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L7 The Photoelectric Effect Page 1 of 1 L7 The Photoelectric Effect We saw in the last lecture that Max Planck introduced the concept of a quantum of light energy or photon when working out the details in his derivation of the equation describing the emission spectrum of a blackbody We mentioned that Einstein took Planck s idea of a light quantum or packet and extended it saying that light was always in the form of these packets of light energy In this lecture we will study the photoelectric e ect This is the effect behind the electric eye in elevator doors for example which opens the doors if something interupts the light beam mnning between the doors This effect is also used in some household smoke detectors In our study of the photoelectric effect we will once again see Einstein enter the picture with his idea of the quantum of light AAAxquot i m A n 139 r Y 1 1 rr F111 Iquot 1A1 nrlnnno Intro Page 1 of 2 The Photoelectric Effect While performing experiments in the late 188039s to continn Maxwell s prediction of electromagnetic waves Heinrich Hertz noticed that the production of an electric discharge between two electrodes was facilitated when light shined onto one of the electrodes It was later shown that this was due to the fact that light can cause electrons to be emitted from a metal surface in a process called the photoelectric effect This effect can be studied using the apparatus shown schematically below K vw jkrwwx e e l 0n 39 Figure 71 In the gure light of frequency f is incident on the emitter E Both the emitter E and the collector C electrodes are encased in an evacuated bulb which keeps the metal surfaces of the emitter and collector from being covered with oxide layers or other contaminants The quantity AV is called the source voltage This voltage has a polarity that is which end is positive and which end is negative which can be switched so that either the emitter can be negative and the collector positive as shown in the diagram m this will correspond to a positive value for the source voltage AV or else the collector can be negative and the emitter positive which corresponds to a negative value for AV It is found experimentally that light shining on a clean metal surface such as that of the emitter in Fig 71 above can result in the emission of electrons from that surface This effect is called the photoelectric effect and the emitted electrons are referred to as photoelectrons The ammeterA in Fig 71 measures any current that ows through the circuit as a result of electrons being emitted by electrode E the small black dots leaving the emitter in the gure above and collected by the collector If AV gt 0 then electrons emitted from the metal surface of the emitter due to light shining on its surface are all immediately repelled by the emitter and attracted to the collector since electrons are negatively charged and are attracted to a positive voltage and are subsequently registered as current by the ammeter A In this case effectively all electrons emitted at E are collected at C and registered by the ammeter We say that the current in this case is saturated On the other hand if AV lt 0 then the charges on E and C actually try to keep the Intro Page 2 of 2 emitted photoelectrons from making it to the collector C The fact that the collector still registers some current for magnitude values of AV that are not too large shows that the electrons must be emitted from the metal anode surface with some amount of kinetic energy KE Potential Page 1 of 1 Electrostatic Potential Difference Before proceeding with our discussion of the photoelectric effect it will be important for us to say a few words about the voltage AV mentioned in the previous section This voltage also called the electrostatic potential difference will be discussed further when we start studying electronics later on in this course For the time being suf ce it to say that the voltage AV tells us about how much work it takes to move a charge from one point to another Remember that charge can be either positive or negative Since electrons are moving from the emitter to the collector in Fig 71 we are interested in the work it takes to move an electron from E to C An electron is a particle which has a negative charge The magnitude of the charge on the electron is denoted by the symbol e and has the value 6 16 x 10 19 C where the unit C stands for coulomb which is the MKS unit of charge The charge on the electron is negative this value e By de nition the electrostatic potential difference between two points such as E and C in Fig 71 or voltage times the charge on a particle gives us the amount of work it will take MS to move the particle from the one point to the other If the potential difference our source voltage in Fig 71 is AV and if the charge is an electron having a charge 6 then the de nition of voltage tells us that the workdone by us is given by Work done byes WeAV 71a The work required to move a charge from one point to another is a result of an electric field in the region being traversed by the charge Remember that an electric eld exerts a force on a charge It then follows that the voltage AV is also a result of this electric eld The work done by the electric eld is negative the work done by us since the work that we do is simply combating the work being done by the electric eld Work done by Electric Field W 6 av 7 1b It is very important when dealing with work done on a charge to keep in mind which work we are interested in Results 1 Page 1 of 2 Results 1 Some interesting and perplexing results were obtained from a study of the photoelectric effect First for a given frequency f and intensity I of the incident light it was found that for positive values of AV the current quickly reached a constant value the saturation value meaning that all electrons emitted by the surface were collected at C When AV was made negative however it was found that the photocurrent as measured by the ammeter A did not immediately drop to zero but rather reached zero when the source voltage reached some value AVO called the stopping potential At this potential difference all electrons being emitted by the emitter were stopped from reaching the collector A graph of the current I as a function of the source voltage AV is shown below I It Saturation Current Stepping Potential AVE 0 av Figure 72 By de nition of the electrostatic potential difference or voltage given in the previous section the work done on the electrons moving to the collector by the electric eld associated with the stopping potential AV AV 0 is equal to from Eq 71b W 6 AV eAVO eAV0 From the work energy theorem net work equals the change in kinetic energvwe then get that eAVOKEf KE1 72 or since the nal kinetic energy equals zero for electrons that are stopped edit 0 RE 73 It thus follows that KEi e AV 0 Since the stopping potential difference is the value of AV that just barely stops all emitted electrons from reaching the collector it must follow that the kinetic energy of the electronsquotmt barely being stopped is the maximum kinetic energy of all emitted electrons Some electrons may be stopped with voltages between 0 Results 1 Page 2 of 2 and 41 V 0 but only the most energetic ones are stopped by the largest negative voltage AVO It then follows from the above that the maximum kinetic energy emitted by a metal surface which has a stopping potential of AVG is given by KEN 6151504 3914 Results 2 Page 1 of 1 Results 2 It was further found that the maximum kinetic energy of emission of the photoelectrons did not depend on the intensity of the incident light increasing the energy ow into the metal surface did not change the maximum kinetic energy of the emitted electronsl but increased linearly with the frequency of the incident radiation f The frequency of incident radiation j below which no photoelectrons are emitted is called the cuto frequency This is the frequency at which the maximum kinetic energy of the emitted electrons equals zero See the graph in Fig 73 below Ki 4 13 SDquot L C N quot 31 1 Figure 73 Einstein Page 1 of 2 Einstein39s Contribution After Planck demonstrated his derivation of the emission spectrum for blackbodies Einstein immediately saw a way to explain the photoelectric effect He said that the energy of light is quantized into packets later called photons whose energy is proportional to the frequency of the radiation where the constant of proportionality is Planck s constant h E hf E The Einstein Relation 75 Recall that h 6 626 x 10 34 J S When light interacts with matter or more speci cally with the electrons in the atoms making up matter the electrons can only absorb one photon s energy at a time This explains why increasing the intensity of the light while holding the li ght s frequency a constant did not increase the kinetic energy of the emitted electrons Also Einstein suggested that the electrons are bound within the material and that it takes a minimum amount of work w for the electrons to be released from the atoms making up the material under consideration the material emitting the electronswthe surface of our emitter in the case of Fig 71 This minimum work for a given material is called the work function for that material w In addition once the electrons have been released from the atoms in the material they may lose additional energy due to collisions with atoms on their way out of the material Einstein claimed that for a given photon energy absorbed by an electron E hf some of the energy goes into removing the electron from its binding atom some of it goes into getting the electron out of the rest of the material and the remainder goes into the kinetic energy of the escaped electron Photon Energy energy to release electron om atom energy to get electron out of metal KE or rearranging KE photon energy energy to release electron from atom energy to remove electron from metal The maximum kinetic energy of the emitted electrons is thus that kinetic energy when the electrons are released from atoms that are at the metal s surface because then there is no extra energy needed to remove the electrons from the metal once they ve been removed from their host atoms KEmax photon energy energy to release electronfrom atom In equation form this becomes KB hf w 75 max Einstein Page 2 of 2 where w is the work function for the metal under consideration The value of Planck s constant k obtained from an examination of the experimental results for the photoelectric effect agreed to within uncertainties with the value obtained by Planck in his study of the blackbody radiation spectrum This was a good independent veri cation of the idea that light acts like particles photons when interacting with matter See the tables of thsicgl constants for values of ze work function for various metals ElectronVolt Page 1 of 1 Units The ElectronVolt For the small energies usually encountered in photoelectriceffect problems a convenient and often used unit is the electron volt eV One electronwolt is de ned to be the amount of energy acquired by an electron as it is accelerated by the electric eld associated with a potential difference of one volt The conversion factor to MKS units is 1eV150x103919J 77 Example 71 Page 1 of 1 Example 71 A photoelectriceffect apparatus is constructed using zinc for the emitter electrode The work function for zinc is 424 eV Light of frequency 35 x 1015 Hz and intensity 27 x 10 3 Wm2 is incident on the emitter which has an area of 18 cm2 a What is the energy of a single photon in the incident light b How much energy does it take to remove one photoelectron from a zinc atom in the emitter 0 Assuming that all of the incident light hits the zinc emitter how many photons are incident on the emitter per second 1 What is the cutoff frequency of the incident radiation f0 below which no photoelectrons are emitted by the zinc emitter e What is the maximum kinetic energy of the photoelectrons emitted from the zinc D What is the stopping potential for the electrons emitted from the zinc electrode Answers a 23 x 10 131 145 eV b 424 eV c 21 x 1017 d 102 x1015 Hz e162x10quot18J101 eV n101v finisher 0171 Page 1 of3 Solution to Example 71 A photoelectriceffect apparatus is constructed using zinc for the emitter electrode The work function for zinc is 424 eV Light of frequency 35 x 10 15 Hz and intensity 27 x 10 3 Wm2 is incident on the emitter which has an area of 18 cm2 We are given that the work function for zinc is w 424 eV 678 x 1019 J using equation 77 the frequency of the incident light is f 35 x 1015 Hz and its intensity if I 27 x 103 Wm2 and the area ofthe emitter plate is A 18 em2 18 x 10 4 m2 watch this area conversionl a What is the energy of a single photon in the incident light From the Einstein relation we have that the energy of a photon of frequency f is E hf23x 10 18J 145 eV photon where h 6626 x 10734 I s is Planck s constant b How much energy does it take to remove one photoelectron from a zinc atom in the emitter The amount of energy it takes to remove an electron from a zinc atom in the emitter is the meaning of the work function for zinc which is given in the problem statement to be w 424 eV 0 Assuming that all of the incident light hits the zinc emitter how many photons are incident on the emitter per second Let the number of photons hitting the emitter each second be denoted N We are given the intensity of the incident light which involves the amount of energy hitting the emitter plate each second If all of this energy is in the form of photons of frequency f and energy B then the energy E hitting the plate each second must be equal to E N E photon photon It therefore follows that I Energyr NEphoton time area t A or solving for N and using t 1 s Ar photon N 21x1017 Sol7l Page 2 of 3 d What is the cutoff frequency of the incident radiation below which no 0 photoelectrons are emitted by the zinc emitter The easiest electrons to remove from a metal are the electrons at the very top surface of the metalwthese electrons only need be removed from the atoms at the surface and do not have to work their way through the bulk of the metal to escape In general the freed electrons escape with some excess kinetic energy It then follows that the least amount of energy that can be given to the emitter to free electrons is the energy required to remove an electron from a zinc atom with no extra energy left over for kinetic energy This energy is just the work function of the metal as given in part b The frequency of this minimum amount of energy is the cut0quot frequency f0 It then results that w hfo or fo111102x1015Hz e What is the maximum kinetic energy of the photoelectrons emitted from the zinc The electrons are given an amount of energy by the photons equal to Ephoton This energy goes into freeing the electrons from the atoms the work function into moving them through the bulk of the emitter metal and then nally into kinetic energy once the electron is freed from the metal the maximum kinetic energy occurs when the electron is freed at the surface of the metal so that no extra energy is lost trying to get the electron to the surface The maximum kinetic energy of the freed electrons is therefore just the energy of the photon which is then given to the electron that absorbs the photon minus the energy it takes to remove the electron from the atom the work function In symbols this becomes KE E w162x10 18J101eV max photon i What is the stopping potential for the electrons emitted from the zinc electrode The stopping potential for the electrons given off by the zinc the socalled photoelectrons is just the voltage required to stop the most energetic electrons This is just basically a statement of the conservation of energy in electronic terms The stopping potential AVO can be obtained from the equation Kme e Av0 where e is just the magnitude of the charge on the electron e 16 x 10 19 C Coulombs Thus avo5 1oiv 3 Note that the value of this answer is the same as that for the maximum kinetic energy of the emitted electrons in units of electronvolts eV This is always the way this will turn out and explains the usefulness of the electron volt as a unit Sample Quiz 7 Page 1 of 1 Sample Quiz 7 1 We have begun studies of blackbody radiation and now the photoelectric effect These are topics in what is known as a analytical chemistry b modern physics c thermal physics d solid state physics e none of these 2 The photoelectric effect is an effect in which electrons are a absorbed by atoms in a gas b moving through space between galaxies c annihilated by antielectrons d emitted from metal surfaces by light e absorbed into crevices in solid material 3 A quantity associated with the electrostatic potential difference is a work b momentum c circular motion d force e velocity 4 The maximum kinetic energy of photoelectrons is the kinetic energy of electrons a after they collide with gas atoms b emitted from very near the surface of a metal c after they collide with large nuclei d deep within large atomic orbitals e none of these 5 A name that has entered into both our study of blackbody radiation and the photoelectric effect is a Bohr b Wien 0 Stefan d Boltzmann e Einstein An ewe rs Homework 7 Page 1 of l Homework 7 Some work functions Sodium 23 eV Aluminum 42 eV Lithium 23 eV Zinc 424 eV Platinum 62 eV Potassium 226 eV 1 Light of wavelength 450 nm is incident on a sodium emitter in a photoelectric effect apparatus a What is the cutoff frequency for the emitter b What is the maximum kinetic energy of the photoelectrons emitted by the surface c What is the stopping potential for the emitted electrons 2 a What is the cutoff frequency for a clean aluminum surface b To what portion of the electromagnetic spectrum does this frequency correspond 3 A beam of blue light of wavelength 500 nm has an intensity of 170 Wmz The light shines onto a clean plate of potassium of area 12 cm2 Assume that all of the incident radiation is incident on the metal plate a What is the cutoff frequency for the potassium surface b How much energy does it take to remove an electron from a potassium atom within the plate c What is the maximum kinetic energy of the photoelectrons emitted by the potassium d What is the stopping potential for the emitted photoelectrons e Assuming that each photon that is incident on the potassium plate results in an emitted photoelectron how many photoelectrons are emitted from the potassium plate per second rim saw 9 re HWAns7 Page 1 of 1 Answers to Homework 7 1 a 555 x1014 Hz b 737 x 1020 J 046 eV c 046 V 2 a 101 x 1015 Hz b ultraviolet 3 a546x1014 Hz b362x10 19J 0356X10 20J022 eV d 022v e 513 x1017