1. The Photoelectric Effect Einstein’s Triumph Graphics courtesy of Physics 2000, University of Colorado

2. Quantum Theory – Early Experiments Cathode rays discovered in late 19 th century Charge to mass ratio e/m measured = E/B 2 r End of tube near + terminal(anode) glowed evB = mv 2 /r eE = evB etc

3. “Discovery” of Electron Cathode rays became known as electrons Most believed they were atoms J.J. Thompson believed they were parts of atoms Millikin measured charge on electron

4. Millikin Oil Drop Experiment Shows that electron charge is quantized. q e = integer x 1.6 x 10 -19 C

5. Closer Look at Millikan Oil Drop Millikan put a charge on a tiny drop of oil and measured how strong an applied electric field had to be in order to stop the oil drop from falling.

6. Planck Distribution Wavelength distribution of “blackbody” radiation

7. Planck E = hf Planck postulated that the radiators or oscillators can only emit electromagnetic radiation in finite amounts of energy. At a given temperature T, there is then not enough thermal energy available to create and emit many large radiation quanta. More large energy quanta can be emitted, however, when the temperature is raised.

8. Do You Know How a Solar Cell Works? Light produces electricity, right? The Photoelectric Effect, first explained correctly by Einstein in 1905 How?

9. Basic Info When light of high enough frequency strikes a metal, electrons are given off

10. Apparatus

11. Simulations of Photoelectric Effect Link #1 Link #2

12. Planck’s E = hf Called quantum hypothesis Needed to explain spectrum of light given off by hot objects Main idea: energy of atomic oscillators is not continuous but finite number of discrete amounts each related to frequency of oscillation by E =nhf h = 6.63 x 10 -34 J-s (Planck’s Constant) Photons act like particles

13. Photoelectric Effect Apparatus When light hits cathode(-) current flows Electrons move toward anode (+) If battery is reversed, electrons can be stopped KE max = eV 0 V 0 is stopping voltage o Light

14. What Wave Theory Predicts If light made brighter –#electrons increases –Maximum KE increases If change frequency, no effect on KE of electrons

15. WRONG! Sorry Maxwell

16. What Photon Theory Predicts Increasing brightness means more photons, not more energy per photon Increasing frequency increases KE max (E=hf) Decreasing frequency below “cutoff” could mean no electrons ejected

17. Two Theories Animated Link

18. Now for the Math… Let hf be incoming energy of the photon Let W 0 be the minimum energy required to eject out through the surface(work function) KE max is the maximum energy of the ejected electron then hf = KE max + W 0 by conservation of energy in a collision

19. How to Analyze KE max can be easily determined by measuring the stopping potential KE max =eV 0 ( =q e V) So let’s plot KE max vs. f

20. What Happens When Light Frequency Increases? KE max = hf - W 0 KE max f f0f0 W0W0 f 0 is called threshold frequency h is the slope

21. Meaning of Threshold(Cutoff) Frequency When f is less than f 0 KE max is negative. There can be no photocurrent The bigger f, the bigger is KE max At cutoff frequency f 0 hf 0 = W 0

22. Problems 1.What stopping voltage is required to stop an electron with KE of 1electron volt? 2.A stopping voltage of 2.5 volts is just enough to stop all photocurrent. What is KE max ? Ans. 1 volt Ans. 2.5 eV

23. Finding Photon Energy What is the energy of a photon of blue light with = 450 nm ? HINT: First find f f = c/ E = hf = hc/  hc/  x10 -34 J-s)(3.0x10 8 m/s)/(4.5 x 10 -7 m) = 4.42x10 -19 J/(1.6)x10 -19 J/eV = 2.76 eV

24. Finding KE max What is the maximum kinetic energy of electrons ejected from a sodium surface whose work function is W 0 = 2.28 eV when illuminated by light of wavelength 410nm? hf = hc/ = 4.85x10 -19 J or 3.03 eV (1243/410) KE max = hf - W 0 = 3.03 eV – 2.28 eV = 0.75 eV

25. Finding Cutoff Frequency or Wavelength What is the cutoff frequency for sodium? What is the longest wavelength for a photo current to flow? hf 0 = W 0 = 2.28 eV = 3.65 x 10 -19 J f 0 = 3.65 x 10 -19 J / 6.63 x 10 -34 J-s = 5.5 x 10 14 Hz 0 = c/f 0 = 3.0 x 10 8 m/s /5.5 x 10 14 Hz = 545 nm ShortcutShortcut-click h =6.63 x 10 -34 J-s W 0 = 2.28 eV

26. Using 1243 Rule The wavelength corresponding to the work function is just 1243/2.28 eV = 545 nm

27. How Can We Measure h Using the Photoelectric Effect? Plot KE max as a function of frequency h is the slope KE max = hf - W 0