Chapter 27 Quantum and Relativistic Physics

Transcription

1 Chapter 27 Quantum and Relativistic Physics GOALS When you have mastered the contents of this chapter you should be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: Planck's constant length contraction Planck's radiation law time dialation Wien's law mass-energy equivalence photon Compton scattering photoelectric effect complementarity principle work function debroglie wave relativistic mass uncertainty principle Quantum and Relativistic Problems Solve problems involving Wien's law, photoelectric effect, Compton scattering, debroglie waves, the uncertainty principle, and relativistic physics formulations. Tunnel Effect Define and explain the physical significance of the tunnel effect. PREREQUISITE You should have mastered the goals of Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, and Chapter 21, Electrical Properties of Matter, before starting this chapter. 222

2 Chapter 27 Quantum and Relativistic Physics OVERVIEW Until the end of the nineteenth century, science viewed the basic quantities of a particle (energy and momentum) as being continuous. With new phenomena being investigated (radiation, photoelectric effect, etc.) it became necessary to alter this basic view to allow each of these quantities to change only in discrete (quantized) amounts. In this chapter, you will read about several of these phenomena and to study the hypotheses which result from the change. SUGGESTED STUDY PROCEDURE This chapter places emphasis on three Chapter Goals: Definitions, Quantum and Relativistic Problems, and Tunnel Effect. When you begin to study this chapter, read these goals carefully. Remember, a discussion of the terms listed under the goal of Definitions can be found in the first section of this Study Guide chapter. Now read text sections As you read, please note the importance of the initial work by Plank on the electromagnetic radiator, and Einstein on the Photoelectric Effect. A summary of Einstein's Theory of Relativity is found in Section Also, be sure that you deal seriously with section 27.9 and work carefully through the examples provided. At the end of the chapter, read the Chapter Summary and complete Summary Exercises Check your answers carefully against those given on page 623. If you have difficulty, refer to the section referencing given. Now do Algorithmic Problems 1-7, and Exercises and Problems 1-4, 6-10 and 20. For more work with the important concepts of this chapter, turn to the Examples section of this Study Guide chapter. Be sure to see the tunnel effect example included in this section. Now you should be prepared to attempt the Practice Test provided at the end of this Study Guide chapter. Be sure to work all the test parts before you check your answers. For more assistance, refer back to the text or to this Study Guide procedure. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 27.1, Quantum & 27.2,27.3,27.4, ,20 Relativistic 27.5,27.6,27.7, Problems 27.8 Tunnel Effect (see example in Study Guide) 223

3 DEFINITIONS PLANCK'S CONSTANT (h) - The proportionally constant for energy quanta equal to 6.64 x Joule/sec. According to Plank's hypothesis, radiation can only have energy given by E n = nhf, where h is Plank's constant. PLANCK'S RADIATION LAW - Predicts the thermal radiation intensity as a function of wavelength for a black body at temperature T ( ø K). WEIN'S DISPLACEMENT LAW - Predicts the relationship between the absolute temperature and wavelength for maximum thermal radiation for a perfect radiator. PHOTON - The quantum of electromagnetic energy given by the product of Planck's constant and the frequency of the radiation. Einstein first suggested the possibility for photons when discussing the photoelectric effect in PHOTOELECTRIC EFFECT - The phenomena that results in electron emission from materials when radiated with electromagnetic radiation. As light is incident in various metal surfaces, electrons are ejected from the metal surface. Study of the cause and effect relationships found in this experiment led to the photon theory for light. WORK FUNCTION - The energy required to free an electron from the surface of a metal. RELATIVISTIC MASS - The special theory of relativity predicts that the mass of an object increases as its speed, v, approaches the speed of light, c. LENGTH CONTRACTION - Another consequence of the special theory of relativity. It predicts that the apparent length of an object (measured along the direction of motion) becomes shorter as the objects speed approaches the speed of light (c). TIME DILATION - The special theory of relativity predicts that moving clocks run slow. MASS-ENERGY RELATION - Einstein's special theory of relativity predicts the famous mass-energy equivalence given by E = mc 2. Another consequence of the special theory of relativity requires that we consider mass and energy as two forms of the same physical quantity. These forms are related through the famous energy-mass equation which involves the speed of light. 224

4 COMPTON EFFECT - The phenomena that results in scattered electrons when photons of incident radiation interact with weakly bound electrons in metals. The loss of energy of the scattered photons (or increase in length) occurs in descrete quantum amounts depending upon the angle of scattering. COMPLEMENTARITY PRINCIPLE - This principle defined by N. Bohr uses both the wave and particle nature of matter and radiation. Is light a particle or a wave? The model we choose depends on the nature of the experiment. This dual description is not a conflict, but these models complement one another. DE BROGLIE WAVES - The matter waves that were postulated by de Broglie to have a wave length equal to Planck's constant divided by the momentum of the particle. These wave characteristics have been detected for small particles such as electrons. Electron diffraction and electron interference experiments have shown that the dual particle-wave model can be extended to include even larger particles. UNCERTAINTY PRINCIPLE - Heisenberg's formulation of the limits of simultaneous knowledge of position and momentum and energy and time in modern quantum theory. The two forms of this principle (Equations and 27.17) express the upper limits of preciseness which we may know the position, velocity and/or energy of an observed particle. ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 27.6 Complementary Principle In the political-economic areas the Marxist-Leninist political- economic model with its emphasis on the collective structure of a society is in sharp contrast to the classical theory of capitalism with its strong political and economic emphasis on individual rights and enterprise. Today it seems likely that most countries have "mixed" economies that combine features of both models. The strict adherents of each model claim the other model as the deadly enemy of themselves and the future of humankind. Can we think of these two models as two aspects of a complementarity principle? SECTION 27.7 The debroglie Wave The thermal neutrons are useful for studying crystals structures since their debroglie wavelength and range are such as to produce good diffraction patterns from crystal lattices. X-rays of the proper wavelength for diffraction from crystals are of greater penetration in solids and do not readily diffract from the lower mass atoms found in organic crystals. Electrons of proper wavelength for diffraction from crystals are of such low energy and penetration that they produce diffraction patterns from surface layers or thin films of matter. 225

5 EXAMPLES QUANTUM AND RELATIVISTIC PROBLEMS 1. Photoelectric Effect. When a copper surface is irradiated with 254 nm light from a mercury arc it is found that the photocurrent can be stopped by a potential of 0.24 volts. (a) What is the energy of the incident photons? (b) What is the energy of the photo electrons? (c) What is the work function in ev of copper? (d) What is the threshold wavelength for copper? What data are given? The incident light has λ = 254 nm. The photoelectrons are stopped by a potential of 0.24 volts. A typical experimental set-up as shown in Figure 27.2 in the textbook is assumed. What physics principles are involved? Einstein's theory of the photoelectric effect, as discussed in Section 27.3 of the text, will be needed. What equations are to be used? E = hf = hc/λ = KE + W (27.6) KE = qv (21.16) Solutions (a) Incident Photon Energy = hf = hc/λ = ((6.63 x J s)(3 X nm/sec))/254 nm = 7.83 x J = 4.89 ev (b) The energy of the photoelectrons is given by the stopping potential, so KE of photoelectrons = 0.24 ev = 3.8 x J. (c) Work function in ev = 4.89eV eV = 4.65eV. It is the difference between the incident photon energy and the kinetic energy of the photoelectrons. (d) Threshold wavelength = hc/ E = hc/4.65ev = [(6.63 x J s) x (1eV/(1.6 x J)) x 3.00 x (10 17 nm/s)] /4.65eV = 267 nm. Thinking about the answer The two wavelengths of light that are mentioned in this problem are in the ultraviolet part of the spectra. Neither of them is visible to humans. Copper is not a photoelectron emitter in visible light. 2. Compton Scattering. A nm photon is scattered through 60 ø by a collision with an electron. (a) What happens? (b) What is the final wavelength of the photon? (c) What is the final velocity of the electron. What data are given? The initial energy of the photon is given as nm and the scattered angle is given as 60 ø. It is assumed that the electron is essentially at rest when struck by the photon. What physics principles are to be used? The conservation of momentum and energy for particles of light was done by Compton, see Section 27.5 in the book. 226

6 What equations are to be used? Wavelength shift λ' - λ = (h / (m o C)) (1 - cosφ) (27.14) Momentum conservation (hf/c) (hf'/c) cosφ = P e cosθ (1) (hf'/c) sinφ = P e sinθ (2) Algebraic solutions Part (b) is found by using equation To find an expression for the velocity of the electron we can combine Equations (1) and (2). To find the angle of electron departure, divide Equation (2) by Equation (1) where λ = c / f, tan θ = (h/λ' sin φ) / (h/λ - ((h cos φ/λ')) tan θ = (λ sin φ) / (λ' - λ cos φ) (3) Similarly to find the magnitude of the electron velocity, square Equations (1) and (2) and add them together, remember sin 2 θ + cos 2 θ = 1 P e 2 sin 2 θ + P e 2 cos 2 θ = (h 2 /λ' 2 ) sin 2 φ + (h/λ - h/λ' cos φ) 2 If the electron does not travel at relativistic speeds, then (m e ) 2 (v e ) 2 P e2 = h 2 /λ' 2 + h 2 /λ 2 [(2h 2 ) /λλ']cos φ (v e ) 2 [h 2 /(m e ) 2 ] (1/λ 2 + 1/λ' 2 - ((2cosφ) / λλ') v e [h/m e ] SQR RT (1/λ 2 + 1/λ' 2 - ((2cosφ) /λλ') (4) Numerical solutions (a) When the photon is treated as a particle then this phenomena can be discussed as if it were a typical classical collision problem. The incident object (a photon) collides with a massive, stationary object (an electron) and both go off after the collision in different directions. Both energy and momentum are conserved for the collision process so if the initial energy of the incident particle and its scattered angle are known all the other results of the collision can be calculated. (b) The calculation of the final wavelength of the scattering photon is straight forward using equation 27.14: λ' = nm + (2.43 x (1 - cos 60 ø )) = 10.4 x x λ' = 1.16 x m. (c) The calculation of the angle and magnitude of the electron velocity after the collision is done using Equations (3) and (4). tan θ = (1.04 x sin 60 ø ) / (1.16 x x cos 60 ø ) tan θ = 1.41 θ = 54.6 ø v e = (6.63 x J s)/(9.1 x kg) SQR RT[1/(1.04 x ) 2 + 1/(1.16 x ) 2 - ((2 cos 60 ø )/(1.04 x )(1.16 x ))] = 7.29 x 10-4 SQR RT(9.25 x x x ) v e = 7.29 x 10-4 SQR RT(2.50 x ) v e = 1.15 x 10 8 m/s Thinking about the answers You can see that the scattered photon has a longer wavelength, hence a lower energy than the incident photon. During the scattering process the photon loses energy and the electron gains energy. In the above problem the incident photon lost about 10% of its energy in the collision. Can you verify this value for the photon energy loss. Since the value obtained for v e is near the speed of light, the classical approximation used to derive Equation (4) is not valid. The momentum equation 27.9 should have been used. 227

7 3. debroglie WAVES. Derive an expression for the wavelength associated with the molecules of a gas whose absolute temperature is T ø K. Then calculate the wavelength associated with oxygen molecules at room temperature (300 ø K). What data are given? The absolute temperature of the gas is T ø K. For the specific case of the oxygen molecules T = 300 ø K. The gas can be assumed to obey the kinetic theory of ideal gases. The mass of the gas molecules must be known, call it m. We can calculate the mass of the oxygen molecule from the periodic chart (p. 754) and Avogadro's number. 1 gram molecular wt. of 0 2 = 32.0 gm and contains 6.02 x molecules. m O2 = 32.0 gm / 6.02 x = 3.20 x 10-2 kg / 6.02 x = 5.32 x kg. What physics principles are involved? The properties of an ideal gas, Section 14.3, must be combined with the debroglie hypothesis, Section What equations are to be used? The kinetic energy of a gas molecule 1/2mv2 = 3/2kT (14.15) debroglie wavelength l = h/p = h/mv if v << c (27.15) Algebraic solution The rms velocity of a gas molecule is found from Equation (½) mv 2 = (3/2) kt (5) v = SQR RT((3kT)/m) So the debroglie wavelength for a gas molecule is given by λ = h/[m SQR RT((3kT)/m)] = h/sqr RT(3mkT) Numerical solution For room temperature oxygen T = 300ø m = 5.32 x 10-26kg lo2 = 6.63 x 10-34J s/(sqr RT)[3(5.32 x 10-26kg)(1.38 x J/moleøK)(300øK)] lo2 = 2.58 x meters Thinking about the answer This wavelength is much smaller than any usual laboratory diffraction grating spacing and will not be easily detected by any usual means. 4. UNCERTAINTY PRINCIPLE. Consider a proton trapped in a nucleus of 6.0 x 10-15m in diameter. Suppose the potential barrier holding the proton in the nucleus is 24 MeV. (a) What is the uncertainty in the momentum of the proton? (b) What is the uncertainty in the energy of the proton. (c) What is the minimum energy the proton needs to escape from the nucleus? (d) What is the lifetime of the proton in this nucleus? What data are given? The uncertainty in the position of the proton is 6.0 x m. The height of the potential barrier is 24 MeV. The classical, nonrelativistic relationships between momentum and energy are good approximations in this problem. The mass of the proton is 1.7 x kg. What physics principles are involved? This problem uses the Heisenberg Uncertainty Principle. 228

8 What equations are to be used? ΔP Δx ò h (27.16) ΔE Δt ò h (27.17) Solutions (a) Uncertainty in momentum h/δx = (6.63 x J s)/(6.0 x m) ΔP 1.1 x 10-19N s (b) Uncertainty in energy = (ΔP) 2 /2m = (1.1 x ) 2 / (2(1.7 x )) = 3.6 x J = 22.4 MeV. (c) Minimum Energy to Escape = 24 MeV MeV = 1.6 MeV (d) Lifetime h / ΔE (6.63x J s) / (3.6 x J) 1.8 x seconds. 5. A Space Ship traveling at 0.6 c passes a "stationary" observer. Both the observer and the spaceship are provided with identical timers, which they start the instant they pass. Each timer is wired to a light, which flashes after ten seconds has elapsed. (a) Which light does the stationary observer believe flashes first? (b) Which light does the spaceship crew believe flashes first? (c) How much time does the stationary observer believe has elapsed on the observers own timer when the one on the spaceship flashes? (d) How much times does the spaceship crew believe has elapsed on the stationary timer when their own spaceship light flashes? (e) How far apart are the two timers when the one on the spaceship flashes according to the stationary observer? (f) How far apart are the two timers when the one on the spaceship flashes according to the spaceship crew? (g) How long does it take the light from the first spaceship light flash to reach the stationary observer, according to the observers timer? (h) How long does it take the light from the first spaceship light flash to reach the stationary observer, according to the spaceship crew? (i) What time does the spaceship crew believe is shown on the stationary timer for the time of flight? (j) From your above results fill out the table below: Reporter Stationary Observer Spaceship Crew Time up to flash of spaceship timer (as computed on the stationary timer) seconds Time to flight of light flash (as computed on the stationary timer) seconds (k) What do the two reporters agree upon? disagree upon? Time light flash from spaceship reaches the stationary timer seconds 229

9 What data are given? The spaceship travels at 0.6c (2.4 x 10 8 m/s) past the observer. Lights are flashed every 10 seconds according to a clock connected to the light flasher. The conditions of special relativity are satisfied. What physics principles are involved? The concept of time dilation is used in conjunction with the classical concepts of distance, time, and velocity. What equations are to be used? Time dilation Δt = Δt o /SQR RT(1 - (v 2 /c 2 )) (27.8) Distance = vt (3.2) Solution (a) The stationary observer knows that moving clocks run slow. So the observer can calculate how long 10 seconds will seem to be in the spaceship, according to the observers clock. SQR RT(1 - (v 2 /c 2 )) = SQR RT(1 - (0.6c/c) 2 ) = SQR RT(1 -.36) = SQR RT(.64) = 0.8 So a spaceship 10.0 seconds would appear to be 10/0.8 or 12.5 seconds on the observers clock. The observer believes the spaceship light flashes after the observers light has already flashed. (b) The spaceship crew believes that moving clocks run slow, but according to them the observer is the one who is moving. Consequently when 10 seconds has passed on their clock the observers clock will show a number smaller than 10 seconds, in fact they compute a time of (0.8) (10 sec) = 8 seconds will show on the observers clock when their light flashes. Which, according to them, occurs a full two seconds on the observers clock before the observers light flashes. So both parties believe the other light flashes after their own light has already flashed. (c) We answered above, 12.5 seconds. (d) We answered above, 8.0 seconds. (e) According to the observer the spaceship has been traveling at 0.6c for 12.5 seconds when the spaceship light flashes, so d = (0.6c) (12.5), d = 7.5c. (f) According to the spaceship crew they have been traveling for 10 seconds when their light flashes so they are d = (0.6c) (10 sec) or 6c from the observer when their light flashes. (g) Since, according to the observer, the light was 7.5c away and light travels at a speed c it will take 7.5 seconds to reach the observer. (h) The spaceship crew has a tough problem to solve. They are 6c away from an object which is going away from them at a speed of 0.6c. So they send a message to that moving object at a speed of c, how long will it take to catch up? Let t' be the time to catch up; then 0.6c t' is the distance the moving object travels in t' seconds, that distance plus 6c will be equal to the distance the message travels which is ct' 0.6ct' + 6c = ct' 6 = 0.4t' t' = 15 seconds. So according to the spaceship crew it will take 15 seconds for their light flash to reach the observer. 230

10 (i) However, the spaceship crew knows that the observers clock runs slow, so the 15 seconds on their clock will appear as (15sec)(0.8) or 12 seconds on the observers clock. (j) We can now fill in the table: (k) The reporters agree on the time of arrival of the flash at the observers clock. They disagree about intermediate computed quantities. 6. Wien's Law - (a) If the absolute temperature of an object is doubled what happens to its emitted radiations? (b) What is a low temperature limit for the emission of visible light? What data are given? (a) The absolute temperature of the emitted body is doubled. (b) Let us take the low energy limit for visible light to be 700 nm. What physics principles are involved? This problem makes use of Planck's Law for black body radiation as expressed in Wien's Law. What equation is to be used? Wien's Law: λ max T = 2.88 x 10-3 m ø K (27.5) Solutions (a) If the absolute temperature is doubled, then the peak wavelength of the light is halved. If the original wavelength is red, the final wavelength would be in the violet, the total spectrum would become whiter. This explains the fact that white hot objects are hotter than red hot objects. (b) λ max 700 nm = 7 x 10-7 m T = (2.88 x 10-3 ø K) / (7x10-7 ) = 4.1 x 10 3 ø K 3800 ø C This is a relatively low temperature for a hot object, about 1600 ø cooler than the sun's surface. 231

11 PRACTICE TEST 1. Einstein's theory of Relativity was a milestone in the development of scientific thought in the 20th century. a. State the two postulates upon which Einstein based his special theory of relativity. Postulate 1: Postulate 2: b. Briefly describe the consequences of observers in one frame of reference measuring time, distance, and mass in a frame which is moving at speeds approaching the velocity of light. 2. Light of wavelength 200 nm falls on an aluminum surface where 4.20 ev's are required to remove electrons. a. What is the kinetic energy of the fastest emitted photoelectrons? b. What is the stopping potential for these photoelectrons? c. What is the cutoff wavelength for aluminum? 3. An alpha particle in a nucleus can be thought of as being trapped in a potential well that has a barrier, height of 8.0 MeV (8.0 x 10 6 ev) and a barrier width and well width of 1.0 x m as shown. a. Calculate the uncertainty in the energy of the alpha particle. (m = 6.7 x kg) b. What minimum energy must the alpha particle have to tunnel through this potential barrier? ANSWERS: 1. Postulate 1: The laws of physics are the same in all inertial reference frames. Postulate 2: The speed of light (c) in a vacuum is measured as a constant by all observers regardless of their relative motion. consequences: length contraction, time dilation, effective mass increase 2. K.E. MAX = 2.00 ev (3.20 x J), 2.00 volt, 295 nm x J (2.0 Mev, 6.0 Mev) 232