The Photoelectric Effect
Inquiry Lab Kit for AP® Physics 2
Materials Included In Kit
Copper sheet, 8-cm square
Zinc foil, 8-cm square
Photon Demonstrator Card
Additional Materials Required
Cardboard box “shield” (optional)
Flashlight or incandescent light
Glass plate (optional)
Ultraviolet lamp, long-wavelength (optional)
Ultraviolet lamp, short-wavelength (optional)
Ultraviolet light may damage the eyes and cause cataracts. Wear safety glasses when using an ultraviolet light source and do not look directly at the light. Wash hands thoroughly with soap and water before leaving the laboratory.
All materials may be saved and stored for future use.
- Enough materials are provided in this kit for 24 students working in groups of two or for 12 groups of students. This investigation can reasonably be completed in two 50-minute class periods.
- The spots under the blue and violet filters should always glow brightly, but sometimes a small glow can be seen under some of the other filters. This is partly due to stray light creeping in around the filters. Try to keep the Demonstrator Card as tightly closed as possible to prevent any extra light from getting to the phosphorescent strip and causing a slight glow. The slight glow can be explained by looking at the transmission curves of the filters. For example, the yellow filter clearly lets through light of wavelengths down to about 450 nm. The phosphorescent strip needs an exciting wavelength of about 480 nm or lower. Therefore, a tiny bit of light with enough energy is actually being transmitted through the yellow filter into the phosphorescent strip.
- The guided-inquiry design and procedure section was developed to subtly lead students through the experimental design process. You may decide to provide less/more information than is included therein. So long as the students are challenged to think critically.
- The electroscope and metal plates must be accurately charged. If a negatively charged rod is brought near a negatively charged electroscope, the leaves will deflect farther. If a negatively charged rod is brought near a positively charged electroscope, the leaves will collapse as the electroscope discharges.
- Note that when depending on atmospheric conditions, sometimes only minimal movement of foil leaves can be observed.
- Rubbing a rubber rod with wool or animal fur gives the best results for charging the electroscope either negatively by conduction or positively by induction. Other materials may also be used—PVC and wool or animal fur may be used to charge the electroscope negatively by conduction, while glass and silk may be used to charge the electroscope positively by conduction.
- When charging the electroscope with the zinc or copper plate attached, it is best to charge the metal sphere on top of the electroscope rather than the zinc or copper plate. When the electroscope is charged, the metal plate will also be charged.
- Check with the biology or earth science teacher at your school for UV lamps. Ultraviolet light is typically used to demonstrate fluorescence in rocks and minerals.
- A combination long-wave/short-wave ultraviolet lamp with a sliding selector (Flinn Catalog No. AP5725) that allows one to select either long-wave or short-wave UV light is the best choice for this experiment. A less expensive short-wave UV lamp is also available from Flinn Scientific (Catalog No. AP6658). The latter will discharge the electroscope, but it should be compared against a standard black light to confirm that only the short-wavelength ultraviolet light is effective.
- It is important to ground yourself both before charging the electroscope and again when the ultraviolet lamp is turned on and shined on the zinc or copper plate. Static charge build-up on the plastic casing of a black light or UV lamp will discharge the electroscope unless the demonstrator is grounded.
- If open windows are available in the science lab, mirrors may also be used to reflect sunlight onto the metal plate. However, it must be direct sunlight. Sunlight transmitted through glass will not possess short-wave UV radiation. The sunlight must be unfiltered.
- The wavelengths of light contained in classroom lights can be compared to those in a UV (black) light using the Demonstrator Card. Perform the demonstration as outlined above using the classroom lights. Then, turn off the lights and shine a black light onto the closed Demonstrator Card for about 30 seconds. When the black light is turned off and the Demonstrator Card is opened, more glowing circles can be observed. Looking at the transmission curves for the filters, explain to the class that many of the filters do not absorb wavelengths between 300 and 400 nm. Therefore, the filters are transmitting these high-energy wavelengths, which are sufficient to excite the phosphorescent strip and make it glow.
Opportunities for Inquiry
Alignment to the Curriculum Framework ofr AP® Physics 2
- A more quantitative version of this experiment would be to measure the potential needed to cause LEDs of different colors to light up. Alternatively or in addition, an interdisciplinary connection with chemistry could be demonstrated to verify the maximum wavelength needed to cause the strip to glow using a spectrophotometer.
- To verify the maximum wavelength needed to cause the strip to glow using a spectrophotometer, cut a narrow strip (so it will just fit in a cuvet) of the phosphorescent strip. First set 0% T with the sample compartment empty. Use an empty cuvet as the blank to set 100% T. Then place the piece of phosphorescent strip in the cuvet, not directly in the spectrophotometer, and measure the absorbance of the strip. Make sure the coated side of the strip is facing the direction from which light comes in the spectrophotometer. For wavelengths of 480 nm and lower, a small glowing spot can be seen on the phosphorescent strip. At higher wavelengths, no glowing is observed. This experiment is best performed in the dark so that the glowing spot is easily seen.
Enduring Understandings and Essential Knowledge
The energy of a system is conserved. (5B)
Electromagnetic radiation can be modeled as waves or as fundamental particles. (6F)
5B8.1: The student is able to describe emission or absorption spectra associated with electronic or nuclear transitions as transitions between allowed energy states of the atom in terms of the principle of energy conservation, including characterization of the frequency of radiation emitted or absorbed.
6F3.1: The student is able to support the photon model of radiant energy with evidence provided by the photoelectric effect.
6F4.1: The student is able to select a model of radiant energy that is appropriate to the spatial or temporal scale of an interaction with matter.
1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain.
1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain.
1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively.
3.2 The student can refine scientific questions.
3.3 The student can evaluate scientific questions.
4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question.
4.2 The student can design a plan for collecting data to answer a particular scientific question.
4.3 The student can collect data to answer a particular scientific question.
5.1 The student can analyze data to identify patterns or relationships.
5.3 The student can evaluate the evidence provided by data sets in relation to a particular scientific question.
6.1 The student can justify claims with evidence.
6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices.
6.3 The student can articulate the reasons that scientific explanations and theories are refined or replaced.
6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
7.1 The student can connect phenomena and models across spatial and temporal scales.
Correlation to Next Generation Science Standards (NGSS)†
Science & Engineering Practices
Asking questions and defining problems
Planning and carrying out investigations
Engaging in argument from evidence
Obtaining, evaluation, and communicating information
Developing and using models
Disciplinary Core Ideas
HS-PS4.A: Wave Properties
HS-PS4.B: Electromagnetic Radiation
Systems and system models
HS-PS4-3. Evaluate the claims, evidence, and reasoning behind the idea that electromagnetic radiation can be described either by a wave model or a particle model, and that for some situations one model is more useful than the other.
Answers to Prelab Questions
- The visible spectrum consists of light with wavelengths that are characterized as red, orange, yellow, green, blue, indigo and violet. Which of these colors has the highest energy photons?
The formula E = hc/λ is derived from the formula E = hv and tells us that energy is inversely proportional to wavelength. Violet light has the highest energy photons because it has the smallest wavelength of all the colors in the visible spectrum.
- Two hypothetical metals, X and Y, have different work functions. X has a higher work function than Y. Which metal will require higher energy light to demonstrate the photoelectric effect?
A metal’s work function corresponds to the minimum energy needed to displace an electron from the metal. Thus, X will require higher energy light to demonstrate the photoelectric effect.
- Explain, with respect to electrostatics, the phrase “like repels like” and how electrons can be transferred between two objects.
An object can be positively charged, negatively charged, or neutral (no charge). Positively charged objects have more protons than electrons, negatively charged objects have more electrons than protons, and neutral objects have equal amounts of protons and electrons. A basic tenet of electrostatics, which describes the interactions between charged particles, is that electrons repel electrons, protons repel protons, and electrons attract protons. So, with respect to this experiment the phrase “like repels like” applies directly to the foil leaves in the electroscope. When the foil leaves are both negatively charged they will deflect each other owing to the fact that they each have an excess of electrons. When the foil leaves are discharged, that is when the excess electrons are transferred out of them, the leaves become neutral and collapse into each other because they no longer repel each other. Electrons can be transferred between two objects mechanically, or by rubbing them together vigorously. One of the objects will give up electrons and one of the objects will accept electrons.
- What is the difference between light intensity and light energy.
The intensity of light corresponds to its brightness whereas the energy of light corresponds to its wavelength, frequency or color. Two lights of different wavelength can be equal in intensity, or brightness, but cannot have equal energies.
- The energy of a quantum of light is proportional to its frequency. Which UV light has higher energy—the long-wave or the short-wave? Explain.
A long-wave UV light has a longer wavelength than a short-wave UV light. A quantum of light refers to a single photon of that light. Because light energy is inversely proportional to wavelength, according to the formulas E = hv and E = hc/λ (v = c/λ), the short-wave UV light will have a higher energy.
- What is a photon? Calculate the energy of a photon of red laser light (λ = 650 nm).
A photon is a single particle, or quantum, of light. The formula E = hc/λ allows for the calculation of light energy given wavelength. The term “h” corresponds to Planck’s constant, 6.626 x 10 –34 J • sec. The term “c” corresponds to the speed of light, 2.998 x 108 m /sec. For red laser light, E = [(6.626 x 10–34 J • sec)( 2.998 x 108 m /sec)]/650 x 10–9 m = 3.056 x 10–19 J
Demonstration of the Photoelectric Effect with an Electroscope
- Rub the rubber rod vigorously with wool and then gently slide the rod against the metal sphere on top of the electroscope. Repeat this process until the foil leaves of the electroscope are fully deflected. The rubber rod and the electroscope must be in direct contact. Explain why the foil leaves move away from each other or repel each other.
Typically, electroscopes are charged by rubbing wool on a rubber pole or glass stirring rod so that electrons are transferred between the two objects. A rubber rod is less willing to give up electrons than a piece of wool. Thus, when the two objects are rubbed together, the rubber rod becomes negatively charged. When the negatively charged rubber rod is brought into proximity with the metal ball on the electroscope, the negative charge on the rod will repel some electrons out of the metal ball and down into both foil leaves, which will diverge away from each other because like charges repel each other. This shows that electrons can be moved mechanically by rubbing two objects together.
- Polish the zinc foil and copper sheet with sandpaper to remove any oxide coating on the zinc surface. Balance the zinc plate against the top of the electroscope so that the plate is almost perpendicular to the ground. The plate should be touching the wire pole and metal sphere and should face in a direction that will make it easy to observe the deflection of the electroscope leaves. Use a piece of transparent tape to hold the zinc plate in place (see Figure 1).
- Shine a flashlight or incandescent lamp on the zinc plate. Explain why the foil leaves do not collapse into each other.
The flashlight or incandescent lamp does not provide light of a high enough energy to overcome the Zn work function and cause it to emit electrons. As a result, the Zn metal does not become positively charged and electrons are not attracted out of the foil leaves towards the positively charged Zn plate. As a result, the foil leaves remain negatively charged and continue to deflect each other.
- If necessary, recharge the electroscope as you did in step 1 by rubbing the rubber rod vigorously with wool and sliding it against the metal sphere on top of the electroscope. Once you observe the foil leaves deflect each other, take the electroscope outside in bright sunshine. If possible, shield the zinc plate from the sunlight by keeping it shaded—place it in a cardboard box “shield” if possible.
- Step away from the electroscope and allow the sunlight to shine on the zinc plate. Explain why the electroscope leaves collapse.
Sunlight shined on Zn causes it to emit electrons (the photoelectric effect) and become positively charged. As a result, electrons present on the foil leaves move toward the positively charged Zn metal. As the foil leaves lose electrons and become neutral, they no longer repel each other and collapse into each other.
- Take the electroscope back inside the classroom and charge the electroscope as you did in step 1 until the foil leaves deflect each other.
- Shine a black light or long-wavelength ultraviolet lamp on the zinc plate. Explain why the foil leaves do not quickly collapse into each other.
Neither the black light nor the long-wavelength ultraviolet light are of high enough energy to cause the Zn plate to emit electrons. It therefore does not become positively charged and cannot draw electrons away from the foil leaves. The foil leaves retain their excess electrons (and excess negative charges) and continue to repel each other.
- Shine a short-wavelength ultraviolet lamp on the zinc plate and record observations. Explain why the foil leaves now collapse into each other.
The equation E = hc/λ indicates that wavelength and energy are inversely related. The short-wavelength light has a higher energy than the long-wavelength light and indeed has a high enough energy to cause the Zn plate to emit electrons and become positively charged so that the electrons in the foil leaves migrate to the Zn plate. The foil leaves consequently become neutral and no longer repel each other. And so the foil leaves collapse into each other.
- Repeat steps 1–8 using a copper plate. Ask students to explain why the electroscope will not discharge (i.e., why the foil leaves will not give up their electrons, become neutral and collapse into each other).
Copper has a higher threshold energy or work function than zinc—none of the light sources used have photons with high enough energy to cause electron emission from the Zn metal. As a result, the charged electroscope will not discharge. That is, the deflected foil leaves will not give up their electrons. Instead, they remain negatively charged and repel each other.
Answers to Questions
Guided-Inquiry Discussion Questions
Using the provided Demonstrator Card, carry out an experiment that confirms energy, not intensity, is the determining factor in the emission of electrons from a metal’s surface. In addition, calculate the approximate wavelength needed to excite the phosphorescent strip.
Review Questions for AP® Physics 2
- Expose the phosphorescent strip provided to the classroom lights for approximately 15 seconds and then darken the room. Explain why the strip glows.
The strip’s glow is due to the emission of photons, analogous to the ejection of electrons from the surface of a metal by light (the photoelectric effect). The phosphorescent material has a critical wavelength (or energy) of light. If a light source is shined on the phosphorescent strip and it contains photons whose energy is greater than the energy needed to cause the strip to glow, it will glow. If the intensity of the source is increased, the glowing of the strip will increase. If, however, a light source is shined on the phosphorescent strip that contains photons whose energy is less than the critical energy for the strip, no glowing will occur no matter how bright the light source.
- Wait for the phosphorescent strip to stop “glowing.” Cover the phosphorescent strip with a piece of dark paper, expose it to light and then darken the room. Explain why the strip does not glow.
The strip does not glow when covered with a piece of dark paper and exposed to the light because all of the light is absorbed by the dark paper and is not transmitted through to the phosphorescent strip, and therefore cannot cause emission of photons from the strip. The dark paper is a filter that actually filters all visible light, not letting any through.
- The filter is composed of plastic films of different colors: black, red, yellow, green, blue and violet. The blue filter transmits blue light but absorbs the other wavelengths, the red filter transmits red light but absorbs the other wavelengths, and so on. Which of the color lights is least likely to cause the strip to glow and which is most likely? Explain.
According to the formula E = hc/λ, a light’s wavelength is inversely proportional to its energy. Given that the phosphorescent strip has a critical energy that must be overcome in order for it to glow, high energy light would be more likely to exceed that critical energy and make the strip glow than low energy light. So, on the visible spectrum, violet light is most likely to make the strip glow, and red light is least likely to make the strip glow owing to their short and long wavelengths, respectively.
- Predict what will happen if the Demonstrator Card is closed so that only “filtered” light is allowed to shine on the phosphorescent strip.
The colored filters are circular and will transmit visible light corresponding to the wavelength of their respective color. Therefore, it is likely that you should observe circular bright spots beneath the higher energy filters, like violet and blue. You should not see bright spots behind the dark paper holding the filters nor behind the lower energy filters, like red and yellow. Again, this is attributable to the idea that the strip has a minimum threshold energy that must be overcome (by a light source) in order to cause it to glow… much like the minimum threshold energy that must be overcome (by a light source) to cause a metal to emit electrons, or demonstrate the photoelectric effect.
- Close the Demonstrator Card tightly making sure that no light can reach the phosphorescent strip from the sides. Paper clip the sides closed. Expose the closed Demonstrator Card to the classroom lights for at least 30 seconds. Darken the classroom. Open the Demonstrator Card. Why does the strip glow only under the blue and violet filters?
Only the blue and violet filters let through light with enough energy to cause the emission of photons from the phosphorescent strip.
- The maximum wavelength needed to excite the phosphorescent strip (and cause it to glow), based on the transmission curves for the filters, is about 480 nanometers (nm). Calculate the minimum energy required of a photon to cause the strip to phosphoresce, or glow.
We know that E = hc/λ, and we know that only blue and violet light provided enough energy to cause the strip to glow. The wavelength region for violet light is 400–425 nm and the wavelength region for blue light is 425–480 nm. The minimum energy required, based on our data, would correspond to light at a maximum wavelength of 480 nm. Plugging that value into the equation, E = hc/λ, we get an approximation of the minimum energy required to exceed the threshold potential of the strip: E = [(6.626 x 10–34 J • sec)(2.998 x 108 m/sec)]/480 x 10–9 m = 3.91 x 10–19 J. This is only an approximation because some high energy light filters are missing from our filter set, including a blue-green filter.
- Describe how to use two light bulbs of different intensity, for example a 40-W bulb and a 100-W bulb, to confirm that energy is the determining factor in inducing the photoelectric effect.
When light is shined on the demonstrator card, the phosphorescent strip glows beneath the blue and violet filters. That is, the blue and violet filters transmit blue and violet light, which have the highest energy on the visible spectrum and are able to excite photons in the phosphorescent strip. Students can use a 40-W bulb and a 100-W bulb and see that in both cases only the blue and violet light cause the strip to glow. That is, even bright (more intense) light filtered through the lower energy filters does not cause the strip to glow. No matter how bright the light only the high-wavelength blue and violet photons exceed the work function of the strip.
- Based on what you know about the photoelectric effect, describe the relationship between the intensity of light and electric current.
Electric current corresponds to the flow of electrons. As the intensity of light is increased and directed at a metal or electroscope with a work function exceeded by the light source, the amount of electrons flowing through the electroscope will increase.
- Photons of which color light have more energy, yellow or green?
Again, the formula E = hc/λ indicates an inverse relationship between wavelength and energy in light. Thus, because yellow light has a higher wavelength than green light, it will have the lower energy.
- If one photon of incident light causes the emission of one electron from sodium, how many electrons are emitted if the intensity of the same light is doubled?
There is a direct relationship between light intensity and electron emission. If intensity is doubled, the number of emitted electrons is doubled. In this case then, the number of electrons emitted doubles to two.
- The work function for Na is 2.36 eV (1 eV = 1.60218 x 10–19 J). What is the lowest frequency of light that will cause photoelectric emission?
E = hv
E = 2.36 eV x (1.60218 x 10–19 J/1 eV) = 3.78 x 10–19 J
3.78 x 10–19 J = (6.626 x 10–34 J • sec) x (v)
v = 5.7065 x 1014 sec–1
AP® Physics 1: Algebra-Based and Physics 2: Algebra-Based Curriculum Framework; The College Board: New York, NY, 2014.