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

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Transcription:

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 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 s Constant In his studies of black-body 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 of tiny bundles of of energy called photons,, each having a well- defined quantum of of energy. Photon E = hf

Energy in Electron-volts Photon energies are so small that the energy is better expressed in terms of the electron-volt volt. One electron-volt (ev) is is the energy of of an electron when accelerated through a potential difference of 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

Example 1: What is the energy of a photon of yellow-green light (( = 555 nm)? First we find f from wave equation: c = f E c f ; E hf hc 34 8 (6.626 x 10 J s)(3 x 10 m/s) -9 555 x 10 m E = 3.58 x 10-19 -19 J Or E = 2.24 ev 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) E hc 9 hc(1 x 10 nm/m) -19 (1.6 x 10 J/eV) -19 (in Joules) ; 1 ev 1.60 x 10 J E(in ev) If is in nm,, the energy in ev is found from: E 1240 Verify the answer in Example 1...

The Photo-Electric Effect Cathode C - Incident light + Anode A Ammeter A 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 is a certain threshold energy, called the work function W, that must be overcome before any electrons can be emitted.

Photo-Electric Equation Cathode C Incident light Anode A hc 1 E W 2 mv 2 - + Ammeter A Threshold wavelength W hc 0 The conservation of of energy demands that the energy of of the incoming light hc/ be equal to to the work function W of of the surface plus the kinetic energy ½mv 2 of 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? hc W K hc hc K 0 = 600 nm A K hc hc 0 1240 1240 450 nm 600 nm ; K = 2.76 ev 2.07 ev K = 0.690 ev Or K = 1.10 x 10-19 -19 J

Stopping Potential A potentiometer is used to vary to the voltage V between the electrodes. Cathode Incident light Anode The stopping potential is that voltage V o that just stops the emission of electrons, and thus equals their original K.E. Photoelectric equation: E hf W ev 0 V Potentiometer V 0 A + - K max = ev o h e f W e

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

Finding Planck s 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. V 0 h e Slope Note that the x-intercept x f o is the threshold frequency. f W e h e V Finding h constant f o Stopping potential Slope x y Frequency

Example 3: In an experiment to determine Planck s 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? V f o Stopping potential Slope x y Frequency V Slope 0 h e h e f W e -15 4.13 x 10 V/Hz h = e(slope) e = (1.6 x 10-19 C)(4.13 x 10-15 V/Hz) Experimental Planck s h = 6.61 x 10-34 -34 J/Hz

Example 4: The threshold frequency for a given surface is 1.09 x 10 15 Hz.. What is the stopping potential for incident light whose photon energy is 8.48 x 10-19 J? Photoelectric Equation: E hf W ev 0 ev E W; W hf 0 0 W = (6.63 x 10-34 Js)(1.09 x 10 15 Hz) =7.20 x 10-19 19 J -19-19 -19 ev0 8.48 x 10 J 7.20 x 10 J 1.28 x 10 J -19 1.28 x 10 J 0-19 V 1.6 x 10 J Incident light Cathode Anode V Stopping potential: A + - V o = 0.800 V

Total Relativistic Energy Recall that the formula for the relativistic total energy was given by: E ( m c ) p c Total Energy, E 2 2 2 0 For a particle with zero momentum p = 0 0: A light photon has m o = 0, but it does have momentum p: E = m o c 2 E = pc

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

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 = ½mv 2 ; p = mv Multiply first Equation by m: mk = ½m 2 v 2 = ½p 2 Momentum from K: p 2 mk

Example 5: What is the de Broglie wavelength of a 90-eV electron? ( (m e = 9.1 x 10-31 kg.) K -19 1.6 x 10 J -17 90 ev 1.44 x 10 J 1 ev Next, we find momentum from the kinetic energy: p 2 - e - 90 ev mk p -31-17 2(9.1 x 10 kg)(1.44 x 10 J) p = 5.12 x 10-24 kg m/s h p h mv -34 h 6.23 x 10 J -24 = 0.122 nm p 5.12 x 10 kg m/s

Summary Apparently, light consists of of tiny bundles of of energy called photons,, each having a well- defined quantum of of energy. Photon E = hf Planck s Equation: E = hf (h = 6.626 x 10-34 J s) The Electron-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

Summary (Cont.) Cathode C Incident light Anode A hc 1 E W 2 mv 2 - + Ammeter A Threshold wavelength W hc 0 If is in nm,, the energy in ev is found from: Wavelength in nm; Energy in ev E 1240

Summary (Cont.) Planck s s Experiment: Cathode V Potentiometer Incident light A + - K max = ev o Anode V V 0 f o h e Slope Stopping potential Slope x Frequency f W e h e y

Summary (Cont.) Quantum physics works for waves or or particles: For a particle with zero momentum p = 0: A light photon has m o = 0, but it does have momentum p: E = m o c 2 E = pc Wavelength of a photon: h p de Broglie Wavelength: h mv

CONCLUSION: Chapter 38B Quantum Physics