General Physics (PHY 2140)

Transcription

1 General Physics (PHY 2140) Lecture 27 Modern Physics Quantum Physics Blackbody radiation Plank s hypothesis Chapter 27 1

2 Quantum Physics 2

3 Introduction: Need for Quantum Physics Problems remained from classical mechanics that relativity didn t explain: Blackbody Radiation The electromagnetic radiation emitted by a heated object Photoelectric Effect Emission of electrons by an illuminated metal Spectral Lines Emission of sharp spectral lines by gas atoms in an electric discharge tube 3

4 Development of Quantum Physics 1900 to 1930 Development of ideas of quantum mechanics Also called wave mechanics Highly successful in explaining the behavior of atoms, molecules, and nuclei Quantum Mechanics reduces to classical mechanics when applied to macroscopic systems Involved a large number of physicists Planck introduced basic ideas Mathematical developments and interpretations involved such people as Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac 4

5 26.1 Blackbody Radiation An object at any temperature is known to emit electromagnetic radiation Sometimes called thermal radiation Stefan s Law describes the total power radiated P Stefan s constant = σ AeT 4 emissivity The spectrum of the radiation depends on the temperature and properties of the object 5

6 Blackbody Blackbody is an idealized system that absorbs incident radiation of all wavelengths If it is heated to a certain temperature, it starts radiate electromagnetic waves of all wavelengths Cavity is a good real-life life approximation to a blackbody 6

7 Blackbody Radiation Graph Experimental data for distribution of energy in blackbody radiation As the temperature increases, the total amount of energy increases Shown by the area under the curve As the temperature increases, the peak of the distribution shifts to shorter wavelengths 7

8 Wien s Displacement Law The wavelength of the peak of the blackbody distribution was found to follow Wein s Displacement Law λ max T = x 10-2 m K λ max is the wavelength at the curve s peak T is the absolute temperature of the object emitting the radiation 8

9 The Ultraviolet Catastrophe Classical theory did not match the experimental data At long wavelengths, the match is good Rayleigh-Jeans law 2 πckt P = 4 λ At short wavelengths, classical theory predicted infinite energy At short wavelengths, experiment showed no energy This contradiction is called the ultraviolet catastrophe 9

10 Planck s Resolution Planck hypothesized that the blackbody radiation was produced by resonators Resonators were submicroscopic charged oscillators The resonators could only have discrete energies E n = n h ƒ n is called the quantum number ƒ is the frequency of vibration h is Planck s constant, h=6.626 x J s Key point is quantized energy states 10

11 QUICK QUIZ A photon (quantum of light) is reflected from a mirror. True or false: (a) Because a photon has a zero mass, it does not exert a force on the mirror. (b) Although the photon has energy, it cannot transfer any energy to the surface because it has zero mass. (c) The photon carries momentum, and when it reflects off the mirror, it undergoes a change in momentum and exerts a force on the mirror. (d) Although the photon carries momentum, its change in momentum is zero when it reflects from the mirror, so it cannot exert a force on the mirror. (a) (b) (c) (d) False False True False p = Ft 11

12 27.2 Photoelectric Effect When light is incident on certain metallic surfaces, electrons are emitted from the surface This is called the photoelectric effect The emitted electrons are called photoelectrons The effect was first discovered by Hertz The successful explanation of the effect was given by Einstein in 1905 Received Nobel Prize in 1921 for paper on electromagnetic radiation, of which the photoelectric effect was a part 12

13 Photoelectric Effect Schematic When light strikes E, photoelectrons are emitted Electrons collected at C and passing through the ammeter are a current in the circuit C is maintained at a positive potential by the power supply 13

14 Photoelectric Current/Voltage Graph The current increases with intensity, but reaches a saturation level for large V s No current flows for voltages less than or equal to V s, the stopping potential The stopping potential is independent of the radiation intensity 14

15 Features Not Explained by Classical Physics/Wave Theory No electrons are emitted if the incident light frequency is below some cutoff frequency that is characteristic of the material being illuminated The maximum kinetic energy of the photoelectrons is independent of the light intensity The maximum kinetic energy of the photoelectrons increases with increasing light frequency Electrons are emitted from the surface almost instantaneously, even at low intensities 15

16 Einstein s Explanation A tiny packet of light energy, called a photon,, would be emitted when a quantized oscillator jumped from one energy level to the next lower one Extended Planck s idea of quantization to electromagnetic radiation The photon s energy would be E = hƒ Each photon can give all its energy to an electron in the metal The maximum kinetic energy of the liberated photoelectron is KE = hƒ Φ Φ is called the work function of the metal 16

17 Explanation of Classical Problems The effect is not observed below a certain cutoff frequency since the photon energy must be greater than or equal to the work function Without this, electrons are not emitted, regardless of the intensity of the light The maximum KE depends only on the frequency and the work function, not on the intensity The maximum KE increases with increasing frequency The effect is instantaneous since there is a one-to-one interaction between the photon and the electron 17

18 Verification of Einstein s Theory Experimental observations of a linear relationship between KE and frequency confirm Einstein s theory The x-intercept is the cutoff frequency f c = Φ h 18

19 27.3 Application: Photocells Photocells are an application of the photoelectric effect When light of sufficiently high frequency falls on the cell, a current is produced Examples Streetlights, garage door openers, elevators 19

20 27.4 The Compton Effect Compton directed a beam of x-rays toward a block of graphite He found that the scattered x-rays had a slightly longer wavelength that the incident x-rays This means they also had less energy The amount of energy reduction depended on the angle at which the x-rays were scattered The change in wavelength is called the Compton shift 20

21 Compton Scattering Compton assumed the photons acted like other particles in collisions Energy and momentum were conserved The shift in wavelength is λ = λ λ o = h mec (1 cosθ) Compton wavelength 21

22 Compton Scattering The quantity h/m e c is called the Compton wavelength Compton wavelength = nm Very small compared to visible light The Compton shift depends on the scattering angle and not on the wavelength Experiments confirm the results of Compton scattering and strongly support the photon concept 22

23 Problem: Compton scattering A beam of 0.68-nm photons undergoes Compton scattering from free electrons. What are the energy and momentum of the photons that emerge at a 45 angle with respect to to the incident beam? 23

24 QUICK QUIZ 1 An x-ray photon is scattered by an electron. The frequency of the scattered photon relative to that of the incident photon (a) increases, (b) decreases, or (c) remains the same. (b). Some energy is transferred to the electron in the scattering process. Therefore, the scattered photon must have less energy (and hence, lower frequency) than the incident photon. 24

25 QUICK QUIZ 2 A photon of energy E 0 strikes a free electron, with the scattered photon of energy E moving in the direction opposite that of the incident photon. In this Compton effect interaction, the resulting kinetic energy of the electron is (a) E 0, (b) E, (c) E 0 E, (d) E 0 + E, (e) none of the above. (c). Conservation of energy requires the kinetic energy given to the electron be equal to the difference between the energy of the incident photon and that of the scattered photon. 25

26 QUICK QUIZ 2 A photon of energy E 0 strikes a free electron, with the scattered photon of energy E moving in the direction opposite that of the incident photon. In this Compton effect interaction, the resulting kinetic energy of the electron is (a) E 0, (b) E, (c) E 0 E, (d) E 0 + E, (e) none of the above. (c). Conservation of energy requires the kinetic energy given to the electron be equal to the difference between the energy of the incident photon and that of the scattered photon. 26

27 27.8 Photons and Electromagnetic Waves Light has a dual nature. It exhibits both wave and particle characteristics Applies to all electromagnetic radiation The photoelectric effect and Compton scattering offer evidence for the particle nature of light When light and matter interact, light behaves as if it were composed of particles Interference and diffraction offer evidence of the wave nature of light 27

28 28.9 Wave Properties of Particles In 1924, Louis de Broglie postulated that because photons have wave and particle characteristics, perhaps all forms of matter have both properties Furthermore, the frequency and wavelength of matter waves can be determined The de Broglie wavelength of a particle is The frequency of matter waves is ƒ = E h λ = h mv 28

29 The Davisson-Germer Experiment They scattered low-energy electrons from a nickel target They followed this with extensive diffraction measurements from various materials The wavelength of the electrons calculated from the diffraction data agreed with the expected de Broglie wavelength This confirmed the wave nature of electrons Other experimenters have confirmed the wave nature of other particles 29

30 Review problem: the wavelength of a proton Calculate the de Broglie wavelength for a proton (m p =1.67x10-27 kg ) moving with a speed of 1.00 x 10 7 m/s. 30

31 Calculate the de Broglie wavelength for a proton (m p =1.67x10-27 kg ) moving with a speed of 1.00 x 10 7 m/s. Given: v = 1.0 x 10 7 m/s Given the velocity and a mass of the proton we can compute its wavelength λ = p h m v p Or numerically, Find: λ p =? λ ps ( J s) ( 31 )( kg m s) = = m 31

32 QUICK QUIZ 3 A non-relativistic electron and a non-relativistic proton are moving and have the same de Broglie wavelength. Which of the following are also the same for the two particles: (a) speed, (b) kinetic energy, (c) momentum, (d) frequency? (c). Two particles with the same de Broglie wavelength will have the same momentum p = mv. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p 2 /2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, Equation tells us that the particles do not have the same frequency. 32

33 QUICK QUIZ 2 A non-relativistic electron and a non-relativistic proton are moving and have the same de Broglie wavelength. Which of the following are also the same for the two particles: (a) speed, (b) kinetic energy, (c) momentum, (d) frequency? (c). Two particles with the same de Broglie wavelength will have the same momentum p = mv. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p 2 /2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, Equation tells us that the particles do not have the same frequency. 33

34 The Electron Microscope The electron microscope depends on the wave characteristics of electrons Microscopes can only resolve details that are slightly smaller than the wavelength of the radiation used to illuminate the object The electrons can be accelerated to high energies and have small wavelengths 34

35 27.10 The Wave Function In 1926 Schrödinger proposed a wave equation that describes the manner in which matter waves change in space and time Schrödinger s wave equation is a key element in quantum mechanics i Ψ = t H Ψ Schrödinger s wave equation is generally solved for the wave function,, Ψ 35

36 The Wave Function The wave function depends on the particle s position and the time The value of Ψ 2 at some location at a given time is proportional to the probability of finding the particle at that location at that time 36

37 27.11 The Uncertainty Principle When measurements are made, the experimenter is always faced with experimental uncertainties in the measurements Classical mechanics offers no fundamental barrier to ultimate refinements in measurements Classical mechanics would allow for measurements with arbitrarily small uncertainties 37

38 The Uncertainty Principle Quantum mechanics predicts that a barrier to measurements with ultimately small uncertainties does exist In 1927 Heisenberg introduced the uncertainty principle If a measurement of position of a particle is made with precision x and a simultaneous measurement of linear momentum is made with precision p, then the product of the two uncertainties can never be smaller than h/4π 38

39 The Uncertainty Principle Mathematically, x p x h 4π It is physically impossible to measure simultaneously the exact position and the exact linear momentum of a particle Another form of the principle deals with energy and time: E t h 4π 39

40 Thought Experiment the Uncertainty Principle A thought experiment for viewing an electron with a powerful microscope In order to see the electron, at least one photon must bounce off it During this interaction, momentum is transferred from the photon to the electron Therefore, the light that allows you to accurately locate the electron changes the momentum of the electron 40

41 Problem: macroscopic uncertainty A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.10%, what is the minimum uncertainty in its position? 41

42 A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.10%, what is the minimum uncertainty in its position? Given: v = 30 m/s δv = 0.10% m = 50.0 g Notice that the ball is non-relativistic. Thus, p = mv, and uncertainty in measuring momentum is ( ) ( δ ) p = m v = m v v ( )( ) = = kg m s kg m s Find: δx =? Thus, uncertainty relation implies 24 h J s x = = π ( ) ( 3 p π kg m s) 32 m 42

43 Problem: Macroscopic measurement A 0.50-kg block rests on the icy surface of a frozen pond, which we can assume to be frictionless. If the location of the block is measured to a precision of 0.50 cm, what speed must the block acquire because of the measurement process? 43