Chapter 27 Quantum And Relativistic Physics

Transcription

1 592 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 dilation 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. PREREQUISITES 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.

2 593 Chapter 27 Quantum And Relativistic Physics 27.1 Introduction The topics we have presented in the proceeding chapters have been in the realm of classical physics. The concepts of mass, velocity, momentum, and energy were applied to macroscopic systems, and we expected the variables of these systems to have continuous ranges of values. It was assumed that classical physics based on Newton's mechanics, Maxwell's electromagnetic theory, and thermodynamics held the answers to all the problems of the physical universe. Some physicists at the end of the nineteenth century believed that the development of physics was complete. On the contrary, the new models developed in physics during the first half of the twentieth century may well signify one of the most exciting periods of human intellectual history. Quantum physics was born and showed that classical physics failed to describe physical phenomena of particles of m dimensions. When velocities approach the speed of light, Einstein's special theory of relativity replaced classical physics as the working theory. In this chapter you will study the foundations of quantum and relativistic physics and some consequences of the new physics Planck's Radiation Law and Its Derivations Perhaps you have noticed the change of color that occurs when you dim the incandescent lights of a room using a dimmer switch. When the switch is completely on, the lights are bright and give off white light. As you dim the lights, you can see that the light given off is not only less bright but is also of a different color. The dim light is red-orange in color. We can conclude that the color, or frequency, of light emitted from an incandescent bulb changes as we change the temperature of the filament in the bulb. This is one phenomenon that classical physics failed to explain, that is, the frequency distribution of the electromagnetic radiation emitted from heated objects. This radiation is emitted in a continuous range of frequencies in a characteristic way that depends on the temperature of the radiation source (Figure 27.1). Classical physics as applied by Lord Rayleigh failed to describe this distribution. In fact, Rayleigh's work predicted the "ultraviolet catastrophe," unlimited energy output at the wavelengths approaching zero.

3 594 Prior to Rayleigh's results Max Planck reported his solution to the problem. He combined classical equations that described each end of the distribution curve with the result being an empirical equation that fit the entire distribution curve. In order to explain his empirical formula, Planck had to postulate that vibrating objects, or oscillators, were responsible for the emission and absorption of thermal radiation. Furthermore, these oscillators could only have discrete energies. They could exist only in energy states that were integral multiples of a fundamental oscillator energy or quantum, when the quantum energy is given by hf, where h is Planck's constant, and f is the frequency of the oscillator. The value of h necessary to make Planck's equation fit the radiation curves shown in Figure 27.1 is h = 6.63 x joule-seconds. According to Planck's postulate, the various oscillators in a radiation source could only have energies given by E n = nhf or E n = nhc/λ (27.1) since fλ = c (see Chapter 16) where n = 1, 2, 3,..., c is the speed of light in meters per second, f is the frequency in cycles per second, and λ is the wavelength in meters. The equation that Planck found to describe the intensity or radiation from a perfect radiator as a function of the temperature of the radiator and the wavelength of the radiation is E λ = [8πhc]/[ λ 5 (exp(hc/λkt) - 1)] (27.2) where c is the speed of light, λ is the wavelength of the radiation, k is Boltzmann's constant, T is the absolute temperature of the radiator and Eλ is the energy intensity (watts/m 2 ) emitted per unit wavelength (m). Planck's theory was not an immediate success as its postulates were quite a break from some of the ideas of classical physics. This was much to the dismay of Planck who had hoped to explain his quantum postulate in classical terms. After all, he had used classical electromagnetic waves coupled to some special oscillators within the radiator. Two important results of Planck's theory were the derivation of the Stefan-Boltzmann law (Equation 11.8) and Wien's law (Equation 27.5). The Stefan- Boltzmann law, which was discussed in Chapter 11, describes the rate at which energy is radiated from a body as a function of the absolute temperature of a body. The mathematical equation is I = P/A = σt 4 watts/m 2 (11.8) where σ is Stefan's constant = 5.67 x 10-8 watts/m 2 - K 4, T is the absolute temperature in degrees Kelvin, and I is the total energy per unit time per unit area emitted by a perfect thermal radiator, or the ratio of the emitted power P divided by area A. For radiators other than these perfect blackbody radiators we can write I = εσt 4 (27.3) where ε is the emissivity of the object (0 < ε <1). If an object of emissivity ε is maintained at a temperature T 1 in a chamber at a temperature of T 2, the net radiation from the object is equal to the difference between

4 595 the radiation emitted from the object and absorbed from the chamber, I = εσ (T T 2 4 ) (27.4) Wien's law is an empirical equation for the relationship between the temperature of a perfect radiator and the wavelength of the radiation of maximum intensity. It can be derived from Planck's law by finding the wavelength for maximum radiation intensity from Equation The temperature of a radiating object and the wavelength of the maximum intensity of the emitted radiation have a constant product, λ max T = 2.88 x 10-3 m- K (27.5) EXAMPLES 1. Given that the sun has its intensity peak output at 500 nm, find its effective blackbody temperature. Using Wien's law we can write: λ max T = 2.88 x 10-3 m- K T = 2.88 x 10-3 (m K)/5.00 x 10-7 m, T = 5760 K. This is the effective surface temperature of the sun. 2. The emissivity of the human skin is close to one. Compare the rate of energy loss of a person (skin area = 1.5 m 2 ) inside when the temperature is 27 C, and outside when the temperature is -23 C. (Assume skin temperature is 32 C.) From Equation 27.4 above you can write, E/ t = IA = εσa(t T 2 4 ) where T 1 is body temperature = 305 K and T 2 is either the inside temperature (300 K) or the outside temperature (250 K), ( E 1 / t)/( E 2 / t) = (σa ( ))/( σa ( )) = The rate of energy lost from the body by radiation is greatly reduced in a warm environment Photoelectric Effect While Heinrich Hertz was investigating the properties of electromagnetic radiation in 1887, he observed that the metallic electrodes of his system were more easily discharged when the electrodes were illuminated by ultraviolet light. This interaction between light and metal surfaces was subsequently studied in more detail and became called the photoelectric effect. The photoelectric effect is explained by a model of light energy interacting with a metal surface to cause electrons to be emitted from the surface. The

5 596 light energy is partially absorbed by the metal and partially transformed into the kinetic energy of the emitted electrons. The photoelectric effect is another example of quantitative experimental results as shown in Figure 27.2 that could not be explained by the models of classical physics. On the basis of the wave model of light, we found in Chapter 19 that the energy in a light beam is proportional to its intensity, and thus an intense red light should be able to provide enough energy to free an electron from the metal surface. The facts did not support this; the observations showed that below some threshold frequency, no matter how intense the light, electrons were not emitted from the surface (Fig. 27.2a). If the intensity of the light is increased, the photoelectric current is increased if the frequency of the light is above the threshold value (Figure 27.2b). The kinetic energy of the emitted photoelectrons (as measured by the voltage it takes to stop them from reaching the anode) is proportional to the frequency of the incident electromagnetic radiation (Figure 27.2a). In 1905 Einstein offered his theory to explain the photoelectric effect. He applied Planck's quantum postulate to the incident light. He argued that the incident light was made up of quanta of light, which are called photons, each of energy hf. The intensity of the light beam is a measure of the number of photons per unit area per second. Einstein reasoned that the quantum interaction is an all or none interaction. If the photons do not have an energy equal to the energy with which electrons are bound to the metal (this binding energy is called the work function), there is no electron emission no matter how intense the light may be. There is no way for the electrons in the metal to store subthreshold photons until the threshold energy is reached. To support this, experiments show no observable time lag between time when light is turned on and when electrons appear. If the photon has energy greater than the work function, the excess energy goes into the kinetic energy of the photoelectron. This is expressed by the following equation:

6 597 energy of the incident photon of light = kinetic energy of the photoelectron + the work function of the metal E = hf = KE + W (27.6) where E is the photon energy, hf =hc/λ, W is the work function of the metal, and KE is the kinetic energy of the photoelectron. The electron volt (1 ev = 1.60 x J) is a convenient energy unit for microscopic physics in this and the remaining chapters and is the energy increase of an electron when it is accelerated through a potential of one volt. EXAMPLE Light of wavelength λ = 400 nm strikes a metal and produces photoelectrons that are brought to rest by 1.30 V. Find the work function of the metal. The energy of the photons is E = hf = hc/λ = (6.63 x J-sec) x (3.00 x 10 8 m/sec)/4.00 x 10-7 m E = 4.97 x J = 3.11 ev where 1 ev = 1.60 x J. The retarding potential of 1.30 V represents the stopping potential for the photoelectrons. Thus the kinetic energy of the photoelectrons is e x Vs = 1.30 ev. From Equation 27.6, we find the work function, W = hf - KE = 3.11 ev ev = 1.81 ev = 2.90 x J 27.4 Relativistic Physics Before we continue our discussion of the quantum properties of light, let us digress to consider some other work that Albert Einstein did during 1905 when he was a 26-yearold patent examiner in the patent office in Berne, Switzerland. Einstein was studying the electrodynamics of moving objects. He wrote a paper in which he proposed the two postulates that form the basis for the special theory of relativity. Einstein's special theory of relativity has some important consequences for the realm of quantum physics. These consequences are derived from Einstein's two fundamental postulates. Postulate 1: The laws of physics are the same in all reference systems at rest or moving at a constant velocity. Postulate 2: The speed of light is always measured to be the same value in a vacuum, and it is independent of the state of motion of the source or observer. We will now give some of the results of Einstein's special theory of relativity along with the physical significance of these results.

7 598 Length Contraction: All moving lengths appear to shrink in the direction of motion. This length contraction is frequently called Lorentz contraction. There is no contraction of lengths perpendicular to the velocity v. This phenomenon seems to contradict everyday experience, but it has been verified many times in modern physics experiments. The equation for the apparent length L in the direction of motion for an object moving with speed v is given by: L = L o SQR RT(1 - v 2 /c 2 ) (27.7) where L o is the length of the object in a system in which it is at rest, v is the speed of the moving object and c is the speed of light. Time Dilation: Moving clocks run slow. Again this result seems to contradict common experience, but it too has been verified many times in modern experimental physics. The equation for a time interval from the point of view of a stationary observer Δt o is: t o = t SQR RT(1 - v 2 /c 2 ) (27.8) where t is the time period shown by the clock moving with the apparatus, v is the speed of the motion of the apparatus, and c is the speed of light. Consider the following experiment which illustrates Lorentz contraction and time dilation. This experiment has been repeated in several ways and in each case there is complete agreement with the consequences of Einstein's postulates. EXAMPLE Muons are a common component of cosmic rays that strike the top of the earth's atmosphere. They are unstable particles that decay with an average lifetime of 2.00 x 10-6 sec when at rest. An experiment was carried out in which the number of muons incident at the top of a mountain 3000 m above sea level was counted. These particles were moving with a speed of 0.998c. In an average lifetime a muon would travel a distance of only about 600 m (~c x 2 x 10-6 sec), and thus no muons would be expected to reach sea level. When the experiment was carried out it was found that most of the muons do reach sea level. The explanation for these observations comes from Einstein's equation as follows: From the point of view of a person in a lab on the mountain, the actual decay time of the muon is slowed down because the muon is a moving clock: its average lifetime moving at speed c was t o = t / SQR RT(1 - v 2 /c 2 ) = 2.00 x 10-6 sec / SQR RT( ) = 31.6 x 10-6 sec and the distance the muons traveled in this time was 0.998c x 31.6 x 10-6 = 9500 m, which is greater than the mountain height, thus explaining why so many muons reached sea level. From the point of view of the moving muon the physical result must be the same, but here we see that the mountain (which is moving at speed 0.998c with respect to the muon) contracts to a height of H = 3000 SQR RT( ) or H = 190 m and thus the 600 m traveled is enough to reach sea level. Here we see an example how equations of

8 599 special relativity give the correct answer in each reference system. Einstein's special theory of relativity has several important modifications for the dynamics of particles. In order to preserve the form of Newton's second law, namely that impulse is equal to the change in momentum (Equation 6.2), the relativistic momentum is given the following definition p = {m 0 /SQR RT(1 - v 2 /c 2 )} v (27.9) where m 0 is the mass of the particle when it is at rest, v is the velocity of the particle, v is the speed, or magnitude of the velocity, of the particle, and c is the speed of light. You may notice that we can write Equation 27.9 in exactly the same form as the usual momentum equation if we invent a quantity m that is equal to the rest mass m 0 divided by SQR RT(1 - v 2 /c 2 ). We call this quantity the relativistic mass m, m = m 0 /SQR RT(1 - v 2 /c 2 ) (27.10) Mass is a measure of inertia. This equation shows us that the inertia of a particle becomes indefinitely large as its speed approaches the speed of light. This means that it becomes more and more difficult to accelerate a relativistic object and that it is physically impossible to accelerate an object to the speed of light. Einstein showed that as a result of his postulates the total energy of a relativistic particle could be expressed in the now famous equation, E = mc 2 E = m 0 c 2 /SQR RT(1 - v 2 /c 2 ) (27.11) where E is the total energy, m is the relativistic mass, and c is the speed of light. If the rest energy of the particle is defined as m 0 c 2, then the kinetic energy of a relativistic particle can be written as KE = E - m 0 c 2 = m 0 c 2 [(1/ SQR RT(1 - v 2 /c 2 )) - 1] (27.12) The mass-energy equivalence formula as expressed in Equation shows that in a fundamental way mass must be considered in a new formulation of the conservation of energy. The difference in masses of objects due to changes in energy in classical physics are too small to be measured, but this mass-energy equivalence is the basis of the nuclear binding energy processes which we will discuss in Chapter 31. EXAMPLE Consider the mass equivalent to a potential energy increase of 40,000 J. (This is about the energy required to lift a 10 kg mass from sea level to the top of the Empire State Building.) The equivalent mass would be given by m = (4.00 x 10 4 J)/(9.00 x (m/sec)) 2 = x kg = 4.44 x g This is a mass much too small to be detected with our best analytical balances.

9 600 In many atomic and nuclear processes the relativistic form of the kinetic energy of a particle should be used. In problem 23 at the end of the chapter you are asked to show that in the classical region of physics (where v is much smaller than c), the kinetic energy reduces to (½) m 0 v Compton Effect The photon model of light gives light particlelike characteristics instead of the wave characteristics we discussed in Chapter 19. In 1922 A. H. Compton reported that he had observed elastic collisions between x-ray photons and electrons. He was able to explain his results by assigning a momentum to the photons given by the equation: p = h/λ = hf/c (27.13) where p is the momentum of a photon, h is Planck's constant, λ is the wavelength of the x rays, c is the speed of light, and f is the x- ray frequency. An experimental set up to illustrate the Compton effect is shown in Figure 27.3a. The intensity of x rays scattered from carbon block is measured as a function of the scattering angle and scattered x-ray wavelength. Loosely bound electrons may be freed from the carbon by the incident x-ray photons. The freed electrons are called photoelectrons. The scattered photons have less energy than the incident photons. Thus, in addition to the unmodified x rays, scattered x rays with a slightly greater wavelength are detected at all angles except the zero scattering angle. Compton observed collisions like the ones shown in Figure 27.3, and he could explain the results by using the conservation of momentum applied to the photon-

10 601 electron collision. The following equations are derived from the application of the conservation of energy and momentum to this collision (Figure 27.3b). MOMENTUM CONSERVATION Resolving the momentum vector into its components, x-components: pλx = pλ'x + p ex hf/c = (hf'/c) cosφ + p e cosθ y-components: pλy = pλ'y - p ey 0 = (hf'/c)sinφ - p e sinθ ENERGY CONSERVATION hf = hf' + KE where KE is the kinetic energy of the electron. By the time of Compton, it was known that the classical equations for the interchange momentum and kinetic energy of a moving object had to be modified for objects traveling at high velocities. The modified equations for the kinetic energy and the momentum for relativistic electrons are given by Equations 27.9 and These equations can be solved to obtain the following equation for the wavelength of the scattered x-ray photon in terms of scattering angle and incident photon wavelength: λ' -λ =( h/m 0 c) (1 - cosφ) (27.14) where λ' is the scattered photon wavelength, λ, the incident photon wavelength, and h/m 0 c, is a constant called the Compton wavelength, which equals 2.43 x m. EXAMPLE Given that the incident photon had a wavelength of 12.4 x m, and the scattered photon recoiled at 180, find the energy transferred to the electron. (Note: when φ = π, then the energy transferred is maximum.) First calculate the energy lost by the photon because of the scattering: λ' - λ =( h/m 0 c) (1 - cos 180 ) = 2h/m 0 c λ' = λ + 2h/m 0 c = ( ) x m = 17.3 x m kinetic energy of the electron = hc/λ - hc/λ' = energy lost by the photon

11 602 energy lost by the photon = (10 12 / /17.3)hc = (6.63 x x 3.00 x 10 8 ) x ((4.9/(12.4 x 17.3)) x ) = 45 x J = 28.1 kev where 1 kev = 1.60 x J Complementarity Principle The Compton effect and the photoelectric effect evoke a particle model for light. The particle model of light is used to describe the interaction of electromagnetic radiation with matter; the classical wave model is used to explain propagation phenomena. We use two models for light, one model is particlelike and the other is wavelike. The aspect of light that we observe in a particular experiment depends on the nature of the experiment. Niels Bohr would cite this as an example of the complementarity principle. The wave and particle models of light complement each other and give us a theory that encompasses the many potentialities of both particle and wave behavior. Bohr applied the complementarity principle to areas outside of physics. Bohr's goal was to discover the interrelationships among all areas of knowledge. He viewed the complementarity principle as an important step toward his goal. In biology, Bohr presented the molecular model of living systems and the teleological approach to the causality model of living systems as complementing components of the organization of the living system. In psychology, the complements may be behaviorism and self-conscious freedom of choice. (What complementary models would you suggest for a political system or organization?) 27.7 The debroglie Wave In 1924 Louis debroglie proposed that material particles such as electrons might have a dual nature. He suggested that the wavelength of a particle should be defined by the same equation that applies to the photon: λ = h/p (27.15) where h is Planck's constant and p is the magnitude of the momentum of the particle. The waves associated with matter are not electromagnetic waves. These waves are probability waves. The intensity of the probability wave at a point is a measure of the probability of finding the particle at the point. The test of this idea would be to do a wave experiment with particles. One experiment that depends entirely on the unique features of waves is a diffraction experiment. Such an experiment was done by Clinton J. Davisson and L. H. Germer in They found that electrons were diffracted by crystals. The results were explained by using Equation for the electron wave diffracted by a crystal lattice with spacing the same order of magnitude as the electron wavelength. The wave model of electrons can be used to explain the operation of the electron microscope. Small electron wavelengths give high resolving power for the electron microscopes.

12 603 EXAMPLES 1. Davisson and Germer used a beam of 54.0-eV electrons in their electron diffraction experiment. Find the wavelength of these electrons. The wavelength l and the momentum p are given by λ = h/p p 2 /2m 0 = KE = ev where e is the charge of the electron, V is the voltage, h is Planck's constant, and m 0 is the electron rest mass of 9.11 = kg. p= (2 m 0 ev) 1/2 λ = h/(2 m 0 ev) 1/2 λ = 6.63 x J-sec / [2 x (9.11 x kg) x 1.60 x C x (54.0 V)] 1/2 = 6.63 x J-sec / 3.97 x kg-m/sec = 1.67 x m = nm 2. Find the speed v of a neutron that has a debroglie wavelength of nm. mv = h/λ = 6.63 x J-sec / 1.00 x m v = 6.63 x / 1.67 x = 3.97 x 10 3 m/sec This is a slow neutron with a kinetic energy of 0.08 ev. This is an energy comparable to thermal energies at room temperature and thus such neutrons are called thermal neutrons. Thermal neutrons are used in neutron diffraction studies to measure the distances between lattice planes in solids. Neutron diffraction is especially useful in studying organic crystals containing hydrogen. (Can you explain why neutrons are useful for such studies while x rays and electrons are not very useful?) 27.8 Uncertainty Principle We have now arrived at an understanding of a property of a system called a particle (e.g., an electron) that can be thought of as a particle or as a wave. At one instant we can think of the electron as a small chunk of matter with an electrical charge e located at some point in space. The next instant we can think of the electron as a cloud of probability waves that extends throughout space. What sense then can we make of the idea of the position of a particle? Werner Karl Heisenberg formulated his now famous uncertainty principle in 1927 in order to clarify the use of the terms position andvelocity when referring to quantum phenomena. The various forms of the uncertainty principle express limits to the preciseness of our knowledge of the magnitudes of pairs of physical variables. Algebraically stated the Heisenberg uncertainty principle can be written as The product of the uncertainty in momentum and position is greater than Planck's constant

13 604 p x x h (27.16) where p x is the uncertainty in the x-component of the momentum, x is the uncertainty in the position of the object in the x- direction and h is Planck's constant. Another equivalent form is: E t h (27.17) Here E is the uncertainty in the energy, and t is the uncertainty in the time. These uncertainty relationships must be understood as a part of our models to explain nature; they are not based on limitations of measuring instruments. As a result of these relationships, the certainties of classical physics are impossible in quantum physics. In fact, quantum theory is a theory based on a probabilistic understanding of events and physical variables. The uncertainty principle is a consequence of this fundamental model of physical reality as it is now formulated by physicists. EXAMPLES 1. Find the uncertainty in velocity of a 1.0-gm particle if its position is measured to within 0.1 mm. m v = p h/ x v h/m x = 6.63 x J-sec / 1.0 x 10-3 kg x 10-4 m = 6.63 x m/sec This is an uncertainty far below any measurement capability. 2. Find the uncertainty in the energy of photons emitted in the time of 10-8 sec. (This is a characteristic time for excitation of many atomic systems.) E t h E 6.63 x J-sec / 10-8 sec = 6.6 x J = 4.1 x 10-7 ev

14 Tunnel Effect One of the results of quantum theory is the prediction of events that are forbidden by classical physics. A potential-energy barrier exists wherever a repelling force acts to restrict a particle from a region of space. Figure 27.4 shows an energy well formed by a finite barrier on one side and an infinite barrier on the other side. (An infinite barrier designates a forbidden region of space.) In this case let us assume we can confine electrons to a region of 1.00 nm using a 2.00-eV energy barrier. This uncertainty in x gives a corresponding uncertainty in momentum of h/ x = 6.63 x J-sec/1.00 x 10-9 m = 6.63 x kg-m/sec The corresponding uncertainty in energy is given by the equation, E = ( p) 2 /2m = [6.63 x kg-(m/sec) 2 ] 2 / 2 x 9.11 x kg = 2.41 x J E = 1.51 ev This uncertainty means that an electron with energy of 0.5 ev or more has an uncertainty in energy sufficient to allow it to get over the top of the 2.00-eV barrier. Since the uncertainty in x is equal to the width of the barrier we cannot say with certainty on which side of the barrier the electron is. In fact, a rigorous quantum mechanical treatment of this problem shows that the electron has a probability of tunneling through the barrier. As the barrier height and/or barrier width is increased, the probability for tunneling through it decreases. This quantum tunneling phenomena is helpful in understanding alpha decay in radioactive nuclei and electron behavior in semiconductors.

15 606 SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of the summary with the number of the section where you can find related content material. Definitions 1. Planck's radiation law provides a correct equation for the graph of radiation as a function of. 2. Wien's law could be used with the spectral distribution of a radiation source to find a. Planck's constant b. velocity of light c. temperature of source d. debroglie wavelength e. source size 3. The photon is defined as the smallest energy packet for a given frequency and is equal to a. h/λ b. hf c. hc/λ d. c/λ e. hc 4. Compton scattering gives experimental evidence that the photon has particle like a. wavelength b. frequency c. velocity d. momentum e. mass 5. The photoelectric effect shows that the electron emitted from a metal has energy that depends on a. metal material b. photoelectric current c. photon energy d. light intensity e. none of these 6. The uncertainty principle states that it is impossible to know exactly both a particle's momentum and a. energy b. position c. velocity d. acceleration e. mass

16 If a proton, which has an electric charge of +e and a rest mass equal to about 1900 times the rest mass of the electron, and an electron are accelerated from rest through the same potential difference, the ratio of proton de Broglie wavelength to electron de Broglie wavelength will be a. one b. zero c. greater than 1 d. less than 1 e. none of these 8. An example of the complementarity principle is a. particle- wave nature of matter b. particle- wave nature of radiation c. momentum-energy of particles d. time-position of particles e. none of these 9. Einstein's mass-energy equivalence formulation predicts a mass change associated with energy change E to be a. E / hƒ b. E / p c. hƒ / E d. E / c 2 e. p / e 10. A long stick is observed to have a length L as it moves past a speed v; its length in its own rest frame is a. L /SQR RT(1 -(v 2 /c 2 )) b. L SQR RT(1 -(v 2 /c 2 )) c. L d. L(V/c) e. L(c/v) 11. A clock moving past an observer at speed v appears to her to have a period t. The period of a clock in a rest frame is a. t /SQR RT(1 -(v 2 /c 2 )) b. t SQR RT(1 -(v 2 /c 2 )) c. t d. t(v/c) e. t(c/v) Quantum and Relativistic Problems 12. The peak wavelength for a 10,000 K source will be a. 2.9 x 10-7 m b x 10-1 m c. 1.2 x m d. 2.4 x m e. 4.8 x m

17 The maximum increase of a photon wavelength undergoing Compton scattering is (in meters) a x b. 0 c. 1.2 x d. 2.4 x e. 4.8 x If a photon has an energy equal to the rest energy of an electron, then its wavelength is a. 2m o c 2 / hƒ b. h / m o c c. hƒ / 2m o c 2 d. hƒ / 2 m o c e If the uncertainty in position of a microbe (10-9 kg) is 10-6 m, the uncertainty in its speed will be (in m/sec) a x b x c x d x e. zero 16. The length of a meter stick moving with speed 0.8c in the direction of its length will appear how long to an observer at rest. a. 80 cm b. 64 cm c. 36 cm d. 125 cm e. 60 cm 17. The period of a clock in its rest frame is one second. When moving at a speed of 0.6c the period will appear to a laboratory observer to be a sec b. 0.6 sec c sec d sec e. 0.8 sec Tunnel Effect 18. The physical significance of the tunnel effect of quantum physics is that particles are found to tunnel through regions where a. PE b. KE c. KE + PE d. KE > 0 e. none of these

18 609 Answers 1. intensity, wavelength (or frequency) (Section 27.2) 2. c. (Section 27.2) 3. b, c (Section 27.2) 4. d (Section 27.5) 5. a, c (Section 27.3) 6. b (Section 27.8) 7. d (Section 27.7) 8. a, b (Section 27.6) 9. d (Section 27.4) 10. a (Section 27.4) 11. b (Section 27.4) 12. a (Section 27.2) 13. e (Section 27.5) 13. e (Section 27.5) 14 b (Sections 27.2, 27.4, 27.7) 15. c (Section 27.8) 16 e (Section 27.4) 17. d (Section 27.4) 18. b (Section 27.9) ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve singleconcept problems. Equations E n = nhf or E n = nhc/λ; n = 1, 2, 3,... (27.1) E λ = [8πhc]/[ λ 5 (exp(hc/λkt) - 1)] (27.2) I = εσt 4 (27.3) λ max T = 2.88 x 10-3 m- K (27.5) E = hf = KE + W (27.6) L = L o SQR RT(1 - v 2 /c 2 ) (27.7) t o = t SQR RT(1 - v 2 /c 2 ) (27.8) p = {m 0 /SQR RT(1 - v 2 /c 2 )} v (27.9) m = m 0 /SQR RT(1 - v 2 /c 2 ) (27.10) E = m 0 c 2 /SQR RT(1 - v 2 /c 2 ) (27.11) p = h/λ = hf/c (27.13) λ' -λ =( h/m 0 c) (1 - cosφ) (27.14) λ = h/p (27.15) p x x h (27.16) E t h (27.17)

19 610 Problems 1. The frequency of a given source of green light is 5.5 x Hz. What is the energy of this radiation, i.e., the energy of each photon? 2. If the absolute temperature of a filament in a light bulb is 2500 K, what is the wavelength for the maximum intensity of the radiation from the hot filament? What is its corresponding energy per photon of radiation? Is this bulb usable as a source of light for reading? 3. Two bodies are exactly alike except in temperature, one is at absolute temperature of T and the other is at absolute temperature of 1.5 T What is the ratio of the rates of energy emission from the two bodies? 4. The value of the work function of potassium is 2.25 ev. What is the longest wavelength that will produce a photoelectron from a potassium surface? 5. What is the momentum of a photon that has a frequency of 5.50 x Hz? 6. What is the wavelength that is associated with you when you are walking at 5 km/hour? What do you think of the prospects of detecting this wavelength? 7. Suppose you know your momentum (as for example in problem 6) to within 0.1 percent. What is the limit upon knowing your exact location at that time? Comment upon this result. Answers x J = 2.3 ev x 10-6 m, 1.73 x J = 1.08 ev 3. 1: nm x kg-m/sec 6. ~10-36 m 7. ~10-32 m EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given in brackets at the end of each exercise. Section A prominent line emitted by mercury street lights is green and has a wavelength of 5460 Å or 546 nm. What is its frequency and its energy in joules and in electron volts? [5.49 x sec-1, 3.64 x J, 2.27 ev]

20 One can express the energy of any photon radiation in terms of electron volts. A very useful rule of thumb relationship is that the product of energy in terms of electron volts and wavelength in nanometers is equal to 1240 ev-nm. Can you verify this? 3. The wavelength of maximum energy of radiation from the sun is 480 nm. What is the frequency of radiation of this wavelength and what is the energy of this photon in electron volts? [6.25 x sec -1, 2.59 ev] 4. Find the temperature of a perfect thermal radiator that has its peak output at 400 nm. How much would this temperature have to be raised to increase the radiated power per unit area by 10 percent? [7200 K, 174 K] Section In studying wave motion, you found that the intensity of a wave depends upon the square of its amplitude. What does the intensity of light mean in terms of quantum concepts? 6. If 5 watts of light of the yellow line of a sodium emission spectrum (589 nm) fall upon a surface, how many photons strike the surface per second? [1.48 x photons/sec] 7. In quantum physics we say energy = h x frequency. What are the dimensions of h? 8. If sodium has a work function of 2.28 ev, what is the longest wavelength that will produce a photoelectron from sodium? If one doubles the intensity of light of 560 nm, will a photoelectron be emitted from sodium? [545 nm] 9. Potassium has a photoelectric work function of 2.24 ev. If it is illuminated by the 436- nm mercury line, what will be the maximum kinetic energy of the photoelectrons? [0.60 ev] 10. It is noted that the threshold wavelength for a certain metal is 500 nm. Find the energy of the photoelectrons emitted when this metal is radiated with 450 nm light. [0.28 ev] 11. The work function of cesium is 2.0 ev. The stopping potential in an experiment is found to be 1.1 V. a. Find the threshold wavelength for cesium. b. Find the wavelength of light used in the experiment. [a. 6.2 x 10-7 m; b. 4.0 x 10-7 m] Section Show that when v/c «1, the relativistic kinetic energy reduces to the classical form, (½) m 0 v A particle with an average lifetime of 10-6 sec at rest moves with a speed of 0.9c. Show that an observer in the lab would measure a lifetime of 2.3 x 10-6 sec. 14. Find the distance the particle in problem 24 will move in the lab. [6.4 x 10 2 m]

21 By solving Equation 27.9 for the magnitude of the velocity, v, eliminate v from Equation to obtain a relationship between total energy, E, momentum, p, and rest energy, m 0 c 2. Show that the following statement is correct: E 2 = p 2 c 2 + m 0 2 c 4. Section Calculate the increase in wavelength of the scattered photon for each of the following scattering angles: a. 0 b. 37 c. 53 d. 90 e. 150 f. 180 [a. 0; b x 10-4 nm; c x 10-4 nm; d x 10-3 nm; e x 10-3 nm; f x 10-3 nm] 17. If a photon behaves as a particle and obeys particle mechanics, what happens if a photon and a particle collide head on? Set up equations obeying conservation of momentum and conservation of kinetic energy for a collision between a photon and an electron. Explain in words what happens. PROBLEMS Each of the following problems may involve more than one physical concept. When appropriate, the answer is given in brackets at the end of the problem. 18. A photon of wavelength 7.2 x m is scattered at π/2 radians. a. Find the energy of the scattered photons. b. Find the energy of the electrons after collision. [a. 129 kev; b kev] 19. Derive the equation for an electron debroglie wavelength after the electron has been accelerated from the rest throughv volts. (Assume nonrelativistic speeds.) 20. Find the energy of an electron with a wavelength equal to the circumference of the Bohr orbit in the hydrogen atom. (Hint: circumference = 2π x 5.3 x m) [13.8 ev] 21. Find the ratio of the de Broglie wavelength of an electron to that of a proton with the same energy. [SQR RT(m p /m e ] 22. The resolving power of a microscope is proportional to the wavelength divided by the numerical aperture. Calculate the ratio of the resolving power of an electron microscope using 10,000-V electrons to that of a microscope using 500-nm light with the same numerical aperture for both. [4.07 x 10 4 ] 23. Find the uncertainty in the momentum and energy of an electron within the nucleus (Δx = m). What does your answer suggest regarding electrons in the nucleus [6.63 x kg - m/sec, 1.51 x ev]

22 Assume the uncertainty in the lifetime of a particle is the time it takes light to travel across the nucleus. Find the uncertainty in the energy of this particle. (This was a calculation used to estimate the rest mass energy of the π meson, the nuclear force field particle.) [1.99 x J, 1.24 x 10 8 ev] 25. Suppose that a 40-g bullet has its position known to within the diameter of the rifle (1 cm). Find its uncertainty in momentum. If it is aimed at a target at a distance that it takes 3 sec to reach, by how much will it miss its target due to the uncertainty principle? [6.63 x kg m/sec, 4.98 x m] 26. If the energy of a particle of mass m is fv where V is the barrier height energy (Figure 24.4), and f is a fraction less than one, find the relation between V and the barrier width (Δx) that will just allow tunneling using the uncertainty principle. [Δx h/sqr RT(2mfV)] 27. Show that if an electron's debroglie wavelength equals the width of a barrier m wide, the electron can just pass over a 145-eV barrier. 28. Use the uncertainty principle to find the energy uncertainty for an electron confined to a space corresponding to a virus D nm in diameter. What physical significance can you give this energy value? [ E = (h/d) 2 /2m = (hc) 2 /2mc 2 D 2 = {(1240 ev nm) 2 /2 x 0.51 x 10 6 ev}(1 /D 2 )] 29. Calculate the ratio of photon energy to electron energy if both the photon and electron have the same wavelength. [E p /E e = v/c = SQR RT(p 2 /(p 2 + m 0 2 c 2 ))] 30. The mass of a body is a function of velocity in accordance with Equation Plot the mass of an electron for velocities of 0.1c to 0.99c. (Rule of thumb: We neglect change in mass for velocities below 0.1c.) 31. What is the mass of an electron accelerated through a potential of a. 100 kv? b. 2 x 10 6 V? [a. 1.19m 0 ; b 4.8 m 0 ] 32. What is the velocity of an electron accelerated through a. 100 kv b. 2 x 10 6 V [a. v 0.54 c; b. v 0.98c]