Chapter 38B - Quantum Physics

1. Chapter 38B - Quantum Physics A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

2. Objectives: After completing this module, you should be able to: • Discuss the meaning of quantum physics and Planck’s constant for the description of matter in terms of waves or particles. • Demonstrate your understanding of the photoelectric effect, the stopping potential, and the deBroglie wavelength. • Explain and solve problems similar to those presented in this unit.

3. Planck’s Equation: E = hf (h = 6.626 x 10-34 J s) Photon E = hf Plank’s Constant In his studies of black-body radiation, Maxwell Planck discovered that electromagnetic energy is emitted or absorbed in discrete quantities. Apparently, light consists of tiny bundles of energy called photons, each having a well-defined quantum of energy.

4. 1 eV = 1.60 x 10-19 J 1 keV = 1.6 x 10-16 J 1 MeV = 1.6 x 10-13 J Energy in Electron-volts Photon energies are so small that the energy is better expressed in terms of the electron-volt. One electron-volt (eV) is the energy of an electron when accelerated through a potential difference of one volt.

5. E = 2.24 eV Or Since 1 eV = 1.60 x 10-19 J Example 1:What is the energy of a photon of yellow-green light (l = 555 nm)? First we find f from wave equation: c = fl E = 3.58 x 10-19 J

6. Useful Energy Conversion Since light is often described by its wavelength in nanometers (nm) and its energy E is given in eV, a conversion formula is useful. (1 nm = 1 x 10-9 m) If lis in nm, the energy in eV is found from: Verify the answer in Example 1 . . .

7. Incident light Cathode Anode A C Ammeter A + - The Photo-Electric Effect When light shines on the cathode C of a photocell, electrons are ejected from A and attracted by the positive potential due to battery. There is a certain threshold energy, called the work function W, that must be overcome before any electrons can be emitted.

8. Incident light Cathode Anode A C Ammeter Threshold wavelength lo A + - Photo-Electric Equation The conservation of energy demands that the energy of the incoming light hc/l be equal to the work function W of the surface plus the kinetic energy ½mv2of the emitted electrons.

9. l = 600 nm A K = 1.10 x 10-19 J Or Example 2:The threshold wavelength of light for a given surface is 600 nm. What is the kinetic energy of emitted electrons if light of wavelength 450 nm shines on the metal? ; K = 2.76 eV – 2.07 eV K = 0.690 eV

10. Incident light Cathode Anode The stopping potential is that voltage Vo that just stops the emission of electrons, and thus equals their original K.E. V A + - Potentiometer Photoelectric equation: Stopping Potential A potentiometer is used to vary to the voltage V between the electrodes. Kmax = eVo

11. The slope of a line: y Slope xo x y x Slope of a Straight Line (Review) The general equation for a straight line is: y = mx + b The x-interceptxooccurs when line crosses x axis or when y = 0. The slope of the line is the rise over the run:

12. Finding h constant Stopping potential V Slope y x fo Frequency Finding Planck’s Constant, h Using the apparatus on the previous slide, we determine the stopping potential for a number of incident light frequencies, then plot a graph. Note that the x-intercept fo is the threshold frequency.

13. Stopping potential V Slope y fo x Frequency Example 3:In an experiment to determine Planck’s constant, a plot of stopping potential versus frequency is made. The slope of the curve is 4.13 x 10-15 V/Hz. What is Planck’s constant? h = e(slope) = (1.6 x 10-19C)(4.13 x 10-15 V/Hz) Experimental Planck’s h = 6.61 x 10-34 J/Hz

14. Incident light Cathode Anode A V + - Stopping potential: Vo= 0.800 V Example 4:The threshold frequency for a given surface is 1.09 x 1015 Hz. What is the stopping potential for incident light whose photon energy is 8.48 x 10-19 J? Photoelectric Equation: W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J

15. Total Energy, E Total Relativistic Energy Recall that the formula for the relativistic total energy was given by: For a particle with zero momentump = 0: E =moc2 A light photon has mo= 0, but it does have momentum p: E =pc

16. Wavelength of a photon: de Broglie Wavelength: Waves and Particles We know that light behaves as both a wave and a particle. The rest mass of a photon is zero, and its wavelength can be found from momentum. All objects, not just EM waves, have wavelengths which can be found from their momentum

17. Momentum from K: Finding Momentum from K.E. In working with particles of momentum p = mv, it is often necessary to find the momentum from the given kinetic energy K. Recall the formulas: K = ½mv2 ; p = mv Multiply first Equation by m: mK =½m2v2= ½p2

18. - e- 90 eV Example 5:What is the de Broglie wavelength of a 90-eV electron? (me = 9.1 x 10-31 kg.) Next, we find momentum from the kinetic energy: p = 5.12x 10-24 kg m/s l = 0.122 nm

19. Planck’s Equation: E = hf (h = 6.626 x 10-34 J s) Photon 1 eV = 1.60 x 10-19 J The Electron-volt: E = hf 1 MeV = 1.6 x 10-13 J 1 keV = 1.6 x 10-16 J Summary Apparently, light consists of tiny bundles of energy called photons, each having a well-defined quantum of energy.

20. Incident light Cathode Anode A C Ammeter Threshold wavelength lo A + - Summary (Cont.) If lis in nm, the energy in eV is found from: Wavelength in nm; Energy in eV

21. Stopping potential V Slope y fo x Frequency Summary (Cont.) Planck’s Experiment: Incident light Cathode Anode V A + - Potentiometer Kmax = eVo

22. Wavelength of a photon: de Broglie Wavelength: Summary (Cont.) Quantum physics works for waves or particles: For a particle with zero momentump = 0: E =moc2 A light photon has mo = 0, but it does have momentum p: E =pc

23. CONCLUSION: Chapter 38BQuantum Physics