Chapter 29 Particles and Waves

Transcription

1 Chapter 29 PARTICLES AND WAVES PREVIEW A photon is the smallest particle of light, and has an energy which is proportional to its frequency. The photon nature of light is the principle behind the photoelectric effect, in which the absorption of photons of a certain frequency causes electrons to be emitted from a metal surface. The Compton effect also verifies the photon nature of light by showing that momentum is conserved in a collision between a photon and an electron. The Heisenberg uncertainty principle states that both the position and momentum of a subatomic particle cannot be measured precisely. Since light waves exhibit particle (photon) properties, de Broglie suggested that particles, such as electrons, can exhibit wave properties. The content contained in all sections of chapter 29 of the textbook is included on the AP Physics B exam. QUICK REFERENCE Important Terms blackbody radiation the radiation emitted by a blackbody, or perfect emitter and absorber of light, due to its temperature Compton effect the interaction of photons with electrons resulting in the increased wavelengths of the photons and kinetic energy of the electrons Compton wavelength of an electron half the maximum wavelength change of a photon in a Compton scattering with an electron de Broglie wavelength the wavelength associated with a moving particle with a momentum mv Heisenberg uncertainty principle the more accurately one determines the position of a subatomic particle, the less accurately its momentum is known photoelectric effect the ejection of electrons from certain metals when exposed to light of a minimum frequency photon the smallest particle of light Planck s constant the quantity that results when the energy of a photon is divided by its frequency 324

2 quantized a quantity that cannot be divided into smaller increments forever, for which there exists a minimum, quantum increment quantum mechanics the study of the properties of matter using its wave properties wave-particle duality under certain circumstances, waves can behave like particles, and particles can behave like waves work function the minimum energy required to release an electron from a metal Equations and Symbols E = hf c = W o KE fλ = hf max h p = λ h λ = mv E = pc o λ λ = = hf W h mc o = ev ( 1 cosθ ) stop where E = energy of a photon c = speed of light = 3 x 10 8 m/s f = frequency of light λ = wavelength of light W o = work function of a photoemissive surface (denoted by φ on the AP Physics exam) h = Planck s constant = 6.63 x J s = 4.14 x ev s f o = threshold frequency of a photoemissive surface KE max = maximum kinetic energy of electrons emitted in the photoelectric effect e = charge on one electron V stop = voltage needed to stop the emission of electrons λ = wavelength of a photon after being scattered by a collision with an electron θ = angle between the scattered photon and electron after they collide p = momentum of a photon m = mass of a moving particle v = speed or velocity Ten Homework Problems Chapter 29 Problems 5, 8, 10, 15, 17, 23, 24, 28, 29,

3 DISCUSSION OF SELECTED SECTIONS The Wave-Particle Duality, Blackbody Radiation and Planck s Constant and Photons and the Photoelectric Effect In prior chapters we treated light as a wave. But there are circumstances when light behaves more like it is made up of individual bundles of energy, separate from each other, but sharing a wavelength, frequency, and speed. The quantum of light is called the photon. light wave c photon c In the late 19 th century, Heinrich Hertz discovered an effect which could not be explained by the wave model of light. He shined ultraviolet light on a piece of zinc metal, and the metal became positively charged. Although he did not know it at the time, the light was causing the metal to emit electrons. This effect of using light to cause electrons to be emitted from a metal is called the photoelectric effect. According to the theory of light at the time, light was considered a wave, and should not be able to knock electrons off of a metal surface. At the turn of the 20 th century, Max Planck showed that light could be treated as tiny bundles of energy called photons, and the energy of a photon was proportional to its frequency. Thus, a graph of photon energy E vs. frequency f looks like this: Energy E E slope = = h f f frequency The slope of this line is a constant that occurs many times in the study of quantum phenomena called Planck s constant. Its symbol is h, and its value is 6.62 x J s (or J/Hz). The equation for the energy of a photon is E = hf or, since hc E = λ c f =, λ 326

4 The energy of a photon is proportional to its frequency, but inversely proportional to its wavelength. This means that a violet has a higher frequency and energy than a red photon. Oftentimes when dealing with small amounts of energy like that of photons or electrons, we may prefer to use a very small unit of energy called the electron-volt (ev). The conversion between joules and electron-volts is 1 ev = 1.6x10 19 J Planck s constant can be expressed in terms of electron-volts as h = 4.14x10 15 ev s In 1905, Albert Einstein used Planck s idea of the photon to explain the photoelectric effect: one photon of energy that is higher than the energy (work function φ) which binds the electron to the metal is absorbed by one electron in the metal surface, giving the electron enough energy be released from the metal. Any energy left over from the photon after the work function has been met becomes the kinetic energy of the electron. Photon E KE KE max max = E φ photon = hf hf 0 where f o is called the threshold frequency, which is the minimum frequency the incoming photon must have to dig the electron out of the metal surface. KE φ e metal Example 1 The metal sodium has a threshold frequency which corresponds to yellow light. Describe what will happen if (a) yellow light is shined on the sodium surface, (b) red light is shined on the metal surface, (c) green light is shined on the metal surface, (d) bright green light is shined on the metal surface. Solution (a) If yellow light is shined on a sodium surface, the yellow photons will be absorbed by electrons in the metal, causing them to be released, but there will be no energy left over for the electrons to have any kinetic energy. 327

5 (b) Red light has a lower frequency and energy than yellow light, and therefore red photons do not have enough energy to release the electrons from the sodium surface. (c) Green light has a higher frequency and energy than yellow light, and therefore a green photon will be absorbed by a sodium electron and the electron will be released from the metal and have kinetic energy. (d) If a brighter (more photons) green light is shined on the surface, more electrons will be emitted, since one photon can be absorbed by one electron. If these electrons are funneled into a circuit, we can use them as current in an electrical device. The photoelectric effect is the principle behind any process in which light produces electricity, such as a solar calculator or an auto-focus camera. The graph of maximum kinetic energy of a photoelectron vs. frequency of light incident on a sodium surface would look like this: Kinetic Energy E slope = = h f R O Y G B V frequency Note that the electrons have no kinetic energy up to the threshold frequency (color), and after that their kinetic energy is proportional to the frequency of the incoming light. Example 2 photon photoemissive surface e + Adjustable Voltage 328

6 Light is shined on a photoemissive surface of work function φ = 2.0 ev and electrons are released with a kinetic energy KE max = 4.0 ev. (a) What voltage, called the stopping voltage V stop, would be necessary to stop the emission of electrons? (b) Determine the energy of each of the incoming photons in ev and in Joules. (c) Determine the frequency of the incoming photons in Hz. (d) If photons of wavelength λ = 2.5x10-7 m were shined on this photoemissive surface, would electrons be emitted from the surface? Justify your answer. Solution (a) Stopping the emission of electrons requires work equal to the maximum kinetic energy of the electrons: KE max = W = q V e stop 19 ( 4.0eV )( 1.6x10 J / ev ) KEmax Vstop = = = 4.0Volts 19 qe 1.6x10 C Thus, it would take 4.0 V to stop electrons with a kinetic energy of 4.0 ev, as we might expect. (b) E photons 6.0eV = KE + φ = 4.0 ev ev max ( 1.6x10 J / ev ) = 9.6x10 J = 6.0 ev 19 E photon 9.6x10 J 15 (c) f = = = 1.5x10 Hz 34 h 6.6x10 J / Hz (d) The wavelength of these incoming photons corresponds to a frequency of 8 c 3x10 m / s 15 f = = = 1.2x10 Hz 7 λ 2.5x10 m Now we need to check to see if this frequency is higher than the threshold frequency f 0 : φ 2.0eV 14 f 0 = = = 4.8x10 Hz. 15 h 4.14x10 ev / Hz The incoming frequency is higher than the threshold frequency, so electrons will be emitted from the metal surface. 329

7 29.4 The Momentum of a Photon and the Compton Effect Since a photon has energy, does it follow that it has momentum? Recall in an earlier chapter that we defined momentum as the product of mass and velocity. But a photon has no mass. It turns out that in quantum physics, photons do have momentum which is inversely proportional to its wavelength. The equation for the momentum of a photon is h p = λ Photons can and do impart momentum to subatomic particles in collisions that follow the law of conservation of momentum. This phenomena was experimentally verified by Arthur Compton in Compton aimed x-rays of a certain frequency at electrons, and when they collided and scattered, the x-rays were measured to have a lower frequency indicating less energy and momentum. The scattering of x-ray photons from an electron with a loss in energy of the x-ray photon is called the Compton effect. It is difficult to understand how a photon, having only energy and no mass, can collide with a particle like an electron and change its momentum, but this has been verified experimentally many times. Example 3 Before After photon c electron c photon electron A photon is fired at an electron which is initially at rest. The photon strikes the electron and reverses its direction, as shown in the diagram representing the photon and electron after the collision. (a) Determine the shift in wavelength of the photon as a result of the collision. The photon is an x-ray with a wavelength of 6.62 x m. (b) Determine i. the initial momentum of the photon before the collision with the electron. ii. the final momentum of the photon after the collision with the electron. (c) Find the final momentum of the electron after the collision. 330

8 Solution (a) The Compton equation gives the shift in wavelength: x10 J / Hz λ λ = ( 1 cosθ ) = 1 cos180 = 31 8 mc 9.11x10 kg 3x10 m / s h 12 ( )( ) [ ] x m 34 h 6.62x10 J / Hz 23 (b) i. p = = = 1.00x10 kgm / s λ x10 m ii. The momentum of the photon after the collision p corresponds to the new wavelength λ : 34 h h 6.62x10 J / Hz 24 p = = = = 9.32x10 kgm / s λ λ + λ 6.62x10 m x10 m The momentum of the photon is given a negative sign, since it reverses direction after the collision. (c) By the law of conservation of momentum, the momentum lost by the photon must have been gained by the electron p = p p = 9.32x10 kgm / s 1.00x10 kgm / s = 1.93x10 kgm / ( ) ( ) s lost by the photon. Thus, the momentum gained by the electron is 1.93 x kgm/s The de Broglie Wavelength and the Wave Nature of Matter In 1924, Louis de Broglie reasoned that if a wave such as light can behave like a particle, having momentum, then why couldn t particles behave like waves? If the momentum of a photon can be found by the equation λ = h p h p =, then the wavelength can be found by λ. De Broglie suggested that for a particle with mass m and speed v, we could write the equation as λ = h mv, and the wavelength of a moving particle could be calculated. This hypothesis was initially met with a considerable amount of skepticism until it was shown by Davisson and Germer in 1927 that electrons passing through a nickel crystal were diffracted through the crystal, producing a diffraction pattern on a photographic plate. Thus, de Broglie s hypothesis that particles could behave like waves was experimentally verified. Nuclear and particle physicists must take into account the wave behavior of subatomic particles in their experiments. We typically don t notice the wave properties of objects moving around us because the masses are large in comparison to subatomic particles and the value for Planck s constant h is extremely small. But the wavelength of any moving mass is inversely proportional to the momentum of the object. 331

9 29.6 The Heisenberg Uncertainty Principle Since a photon is the smallest and most unobtrusive measuring device we have available to us, and even a photon has too large of a momentum to make accurate measurements of the speed and position of subatomic particles, we must admit to an uncertainty that will always exist in quantum measurements. This limit to accuracy at this level was formulated by Werner Heisenberg in 1928 and is called the Heisenberg uncertainty principle. It can be stated like this: There is a limit to the accuracy of the measurement of the speed (or momentum) and position of any sub-atomic particle. The more accurately we measure the speed of a particular particle, the less accurately we can measure its position, and vice-versa. CHAPTER 29 REVIEW QUESTIONS For each of the multiple-choice questions below, choose the best answer. 1. The smallest, discrete value of any quantity in physics is called the (A) atom (B) molecule (C) proton (D) electron (E) quantum 2. The smallest discrete value of electromagnetic energy is called the (A) photon (B) proton (C) electron (D) neutron (E) quark 3. Which of the following photons has the highest energy? (A) x-ray (B) ultraviolet light (C) green light (D) microwave (E) radio wave 4. The threshold frequency of zinc for the photoelectric effect is in the ultraviolet range. Which of the following will occur if x-rays are shined on a zinc metal surface? (A) No electrons will be emitted from the metal. (B) Electrons will be released from the metal but have no kinetic energy. (C) Electrons will be released from the metal and have kinetic energy. (D) Electrons will be released from the metal but then will immediately be recaptured by the zinc atoms. (E) Electrons will simply move from one zinc atom in the metal to another zinc atom in the metal. 332

10 5. A metal surface has a threshold frequency for the photoelectric effect which corresponds to green light. If blue light is shined on this metal, (A) no electrons will be emitted from the metal. (B) the number of emitted electrons is proportional to the brightness (intensity) of the blue light. (C) the electrons will have no kinetic energy. (D) more electrons will be emitted than if green light were shined on the metal. (E) electrons will be emitted from the metal, but since the light is not green, only a few electrons will be released. 6. Light is shined on a metal surface which exhibits the photoelectric effect according to the graph shown. What color(s) correspond to the threshold frequency of the metal? (A) red only (B) red and orange (C) red, orange, yellow, and green (D) blue only (E) blue, indigo, and violet KE R O Y G B V 7. Which of the following is true of the momentum of a photon? (A) It is proportional to the wavelength of the photon. (B) It is inversely proportional to the wavelength of the photon. (C) It is inversely proportional to the square of the wavelength of the photon. (D) It is proportional to the mass of the photon. (E) It is equal to the energy of the photon. 8. The Heisenberg uncertainty principle implies that (A) Electrons are too small to be studied. (B) Every photon is exactly the same size. (C) The more you know about the momentum of an electron, the less you can know about its position. (D) The more you know about the energy of a photon, the less you can know about its frequency. (E) You cannot state with accuracy the number of electrons in an atom. 9. Which of the following statements is true for the de Broglie wavelength of a moving particle? (A) It is never large enough to measure (B) It is proportional to the speed of the particle. (C) It is inversely proportional to the momentum of the particle. (D) It is equal to Planck s constant. (E) It has no effect on the behavior of electrons. 333

11 10. When a photon transfers momentum to an electron, the wavelength of the photon (A) increases (B) decreases (C) remains the same (D) is equal to the wavelength of the electron (E) is always in the x-ray range Free Response Question Directions: Show all work in working the following question. The question is worth 15 points, and the suggested time for answering the question is about 15 minutes. The parts within a question may not have equal weight. 1. (15 points) photon photoemissive surface e + Adjustable Voltage Light of a certain wavelength is shined on a photoemissive surface, ejecting electrons as shown above. The graph below shows the maximum kinetic energy of each electron (x J) vs. frequency of the incoming light (x Hz). (a) On the graph below, draw the line that is your estimate of the best straight-line fit to the data points. (b) Using your graph, find a value for Planck s constant, and briefly explain how you found the value. (c) From the graph, estimate the threshold frequency of the photoemissive surface. 334

12 (d) Photons of frequency 7.0 x Hz are shined on the metal surface. Determine the i. kinetic energy of the emitted electrons ii. speed of the emitted electrons iii. de Broglie wavelength of the emitted electrons. ANSWERS AND EXPLANATIONS TO CHAPTER 29 REVIEW QUESTIONS Multiple Choice 1. E The quantum is the smallest discrete value of any quantity, such as the electron for charge and the photon for light. 2. A A photon is the smallest bundle of light energy. 3. A The x-ray has the highest frequency of the choices, and since energy is proportional to frequency, has the highest energy as well. 335

13 4. C Since the frequency of x-rays is higher than the ultraviolet threshold frequency, electrons will be emitted from the metal and have kinetic energy left over. 5. B After the threshold frequency is met, the number of photons (brightness) dictate how many electrons are emitted, since one photon can release one electron. Thus, a brighter light will release more electrons. 6. D The electrons begin being released when blue light is shined on the metal, so blue has the threshold (minimum) frequency for this metal. 7. B Since the equation for the momentum of a photon is p = h/λ, the momentum is inversely proportional to the wavelength of the photon, implying that a photon with a shorter wavelength has a higher momentum than one with a longer wavelength. 8. C Heisenberg s uncertainty principle states that you have to sacrifice your knowledge of the position of any subatomic particle to know its momentum accurately, and vice-versa. 9. C According to the equations for the de Broglie wavelength, the higher the momentum of the particle, the shorter its wavelength. 10. A A decrease in the photon s momentum corresponds to an increase in the photon s wavelength, since momentum and wavelength are inversely proportional to each other. Free Response Question Solution (a) 3 points The best-fit straight line represents the average of the data points, and therefore there would be some data points above the line and some below the line. The best-fit line does not necessarily connect the first and last points, and does not necessarily pass through any particular data point. You should always use a straight-edge to draw a best-fit line that you know to be straight. 336

14 (b) 4 points Planck s constant is equal to the slope of the graph. Let s choose two convenient points on the line to find the slope, (4.0 x Hz, 0) and (8.0 x Hz, 20.0 x J). Then the slope would be 20 KE 20.0x10 J 0 34 h = = = 5.7x10 J / Hz f 8.0x10 Hz 4.5x10 Hz This value for h is close to the actual value of h within a reasonable margin of error. (c) 2 points The threshold frequency can be found by marking the place where the graph crosses the frequency axis. From the graph, f 0 = 4.5 x Hz. (d) 6 points i. From the graph, the KE associated with the frequency of 7.0 x Hz is 14 x J, or 1.4 x J. 1 2 KE = mv 2 ii. 19 2KE 2( 1.4x10 J ) 5 v = = = 5.5x10 m / s 31 m 9.1x10 kg 34 h 6.6x10 J / Hz 9 iii. λ = = = 1.3x10 m mv 31 5 ( 9.1x10 kg)( 5.5x10 m / s) 337