Chapter 9: Quantization of Light

Transcription

1 Chapter 9: Quantizatin Light 9.1 Planck s Quantum Thery Distinguish between Planck s quantum thery and classical thery energy The undatin the Planck s quantum thery is a thery black bdy radiatin. Black bdy is deined as an ideal system that absrbs the entire radiatin incident n it. The electrmagnetic radiatin emitted by the black bdy is called black bdy radiatin. Frm the black bdy experiment, the distributin energy in black bdy depends nly n the temperature. I the temperature increases thus the energy the black bdy increases and vice versa. The spectrum electrmagnetic radiatin emitted by the black bdy (experimental result) is shwn in igure Failed t explain the shape The Rayleigh-Jeans and Wien s theries ailed t it the experimental curve because this tw theries based n classical ideas which are: i. Energy the EM radiatin des nt depend n its requency r wavelength. ii. Energy the EM radiatin is cntinuusly. In 1900, Max Planck prpsed his thery that is it with the experimental curve in igure abve at all wavelengths knwn as Planck s quantum thery. The assumptins made by Planck in his thery are: i. The EM. radiatin emitted r absrb by the black bdy nt in a cntinuus stream waves but in discrete little bundles (separate) packets energy r quanta, knwn as phtn. This means the energy e.m. radiatin is quantized, nt all values energy are pssible ii. The energy size the radiatin depends n its requency. 1

2 Accrding t Planck s assumptins, the quantum E the energy r radiatin requency is given by E h Since the speed electrmagnetic wave in a vacuum is c, then equatin can als be written as E hc Planck s quantum thery Use Einstein s equatin r a phtn energy In 1905, Albert Einstein prpsed that light cmes in bundle energy (light is transmitted as tiny particles), called phtns. Phtn is deined as a particle with zer mass cnsisting a quantum electrmagnetic radiatin where its energy is cncentrated. (Quantum means ixed amunt ) In equatin rm, phtn energy (energy phtn) is E h. Unit phtn energy is Jule (J) r electrn-vlt (ev). The electrn-vlt (ev) is a unit energy that can be deined as the kinetic energy gained by an electrn in being accelerated by a ptential dierence (vltage) 1 vlt. 1eV J Phtns travel at the speed light in a vacuum. They are required t explain the phtelectric eect and ther phenmena that require light t have particle prperty. Example Questin A phtn the green light has a wavelength 740 nm. Calculate a. the phtn s requency b. the phtn s energy in jule and electrn-vlt (Given the speed light in the vacuum, c = m s 1 and Planck s cnstant, h = J s) Slutin 2

3 9.2 The Phtelectric Eect Explain the phenmenn phtelectric eect The phtelectric eect is deined as the emissin electrn rm the surace a metal when the EM radiatin (light) higher requency strikes its surace. Figure belw shws the emissin the electrn rm the surace the metal ater shining by the light. Phtelectrn is deined as an electrn emitted rm the surace the metal when EM radiatin (light) strikes its surace Describe and sketch diagram the phtelectric eect experiment set-up When a mnchrmatic light knwn requency (r wavelength) shines n the cathde, phtelectrns are emitted. These phtelectrns are attracted t the ande and give rise t a phtelectric current r phtcurrent I which is detected by the galvanmeter. When the psitive vltage (ptential dierence) is increased, mre phtelectrns reach the ande, hence the phtelectric current als increase. As psitive vltage becmes suiciently large, the phtelectric current reaches a maximum cnstant value I m, called saturatin current (the maximum cnstant value phtcurrent when all the phtelectrns have reached the ande). I the psitive vltage is gradually decreased, the phtelectric current I als decrease slwly. Even at zer vltage there are still sme phtelectrns with suicient energy reach the ande and the phtelectric current lws is I 0. 3

4 Reversing pwer supply terminal When the vltage is made negative by reversing the pwer supply terminal as shwn in igure belw, the phtelectric current decreases even urther t very lw values since mst phtelectrns are repelled by ande which is nw negative. As the ptential the ande becmes mre negative, less phtelectrns reach the ande thus the phtelectric current drps until its value equals zer (n phtelectrns have suicient kinetic energy t reach the cllectr). The electric ptential at this mment is called stpping ptential (vltage) V s (the minimum value reverse ptential (vltage) when there are n phtelectrns reaching the ande). By using cnservatin energy : (lss KE phtelectrn = gain in PE) K U max mvmax ev S At stpping vltage The variatin phtelectric current I as a unctin the vltage V can be shwn thrugh the graph in igure belw. 4

5 9.2.3 Deine threshld requency, wrk unctin and stpping ptential Use Einstein s phtelectrn equatin Accrding t Einstein s thery, an electrn is ejected/ emitted rm the target metal by a cllisin with a single phtn. In this prcess, all the phtn energy (E = h ) is transerred t the electrn n the surace metal target. Since electrns are held in the metal by attractive rces, sme minimum energy, W (wrk unctin, which is n the rder a ew electrn vlts r mst metal) is required just enugh t get an electrn ut thrugh the surace. I the requency the incming light is s lw that is h < W, then the phtn will nt have enugh energy t eject any electrn at all. I h > W, then electrn will be ejected and energy will be cnserved (the excess energy appears as kinetic energy the ejected electrn). I h = W, then electrn will be ejected but the kinetic energy is equal t zer. This is summed up by Einstein s phtelectric equatin: h 1 2 mv max E K max W r W h Einstein s phtelectric equatin ev s W 5

6 Wrk unctin W 0 a metal Is deined as the minimum energy e.m. radiatin required t emit an electrn rm the surace the metal. It depends n the metal used. Equatin : W E min h is knwn as threshld requency Is deined as the minimum requency EM radiatin required t emit an electrn rm the surace the metal. I the requency the incident radiatin is less than the threshld requency ( < 0 ) then electrns wuld nt be remved rm the metal surace. Since c, then c is knwn as threshld wavelength Example Is deined as the maximum wavelength EM radiatin required t emit an electrn rm the surace the metal. I the wavelength the incident radiatin is greater than the threshld wavelength ( > ) then electrns wuld nt be remved rm the metal surace. Questin Slutin The wrk unctin r a silver surace is W = 4.74 ev. Calculate the a) minimum requency that light must have t eject electrns rm the surace b) maximum wavelength that light must have t eject electrns rm the surace (Given c = m s -1, h = J s, 1 ev= J, m e = kg, e = C) 6

7 Questin What is the maximum kinetic energy electrns ejected rm calcium by 420 nm vilet light, given the wrk unctin r calcium metal is 2.71 ev? (Given c = m s -1, h = J s, 1 ev= J, m e = kg, e = C) Slutin Sdium has a wrk unctin 2.30 ev. Calculate a. its threshld requency, b. the maximum speed the phtelectrns prduced when the sdium is illuminated by light wavelength 500 nm, c. the stpping ptential with light this wavelength In a phtelectric eect experiment it is bserved that n current lws unless the wavelength is less than 570 nm. Calculate a. the wrk unctin this material in electrn-vlts b. the stpping vltage required i light wavelength 400 nm is used Exercise Questin The energy a phtn rm an electrmagnetic wave is 2.25 ev a. Calculate its wavelength. b. I this electrmagnetic wave shines n a metal, phtelectrns are emitted with a maximum kinetic energy 1.10 ev. Calculate the wrk unctin this metal in jules. Answer: 553 nm, J In an experiment phtelectric eect, n current lws thrugh the circuit when the vltage acrss the ande and cathde is V. Calculate a. the wrk unctin b. the threshld wavelength the metal (cathde) I it is illuminated by ultravilet radiatin requency 1.70 x Hz. Answer: J, m 7

8 9.2.4 Explain by using graph and equatins the bservatins phtelectric eect experiment in terms the dependence i. kinetic energy phtelectrn n the requency light ii. phtelectric current n intensity incident light iii. wrk unctin and threshld requency n the types metal surace Generally, Einstein s phtelectric equatin: E K max W K K max max E W h W y mx c Kmax ev s V s y h W e e mx c K max V s Gradient = h Gradient = W W, K max, V s Variatin stpping vltage V s with requency the radiatin r dierent metals but the intensity is ixed: V s W 01 W 02 W 03 Since, Dierent threshld requency r dierent metal 8

9 Variatin phtelectric current I with vltage V r the radiatin dierent intensities but its requency and metal are ixed. When intensity is increased, the maximum current attained is higher shwing that mre electrns are emitted. Light intensity number phtns V s remains the same shws that the K max phtelectrn independent intensity light Extra Knwledge Classical physics Quantum physics Energy Light intensity Time Area Since Light intensity light intensity number phtn Number phtn Time Area Light intensity, number phtns, number electrns, current. I light intensity, phtelectric current Variatin phtelectric current I with vltage V r the radiatin dierent requencies but its intensity and metal are ixed. V s h e W e, V s V V s2 s1 2 1 Variatin phtelectric current I with vltage V r the dierent metals but the intensity and requency the radiatin are ixed. V s h e W e, V s W V V s2 s1 W02 W 01 9

10 Example Questin Slutin Use the graph abve t ind the value a. wrk unctin and b. the threshld wavelength Based n the graph, r the light requency 6.00 x Hz, calculate a. the threshld requency b. the maximum kinetic energy the phtelectrn c. the maximum velcity the phtelectrn V s W 02 W Explain why a. the graphs are parallel b. the visible light cause phtemissin rm caesium but nt rm zinc 10

11 Exercise Questin In an experiment n the phtelectric eect, the llwing data were cllected. Wavelength EM radiatin, (nm) Stpping ptential, V S (V) a. Calculate the maximum velcity the phtelectrns when the wavelength the incident radiatin is 350 nm. b. Determine the value the Planck cnstant rm the abve data. Answer: 7.73 x 10 5 m s -1, 6.72 x J s A phtcell with cathde and ande made the same metal cnnected in a circuit as shwn in the igure belw. Mnchrmatic light wavelength 365 nm shines n the cathde and the phtcurrent I is measured r varius values vltage V acrss the cathde and ande. The result is shwn in the graph. a. Calculate the maximum kinetic energy the phtelectrn. b. Deduce the wrk unctin the cathde. c. I the experiment is repeated with mnchrmatic light wavelength 313 nm, determine the new intercept with the V-axis r the new graph. Answer: J, J, 1.57 V 11

12 9.2.5 Explain ailure classical thery t justiy phtelectric eect OBSERVATIONS the phtelectric eects experiment 1. Electrns are emitted immediately 2. Stpping ptential des nt depend n the intensity light. 3. Threshld requency light is dierent r dierent target metal. 4. Number electrns emitted the phtelectrn current depends n the intensity light. SUMMARY : Cmparisn between classical physics and quantum physics abut phtelectric eect experiment Feature Classical physics Quantum physics Threshld requency An incident light any requency can eject electrns (des nt has threshld requency/ independent requency), as lng as the beam has suicient intensity. T eject an electrn, the incident light must have a requency greater than a certain minimum value, (threshld requency), n matter hw intense the light. Maximum kinetic energy phtelectrns Depends n the light intensity. Depends nly n the light requency. Emissin phtelectrns There shuld be sme delays t emit electrns rm a metal surace. Electrns are emitted spntaneusly. Energy light Depends n the light intensity. Depends nly n the light requency. Experimental bservatins deviate rm classical predictins based n Maxwell s EM thery. Hence the classical physics cannt explain the phenmenn phtelectric eect. The mdern thery based n Einstein s phtn thery light can explain the phenmenn phtelectric eect. It is because Einstein pstulated that light is quantized and light is emitted, transmitted and reabsrbed as phtns. 12