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

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.

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

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. 1 eV = 1.60 x 10-19 J 1 keV = 1.6 x 10-16 J 1 MeV = 1.6 x 10-13 J

First we find f from wave equation: c = fl 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 E = 2.24 eV Or Since 1 eV = 1.60 x 10-19 J

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 l is in nm, the energy in eV is found from: Verify the answer in Example 1 . . .

The Photo-Electric Effect Cathode Anode Incident light Ammeter + - A C 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.

Photo-Electric Equation Cathode Anode Incident light Ammeter + - A C Threshold wavelength lo 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 ½mv2 of the emitted electrons.

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? A l = 600 nm ; K = 2.76 eV – 2.07 eV K = 0.690 eV Or K = 1.10 x 10-19 J

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

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

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. fo Stopping potential Frequency V Finding h constant y x Slope Note that the x-intercept fo is the threshold frequency.

h = e(slope) = (1.6 x 10-19C)(4.13 x 10-15 V/Hz) 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? fo Stopping potential Frequency V y x Slope 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

Photoelectric Equation: 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? A Cathode Anode Incident light + - V Photoelectric Equation: W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J Stopping potential: Vo = 0.800 V

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

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. Wavelength of a photon: All objects, not just EM waves, have wavelengths which can be found from their momentum de Broglie Wavelength:

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 Momentum from K:

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

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

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

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

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

CONCLUSION: Chapter 38B Quantum Physics