaning that the graph of ๐ธmax against ๐œ† does not have a linear slope. The general shape of the graph of electron kinetic energy against photon wavelength is drawn below.

Notice that no photoelectrons are emitted when photon wavelength exceeds a certain value. This is because as we increase the wavelength of the incident light, we simultaneously decrease its frequency (and therefore energy). Let us practice using this relationship in a couple of examples.

Example 4: Determining Work Function Using a Graph of Electron Energy versus Photon Wavelength

A tunable laser is used to illuminate the surface of a metal with different wavelengths of light. When the wavelength of the light is shorter than a certain value, electrons are emitted from the surface of the metal. The graph shows the maximum kinetic energy of the electrons emitted against the wavelength of the photons.

  1. What is the maximum wavelength of light for which electrons will be emitted from the surface of the metal?
  2. What is the work function of the metal? Use a value of 4.14ร—10๏Šฑ๏Šง๏Šซ eVโ‹…s for the Planck constant. Give your answer in electron volts to two decimal places.

Answer

Part 1

To begin, let us recall the formula for maximum photoelectron kinetic energy against incident photon wavelength: ๐ธ=โ„Ž๐‘๐œ†โˆ’๐‘Š.max

There is an inverse relationship between photon energy and wavelength. Thus, above a certain threshold wavelength, photons do not have enough energy to overcome the work function barrier and induce the photoelectric effect.

This point is visible on the graph where ๐ธ=0max. The wavelength at this point represents the maximum wavelength of light for which electrons will be ejected from the surface. This point is located on the horizontal axis at ๐œ†=300nm.

Thus, the maximum wavelength of incident light that will cause electrons to be emitted from the surface of the metal is 300 nm.

Part 2

Recall that the formula for work function given incident photon wavelength is ๐‘Š=โ„Ž๐‘๐œ†โˆ’๐ธ.max

To find the work function of the metal, we can substitute the graphโ€™s horizontal intercept value into this equation. We must convert nanometers into meters, so this threshold wavelength value is 300=300ร—10nmm๏Šฑ๏Šฏ. At this wavelength of incident light, electron kinetic energy equals zero, so we will eliminate ๐ธmax. Further, we substitute in the values for the Planck constant and the speed of light, and we can calculate the work function: ๐‘Š=๏€น4.14ร—10โ‹…๏…๏€ป3.0ร—10๏‡300ร—10=4.14.๏Šฑ๏Šง๏Šซ๏Šฎ๏Šฑ๏ŠฏeVsmeVms

Thus, we have found that the work function of the metal is 4.14 eV.

Example 5: Calculating Properties of an Experimental Photoelectric Effect Apparatus

The diagram shows an electrical circuit. The circuit contains an anode and cathode in a vacuum chamber. The anode and cathode are connected to an ammeter and battery in series. The cathode is made of nickel.

  1. Light of different wavelengths is used to illuminate the nickel cathode. When the wavelength of the light is shorter than 248 nm, the ammeter shows a reading of 12.8 mA. What is the work function of nickel? Use a value of 4.14ร—10๏Šฑ๏Šง๏Šซ eVโ‹…s for the Planck constant. Give your answer to two decimal places.
  2. Initially, the laser used to illuminate the cathode had a power output of 64 mW. If this were increased to 128 mW, what would the current in the circuit be? Give your answer to one decimal place.

Answer

Part 1

Let us begin by recalling the formula for the work function given incident photon wavelength, ๐‘Š=โ„Ž๐‘๐œ†โˆ’๐ธ.max

We know that when the incident light has great enough energy, electrons will be emitted from the copper surface, causing the ammeter to detect a current.

Here we know that the ammeter detects a current only when the wavelength of the incident light is lower than 248 mA. At this threshold wavelength value, which we will call ๐œ†๏Šฆ, the incident photons have just enough energy to overcome the work function barrier. Thus, there will not be kinetic energy left over for the photoelectrons, meaning ๐ธ=0max, so the formula becomes ๐‘Š=โ„Ž๐‘๐œ†.๏Šฆ

To calculate the work function, let us substitute in the values for the Planck constant, the speed of light, and the threshold wavelength: ๐‘Š=๏€น4.14ร—10โ‹…๏…๏€ป3.0ร—10๏‡248ร—10=5.01.๏Šฑ๏Šง๏Šซ๏Šฎ๏Šฑ๏ŠฏeVsmeVms

Thus, we have found that the work function of nickel is 5.01 eV.

Part 2

The power of the laser gives an amount of energy per second. Photons carry the energy of the laser beam, so if the laser is turned up to twice the amount of energy per second, it is emitting twice as many photons per second. Recall that one incident photon interacts with one electron on the metal surface. Thus, with twice as many photons incident on the surface, there will be twice as many electrons receiving energy and leaving the surface.

Therefore, if the power of the laser is doubled, the current is doubled as well. Since the ammeter initially detected a current of 12.8 mA, it will now detect twice this value.

Thus, the current in the circuit would be 25.6 mA.

Let us finish by summarizing some important concepts.

Key Points

  • The photoelectric effect is the phenomenon of removing electrons from a metal surface by shining light on it. A photoelectron is an electron emitted from the surface after receiving energy from an incident photon.
  • The work function of a material is the minimum amount of energy needed to remove an electron from its surface, and its value can be found from the graph of electron kinetic energy against photon energy.
  • The energy of light is proportional to its frequency and inversely proportional to its wavelength.
  • We can relate the work function, ๐‘Š, and maximum electron energy, ๐ธmax, given frequency, ๐‘“, using the formula ๐ธ=โ„Ž๐‘“โˆ’๐‘Šmax, where โ„Ž is the Planck constant.
  • We can relate the work function, ๐‘Š, and maximum electron energy, ๐ธmax, given wavelength, ๐œ†, using the formula ๐ธ=โ„Ž๐‘๐œ†โˆ’๐‘Šmax, where โ„Ž is the Planck constant and ๐‘ is the speed of light.

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