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OpenStax College Physics Instructor Solutions Manual Chapter 29 CHAPTER 29: INTRODUCTION TO QUANTUM PHYSICS 29.1 QUANTIZATION OF ENERGY 13 A LiBr molecule oscillates with a frequency of 1.7 × 10 Hz (a) What is the difference in energy in eV between allowed oscillator states? (b) What is the approximate value of n for a state having an energy of 1.0 eV? Solution (a) ∆E = hf = ( 6.63 × 10 −34 J ⋅ s )(1.7 × 1013 s −1 ) = 1.127 × 10 −20 J −2 −2 (1.127 × 10 −20 J ) 1.601×eV = 7.04 × 10 eV = 7.0 × 10 eV −19 10 J so that (b) Using the equation E = nhf , we can solve for n : ( ) ( E 1.0 eV) 1.60 ×10 −19 J/eV n= − = − = 13.7 = 14 −34 13 −1 hf 6.63 ×10 J ⋅ s 1.7 ×10 s ( Solution )( ) The difference in energy between allowed oscillator states in HBr molecules is 0.330 eV What is the oscillation frequency of this molecule? ∆E = hf ⇒ f = ∆E ( 0.330 eV) (1.60 ×10 −19 J/eV) = = 7.96 ×1013 Hz h 6.63 ×10 −34 J ⋅ s A physicist is watching a 15-kg orangutan at a zoo swing lazily in a tire at the end of a rope He (the physicist) notices that each oscillation takes 3.00 s and hypothesizes that the energy is quantized (a) What is the difference in energy in joules between allowed oscillator states? (b) What is the value of n for a state where the energy is 5.00 J? (c) Can the quantization be observed? Solution 1 = = 0.333sec−1 T 3.00 s (a) so that ∆E = hf = ( 6.63 × 10 −34 J ⋅ s )( 0.333sec−1 ) = 2.21 ×10 −34 J E E 5.00 J n= − = − = − = 2.26 × 1034 − 34 hf ∆E 2.210 × 10 J (b) (c) No, ∆E is much too small and n is much too large f = 29.2 THE PHOTOELECTRIC EFFECT What is the longest-wavelength EM radiation that can eject a photoelectron from OpenStax College Physics Instructor Solutions Manual Chapter 29 silver, given that the binding energy is 4.73 eV? Is this in the visible range? Solution hc = BE ⇒ λ hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) λ= = = 2.63 × 10 −7 m = 263 nm BE ( 4.73 eV) (1.60 ×10 −19 J/eV) KE = hf − BE; KE = ⇒ hf = No, this is UV Find the longest-wavelength photon that can eject an electron from potassium, given that the binding energy is 2.24 eV Is this visible EM radiation? Solution hc = BE ⇒ λ hc ( 6.63 ×10 −34 J ⋅ s )( 3.00 ×108 m/s ) λ= = = 5.55×10 −7 m = 555 nm BE ( 2.24 eV) (1.60 ×10 −19 J/eV) Yes, it is visible as green light What is the binding energy in eV of electrons in magnesium, if the longest-wavelength photon that can eject electrons is 337 nm? Solution KE = hf − BE; KE = ⇒ hf = hc ⇒ λ hc 6.63 ×10 −34 J ⋅ s 3.00 ×108 m/s BE = = λ 3.37 ×10 −9 m (1 eV) = 5.902 ×10 −19 J × = 3.69 eV 1.60 ×10 −19 J KE = hf − BE; KE = ⇒ BE = hf = ( )( ( ( ) ) ) Calculate the binding energy in eV of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm Solution The longest wavelength corresponds to the shortest frequency, or the smallest energy Therefore, the smallest energy is when the kinetic energy is zero From the equation KE = hf − BE = , we can calculate the binding energy (writing the frequency in hc ⇒ λ hc 6.63 ×10 −34 J ⋅ s 3.00 ×108 m/s BE = = λ 3.04 ×10 −7 m BE = hf = ( terms of the wavelength): ( )( ) ) 1.000 eV = 6.543×10 −19 J × = 4.09 eV −19 1.60 ×10 J OpenStax College Physics Solution Instructor Solutions Manual Chapter 29 What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-nm EM radiation, given that the binding energy is 2.28 eV? hf = hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) = λ ( 4.50 × 10−7 m) 1.000 eV = 4.42 × 10 −19 J × = 2.7625 eV −19 1.60 × 10 J KE = hf − BE = 2.7625 eV − 2.28 eV = 0.483 eV Solution UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV What is the maximum kinetic energy of the ejected photoelectrons? hf = hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) = λ (1.20 × 10 −7 m ) 1.000 eV = 1.6575 × 10 −18 J × = 10.359 eV −19 1.602 × 10 J KE = hf − BE = 10.359 eV − 4.82 eV = 5.54 eV 10 Solution Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal What is the binding energy of electrons to sodium metal? hf = hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) = λ ( 4.00 × 10−7 m ) 1.000 eV = 4.9725 × 10 −19 J × = 3.1078 eV −19 1.60 × 10 J KE = hf − BE ⇒ BE = hf − KE = 3.1078 eV − 0.86 eV = 2.25 eV 11 Solution UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV electrons What is the binding energy of electrons to uranium metal? hf = hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 10 m/s ) = λ (3.00 × 10 −7 m ) 1.000 eV = 6.63 × 10 −19 J × = 4.14 eV −19 1.602 × 10 J KE = hf − BE ⇒ BE = hf − KE = 4.14 eV − 0.500 eV = 3.64 eV 12 What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium metal, given that the binding energy is 2.71 eV? What type of EM radiation is this? OpenStax College Physics Solution Instructor Solutions Manual Chapter 29 KE = hf − BE ⇒ 1.60 × 10 −19 J = 7.536 × 10 −19 J hf = BE + KE = 2.71 eV + 2.00 eV = 4.71 eV × eV −34 hc hc 6.63 × 10 J ⋅ s 3.00 × 10 m/s hf = ⇒λ = = = 2.639 × 10 −7 = 264 nm λ hf 7.536 × 10 −19 J This is ultraviolet radiation ( )( ( ) ) 13 Find the wavelength of photons that eject 0.100-eV electrons from potassium, given that the binding energy is 2.24 eV Are these photons visible? Solution KE = hf − BE and c = λf so λ= hc ( 6.63 ×10 −34 J ⋅ s)(3.00 ×108 m/s ) × 1.000 eV = −19 BE + KE 2.24 eV + 0.100 eV 1.60 × 10 J = 5.313× 10 −7 m = 531nm Yes, these photons are visible 14 Solution What is the maximum velocity of electrons ejected from a material by 80-nm photons, if they are bound to the material by 4.73 eV? hc − BE λ 1.60 × 10 −19 J 6.63 × 10 −34 J ⋅ s 3.00 × 108 m/s ( ) = − 4.73eV 8.00 × 10 −8 m eV KE = hf − BE = ( ( )( ) ) = 1.7295 × 10 −18 J ( ) 1 2KE ( ) 1.729 × 10 −18 J KE = mv ⇒ v = = = 1.95 × 10 m/s −31 m 9.11 × 10 kg 15 Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420nm photons Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device? OpenStax College Physics Solution Instructor Solutions Manual hc − BE λ ( 6.63 × 10 −34 J ⋅ s)(3.00 × 108 m/s ) − ( 2.71eV ) 1.60 × 10 −19 = eV ( 4.20 × 10 −7 m) Chapter 29 KE = hf − BE = J = 3.997 × 10 − 20 J 1 2KE ( ) ( 3.997 × 10 −20 J ) KE = mv ⇒ v = = = 2.962 × 10 m/sec −31 m 9.11 × 10 kg d 2.50 × 10 −2 m t= = = 8.44 × 10 −8 s v 2.962 × 10 m/sec 16 A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.31 eV? Solution (a) Eγ = hf = ( )( ) hc 6.63 × 10 −34 J ⋅ s 3.00 × 10 m/s = = 4.9725 × 10 −19 J λ 4.00 × 10 −7 m ( ) 2.00 ×10 −3 J/s = 4.02 ×1015 /s −19 4.9725 ×10 J KE = hf − BE = Eγ − BE N= 1.60 × 10 −19 J = 1.3245 × 10 −19 J = 4.9725 × 10 J − ( 2.28 eV) eV 15 −19 P = NKE = 4.022 × 10 /s 1.3245 × 10 J = 5.33 × 10 −4 W = 0.533 mW −19 (b) 17 ( )( ) (a) Calculate the number of photoelectrons per second ejected from a 1.00-mm area of sodium metal by 500-nm EM radiation having an intensity of 1.30 kW/m (the intensity of sunlight above the Earth’s atmosphere) (b) Given that the binding energy is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons Where does the other power go? How can it be recovered? Solution (a) First calculate n, the number of photons per second hitting a square-meter area Each photon has energy ( 6.626 ×10 −34 J ⋅ s)( 2.998 ×108 m/s ) = 3.973 ×10−19 J ⇒ Eγ = (5.00 ×10 −7 m ) 1.30 ×10 J/s ⋅ m = 3.272 ×10 21 / s ⋅ m −19 3.973 ×10 J The number of ejected electrons per second equals the number of photons per second hitting the sodium metal, or N = nA : n= OpenStax College Physics Instructor Solutions Manual ( )( Chapter 29 ) 1m 15 N = 3.272 × 10 /s ⋅ m 1.00 nm = 3.27 × 10 /s 1000 nm hc KE = hf − BE = − BE λ 1.60 × 10 −19 J = 3.204 × 10 − 20 J = 3.973 × 10 −19 J − ( 2.28 eV) eV (b) 21 2 ( )( ) P = N ⋅ KE = 3.272 × 1015 /s 3.204 × 10 − 20 J = 1.05 × 10 − W = 0.105 mW (c) The other energy goes to the sodium metal to free the electrons (binding energy) This lost power can be recovered by the spontaneous absorption of electrons by sodium metal For each electron absorbed an energy of 2.28eV will be released 18 Unreasonable Results Red light having a wavelength of 700 nm is projected onto magnesium metal to which electrons are bound by 3.68 eV (a) Use KE e = hf – BE to calculate the kinetic energy of the ejected electrons (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? Solution ( (a) )( ) hc 6.63 × 10 −34 J ⋅ s 3.00 × 108 m/s = λ 7.00 × 10 −7 m eV = 1.776 eV = 2.841× 10 −19 J −19 1.602 × 10 J hf = ( )( ) and since KE = hf − BE = 1.776 eV − 3.68 eV = − 1.90 eV (b) Negative kinetic energy is impossible (c) The assumption that the photon can knock the electron free is unreasonable 19 Unreasonable Results (a) What is the binding energy of electrons to a material from which 4.00-eV electrons are ejected by 400-nm EM radiation? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? Solution (a) We want to use the equation KE = hf − BE to determine the binding energy, so we first need to determine an expression of hf Using E = hf , we know: ( )( ) hc 6.626 × 10 −34 J ⋅ s 2.998 × 108 m/s = λ 4.00 × 10 −7 m eV = 3.100 eV = 4.966 × 10 −19 J −19 1.602 × 10 J and since KE = hf − BE : BE = hf − KE = 3.100 eV − 4.00 eV = − 0.90 eV hf = ( )( ) (b) The binding energy is too large for the given photon energy (c) The electron’s kinetic energy is too large for the given photon energy; it cannot be OpenStax College Physics Instructor Solutions Manual Chapter 29 greater than the photon energy 29.3 PHOTON ENERGIES AND THE ELECTROMAGNETIC SPECTRUM 20 Solution What is the energy in joules and eV of a photon in a radio wave from an AM station that has a 1530-kHz broadcast frequency? E = hf = ( 6.63 × 10 −34 J/s )(1.53 × 10 s −1 ) = 1.0144 × 10 −27 J = 1.01 × 10 −27 J eV = 1.0144 × 10 −27 J = 6.340 × 10 −9 eV = 6.34 × 10 −9 eV −19 1.60 × 10 J 21 (a) Find the energy in joules and eV of photons in radio waves from an FM station that has a 90.0-MHz broadcast frequency (b) What does this imply about the number of photons per second that the radio station must broadcast? E = hf = ( 6.63 × 10 −34 J/s)( 9.00 × 108 s −1 ) = 5.97 × 10 −26 J Solution eV −7 = 5.97 × 10 − 26 J = 3.73 × 10 eV −19 1.60 × 10 J (a) (b) This implies that a tremendous number of photons must be broadcast per second In order to have a broadcast power of, say 50.0 kW, it would take 5.00 × 10 J/s = 8.38 × 10 29 photon/sec −26 5.97 × 10 J/photon 22 Solution 23 Calculate the frequency in hertz of a 1.00-MeV γ -ray photon E = hf = f = E 1.00 × 10 eV = = 2.42 × 10 20 Hz −15 h 4.14 × 10 eV ⋅ s (a) What is the wavelength of a 1.00-eV photon? (b) Find its frequency in hertz (c) Identify the type of EM radiation Solution (a) E= hc hc 1240 eV ⋅ nm ⇒λ= = = 1240 nm = 1.24 µm λ E 1.00 eV c 3.00 ×108 m/s = = 2.42×1014 Hz −6 λ 1.24 ×10 m (b) (c) The radiation is in the infrared part of the spectrum f = 24 Do the unit conversions necessary to show that hc = 1240 eV ⋅ nm , as stated in the text OpenStax College Physics Instructor Solutions Manual Chapter 29 Solution hc = ( 6.63 × 10 25 Confirm the statement in the text that the range of photon energies for visible light is 1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm Solution 26 − 34 10 nm 1.00 eV J ⋅ s )( 3.00 × 10 m/s ) = 1240 eV ⋅ nm −19 m 1.60 × 10 J Emin = hc 1240 eV ⋅ nm = = 1.63 eV λ max 760 nm Emax = hc 1240eV⋅ nm = = 3.26 eV λ 380 nm 13 (a) Calculate the energy in eV of an IR photon of frequency 2.00 × 10 Hz (b) How many of these photons would need to be absorbed simultaneously by a tightly bound molecule to break it apart? (c) What is the energy in eV of a γ ray of frequency 3.00 × 1020 Hz ? (d) How many tightly bound molecules could a single such γ ray break apart? Solution (a) E = hf = ( 6.63 ×10 −34 J ⋅ s )( 2.00 ×1013 s −1 ) 1eV = 0.0829 eV 1.60 ×10 -19 J 10.0 eV = 121 −1 (b) 8.29 × 10 eV −16 20 −1 (c) E = hf = ( 4.144 ×10 eV)(3.00 ×10 s ) = 1.242 × 10 eV = 1.24 MeV 1.242 ×10 eV = 1.24 ×105 10.0 eV (d) 27 −15 Prove that, to three-digit accuracy, h = 4.14 × 10 eV ⋅ s , as stated in the text Solution 1.00 eV h = ( 6.63 × 10 −34 J ⋅ s ) −19 1.60 × 10 28 (a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV accelerating potential, such as a color TV? (b) What is their frequency? Solution (a) (b) 29 ( )( −15 −15 = 4.144 × 10 eV ⋅ s = 4.14 × 10 eV ⋅ s J qv = 1.60 ×10 −19 C 2.50 ×10 V E = hf ⇒ f = eV ) 1.601.00 ×10 −19 J = 2.50 ×10 eV E 2.5.0 ×10 eV = = 6.04 ×1018 Hz h 4.14 ×10 −15 eV ⋅ s What is the accelerating voltage of an x-ray tube that produces x rays with a shortest wavelength of 0.0103 nm? OpenStax College Physics Solution qV = E = Instructor Solutions Manual ( )( )( Chapter 29 ) hc hc 6.63 × 10 −34 J ⋅ s 3.00 × 10 m/s ⇒V = = = 1.21 × 10 V = 121 kV −19 −11 λ qλ 1.60 × 10 C 1.03 × 10 m ( ) 30 (a) What is the ratio of power outputs by two microwave ovens having frequencies of 950 and 2560 MHz, if they emit the same number of photons per second? (b) What is the ratio of photons per second if they have the same power output? Solution (a) Let N be the number of photons per second P N hf N f 2560 MHz ⇒ P = NEγ = Nhf = 1 = 1 = = 2.69 P2 N hf N f 950 MHz since N = N P1 N hf N f 950 MHz = 1 = = = = 0.371 f1 2560 MHz (b) P2 N hf N since P1 = P2 31 How many photons per second are emitted by the antenna of a microwave oven, if its power output is 1.00 kW at a frequency of 2560 MHz? Solution Eγ = hf = 6.63 × 10 −34 2.56 × 10 s −1 = 1.697 × 10 −24 J ( )( ) 1.00 × 103 J/s N= = 5.89 × 10 26 photons/second − 24 1.697 × 10 J/photon 32 Some satellites use nuclear power (a) If such a satellite emits a 1.00-W flux of γ rays having an average energy of 0.500 MeV, how many are emitted per second? (b) These γ rays affect other satellites How far away must another satellite be to only receive one γ ray per second per square meter? ( Solution )( N= (a) 1.00 × 103 J/s = 1.25 × 1013 photons/s −14 8.00 × 10 J/photon Φ N = photons/m = (b) 33 Solution ) Eγ = 5.00 ×10 eV 1.60 × 10 −19 J/eV = 8.00 × 10 −14 J N r = 4π Φ N N ⇒ 4π r 1.25 × 1013 photon/s = 4π photon/s ⋅ m ( ) = 9.97 × 10 m = 997 km (a) If the power output of a 650-kHz radio station is 50.0 kW, how many photons per second are produced? (b) If the radio waves are broadcast uniformly in all directions, find the number of photons per second per square meter at a distance of 100 km Assume no reflection from the ground or absorption by the air (a) ( )( ) Eγ = hf = 6.63 ×10 −34 Js 6.50 ×10 s −1 = 4.31×10 −28 J OpenStax College Physics N= Instructor Solutions Manual Chapter 29 5.00 × 10 J/s = 1.16 × 10 32 photon/s = 1.16 × 10 32 photon/s 4.31× 10 −28 J/photon Then, (b) To calculate the flux of photons, we assume that the broadcast is uniform in all directions, so the area is the surface area of a sphere giving: N 1.16 × 10 32 photons/s ΦN = = = 9.23 × 10 20 photons/s ⋅ m 2 4π r 4π 1.00 × 10 m ( 34 Solution How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV E γ = ( 7.50 × 10 eV)(1.60 × 10 −19 J/eV) = 1.20 × 10 −14 J N= 35 Solution 36 ) 1.00 J/s = 8.33 × 1013 photon/s −14 1.20 × 10 J/photon (a) How far away must you be from a 650-kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space (b) Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts N N ΦN = ⇒ r = 4π r 4π Φ N 1 1.161 × 1032 photon/s 15 = = 3.04 × 10 m 4π photon/s ⋅ m (a) (b) The distance calculated in part (a) is approximately 1/3 ly Therefore, if radio stations from intelligent life in other solar systems are to be detected, their broadcasts would have to have substantial power outputs Also, since there are stray radio waves in outer space, their signals would have to be large compared to the background radio waves This means it is rather unlikely for us to detect intelligent life by detecting their radio broadcasts ( ) Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.) OpenStax College Physics Solution ΦN = Instructor Solutions Manual 500 ( −3 π 1.50 ×10 m ) = 7.073 ×10 photons/s ⋅ m N N ΦN = ⇒ r = 4π r 4πΦ N = 1.81 ×10 m = 181 km Chapter 29 2.92 ×1019 photon/s = 4π 7.073 × 10 photon/s ⋅ m ( ) 29.4 PHOTON MOMENTUM 38 (a) Find the momentum of a 4.00-cm-wavelength microwave photon (b) Discuss why you expect the answer to (a) to be very small Solution h 6.63 × 10 −34 J ⋅ s = = 1.66 × 10 −32 kg ⋅ m/s −2 λ 4.00 × 10 m (a) (b) The wavelength of microwave photons is large, so the momentum they carry is very small 39 (a) What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? (b) What is its energy in MeV? p= Solution (a) (b) 40 p= h 6.63 ×10 −34 J ⋅ s = = 6.63 ×10 −23 kg.m/s −11 λ 1.00 ×10 m E= hc MeV = pc = 6.63 × 10 −23 kg ⋅ m/s 3.00 × 10 m/s = 0.124 MeV −13 λ 1.602 × 10 J ( )( ) −29 (a) What is the wavelength of a photon that has a momentum of 5.00 × 10 kg ⋅ m/s ? (b) Find its energy in eV Solution h h 6.63 × 10 −34 J.s ,λ= = = 1.326 × 10 −5 m = 13.3 àm 29 p 5.00 ì 10 kg ⋅ m/s p= (a) Using (b) Using E = pc = 5.00 × 10 −29 kg ⋅ m/s 3.00 × 108 m/s ( )( ) eV = 1.50 × 10 − 20 J × = 9.38 × 10 -2 eV −19 1.60 × 10 J 41 −21 (a) A γ -ray photon has a momentum of 8.00 × 10 kg ⋅ m/s What is its wavelength? (b) Calculate its energy in MeV Solution (a) λ= h 6.63 ×10 −34 J ⋅ s = = 8.29 ×10 −14 m − 21 p 8.00 ×10 kgm/s OpenStax College Physics Instructor Solutions Manual Chapter 29 E = pc = (8.00 × 10 −21 kgm/s)( 3.00 × 10 m/s ) (b) 42 MeV = 2.40 × 10 −12 J × = 15.0 MeV −13 1.602 × 10 J (a) Calculate the momentum of a photon having a wavelength of 2.50 μm (b) Find the velocity of an electron having the same momentum (c) What is the kinetic energy of the electron, and how does it compare with that of the photon? Solution (a) (b) pλ = pe = me ve ⇒ ve = KE e (c) = 2.06 ×10 Repeat the previous problem for a 10.0-nm-wavelength photon Solution (a) (b) pγ = h 6.63 ×10 −34 J ⋅ s = = 6.63 ×10 −26 kg.m/s λ 1.00 ×10 −8 m ve = pe 6.63 × 10 −34 kg ⋅ m/s = = 7.28 × 10 m/s me 9.11 × 10 −31 kg 2 KE e = mve = ( 0.5 ) ( 9.11 ×10 −31 kg )( 7.277 ×10 m/s ) = 2.41×10 −21J E γ = p γ c = ( 6.63 ×10 −26 kg ⋅ m/s )( 3.00 ×10 m/s ) = 1.99 ×10 −17 J; Eγ (c) 44 pe 2.652 × 10 −28 kgm/s = = 291 m/s me 9.11 × 10 −31 kg 2 KE e = mve = ( 0.5 ) ( 9.11 ×10 −31 kg )( 2.911 ×10 m/s ) = 3.86 ×10 −26 J Eγ = pγ c = ( 2.652 ×10 −28 kgm/s )( 3.00 ×10 m/s ) = 7.96 ×10 −20 J; Eγ 43 h 6.63 ×10 −34 Js = = 2.65 ×10 −28 kgm/s −6 λ 2.50 ×10 m KE e = 8.25 ×10 (a) Calculate the wavelength of a photon that has the same momentum as a proton moving at 1.00% of the speed of light (b) What is the energy of the photon in MeV? (c) What is the kinetic energy of the proton in MeV? ( Solution )( )( ) p p = mv p = 1.67 × 10 −27 kg 1.00 × 10 −2 3.00 × 10 m/s = 5.01 × 10 −21 kg ⋅ m/s pγ = (a) h h 6.63 × 10 −34 J ⋅ s ⇒λ= = = 1.32 × 10 −13 m − 21 λ p γ 5.01 × 10 kg ⋅ m/s OpenStax College Physics Instructor Solutions Manual Chapter 29 Eγ = pγ c = ( 5.01× 10 −21 kg ⋅ m/s )( 3.00 × 108 m/s ) (b) MeV = 1.503 × 10 −12 J × = 9.39 MeV −13 1.60 × 10 J ( )[ ( )( KE = mv = ( 0.5 ) 1.67 × 10 − 27 kg 1.00 × 10 − 3.00 × 108 m/s (c) = 7.515 × 10 −15 J = 4.70 × 10 − MeV 45 )] (a) Find the momentum of a 100-keV x-ray photon (b) Find the equivalent velocity of a neutron with the same momentum (c) What is the neutron’s kinetic energy in keV? Solution (a) E = pc ⇒ p = v= (b) E (1.00 ×10 eV)(1.60 ×10 −19 J/eV) = = 5.33×10 −23 kg ⋅ m/s c 3.00 ×10 m/s p 5.333× 10 −23 kg ⋅ m/s = = 3.19 × 10 m/s − 27 m 1.67 × 10 kg ( mv = ( 0.5 ) 1.67 × 10 −27 kg keV = 8.516 × 10 −19 J × −16 1.60 × 10 KE = (c) )(3.194 × 10 m/s ) −3 = 5.32 × 10 keV J 46 Take the ratio of relativistic rest energy, E = γmc , to relativistic momentum, p = γmu , and show that in the limit that mass approaches zero, you find E / p = c Solution E γmc c = = E = γmc and p = γmu so p γmu u As the mass of particle approaches zero, its velocity u will approach c so that the E c2 limm→ = =c p c ratio of energy to momentum approaches , which is consistent with the equation 48 p= E c for photons Unreasonable Results A car feels a small force due to the light it sends out from its headlights, equal to the momentum of the light divided by the time in which it is emitted (a) Calculate the power of each headlight, if they exert a total force of 2.00 × 10 −2 N backward on the car (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? OpenStax College Physics Solution Instructor Solutions Manual Chapter 29 ∆p E / c E = ⇒ = P = Fc ∆t ∆t ∆t (a) P = 1.00 × 10 −2 N 3.00 × 108 N = 3.00 × 10 W F= ( )( ) (b) This is far too much energy for a car headlight (c) The force assumed is much too large, although it would have only a small effect on the car 29.6 THE WAVE NATURE OF MATTER 49 At what velocity will an electron have a wavelength of 1.00 m? Solution h h h 6.63 ×10 −34 J ⋅ s λ= = ⇒v = = = 7.28 ×10 −4 m −31 p mv mλ 9.11 ×10 (1.00 m ) 50 What is the wavelength of an electron moving at 3.00% of the speed of light? Solution 51 Solution 52 Solution 53 Solution ( λ= ) h 6.63 ×10 −34 J ⋅ s = = 8.09 ×10 −11 m −2 mv 3.00 ×10 3.00 ×10 m/s ( )( ) At what velocity does a proton have a 6.00-fm wavelength (about the size of a −15 nucleus)? Assume the proton is nonrelativistic (1 femtometer = 10 m ) λ= h h h 6.63 ×10 −34 J ⋅ s = ⇒v = = = 6.62 ×10 m/s p mv mλ 1.67 ×10 −27 kg 6.00 ×10 −15 m ( )( ) What is the velocity of a 0.400-kg billiard ball if its wavelength is 7.50 cm (large enough for it to interfere with other billiard balls)? λ= h h h 6.63 ×10 −34 J ⋅ s = ⇒v= = = 2.21 ×10 −32 m/s −2 p mv mλ ( 0.400 kg ) 7.50 ×10 m ( ) Find the wavelength of a proton moving at 1.00% of the speed of light λ= h h 6.63 × 10 −34 J ⋅ s = = = 1.32 × 10 -13 m p mv 1.67 × 10 − 27 kg 0.01 × 3.00 × 10 m/s ( )( ) 54 Experiments are performed with ultracold neutrons having velocities as small as 1.00 m/s (a) What is the wavelength of such a neutron? (b) What is its kinetic energy in eV? Solution h h 6.63 × 10 −34 J ⋅ s λ= = = = 3.956 × 10 −7 m = 396 nm −27 p mv 1.675 × 10 kg (1.00 m/s ) (a) ( ) OpenStax College Physics Instructor Solutions Manual ( Chapter 29 ) 2 mv = × 1.675 × 10 − 27 kg (1.00 m/s ) 2 eV −9 = 8.375 × 10 − 28 J × = 5.23 × 10 eV −19 1.602 × 10 J KE = (b) 55 (a) Find the velocity of a neutron that has a 6.00-fm wavelength (about the size of a nucleus) Assume the neutron is nonrelativistic (b) What is the neutron’s kinetic energy in MeV? Solution h h h 6.63 ×10 −34 J ⋅ s = ⇒v = = = 6.62 ×10 m/s −27 −15 p mv m λ 1.67 × 10 kg 6.00 × 10 m (a) KE = mv = ( 0.5 ) 1.67 ×10 − 27 kg 6.62 × 10 m/s MeV = 3.659× 10 −12 J × = 22.9 MeV −13 1.60 × 10 J (b) λ= ( )( ( )( ) ) 56 What is the wavelength of an electron accelerated through a 30.0-kV potential, as in a TV tube? Solution 2qV KE = qV = mv ⇒ v = m h h h 6.63 ×10 −34 J ⋅ s λ= = = = p mv ( 2qVm ) ( 2) (1.60 ×10 −19 C )(3.00 ×10 V )( 9.11 ×10 −31 kg ) [ ] = 7.09 ×10 −12 m 57 What is the kinetic energy of an electron in a TEM having a 0.0100-nm wavelength? Solution h2 (6.63 × 10 −34 J ⋅ s) = × keV / 1.60 × 10 −16 J = 15.1 keV 2mλ (2) 9.11 × 10 −31 kg (10 −11 m) Note that in the previous problem the electron with KE of 30 KeV was found to have a wavelength of 0.007 nm To obtain this λ of 0.01 nm, the electron would have to KE = ( ) accelerate through a potential difference of (0.007 / 0.010) × 30 kV = 15 kV , which is consistent with the above result 58 (a) Calculate the velocity of an electron that has a wavelength of 1.00 μm (b) Through what voltage must the electron be accelerated to have this velocity? Solution h h h 6.63 ×10 −34 J ⋅ s λ= = ⇒v= = = 7.28 ×10 m/s −31 −6 p mv mλ 9.11 ×10 kg 1.00 ×10 m (a) ( )( ) OpenStax College Physics Instructor Solutions Manual Chapter 29 mv ( 9.11 × 10 −31 kg )( 7.278 × 10 m/s ) qV = mv ⇒ V = = = 1.51 × 10 −6 V −19 2 q ( ) ( ) 1.60 × 10 C (b) 59 The velocity of a proton emerging from a Van de Graaff accelerator is 25.0% of the speed of light (a) What is the proton’s wavelength? (b) What is its kinetic energy, assuming it is nonrelativistic? (c) What was the equivalent voltage through which it was accelerated? Solution λ= (a) (b) (c) h h 6.63 × 10 −34 J ⋅ s = = mv m( 0.250c ) 1.67 × 10 − 27 kg ( 0.250) 3.00 × 108 m/s ( = 5.29 × 10 KE = −15 ) ( ) m = 5.29 fm )[ ( ( mv = ( 0.5 ) 1.67 × 10 − 27 kg ( 0.250 ) 3.00 × 10 m/s KE = qV ⇒ V = )] = 4.70 × 10 −12 J KE 4.697 × 10 −12 J = = 2.94 × 10 V = 29.4 MV q 1.60 × 10 −19 C 60 The kinetic energy of an electron accelerated in an x-ray tube is 100 keV Assuming it is nonrelativistic, what is its wavelength? Solution KE = 100keV = 1.00 × 10 1.60 × 10 −19 J = 1.60 × 10 −14 J ( KE = 2KE mv ⇒ v = m )( ) ⇒ mv = ( 2KE m ) h h h 6.63 ×10 −34 J ⋅ s λ= = = = p mv ( 2KE m ) 12 ( 2) (1.60 ×10−19 C)( 9.11 ×10 −31 kg ) [ 61 Solution ] = 3.88 × 10 −12 m Unreasonable Results (a) Assuming it is nonrelativistic, calculate the velocity of an electron with a 0.100-fm wavelength (small enough to detect details of a nucleus) (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? λ= h h h 6.63 ×10 −34 kg.m/s = ⇒v= = = 7.28 ×1012 m/s −31 −16 p mv mλ 9.11 ×10 kg 1.00 ×10 m ( )( ) (a) (b) It is many times faster than the speed of light (an impossibility) (c) The non-relativistic assumption is unreasonable at the given wavelength 29.7 PROBABILITY: THE HEISENBERG UNCERTAINTY PRINCIPLE 62 (a) If the position of an electron in a membrane is measured to an accuracy of 1.00 μm , what is the electron’s minimum uncertainty in velocity? (b) If the electron OpenStax College Physics Instructor Solutions Manual Chapter 29 has this velocity, what is its kinetic energy in eV? (c) What are the implications of this energy, comparing it to typical molecular binding energies? Solution h h ⇒ Δp = = mΔv ⇒ 4π 4π∆x h 6.63 × 10 −34 J ⋅ s Δv = = = 57.9 m/s 4π∆x 4π( 9.11 × 10 −31 kg )(1.00 × 10 −6 m ) ΔxΔp = (a) ( ) KE = mv = 0.5 × 9.11 × 10 −31 kg ( 57.91 m/s ) eV −9 = 1.528 ×10 −27 J × = 9.55 ×10 eV −19 1.60 × 10 J (b) (c) From Table 29.1 we see that typical molecular binding energies range from about 1eV to 10 eV; therefore the result in part (b) is approximately orders of magnitude smaller than typical molecular binding energies 63 (a) If the position of a chlorine ion in a membrane is measured to an accuracy of 1.00 μm , what is its minimum uncertainty in velocity, given its mass is 5.86 × 10−26 kg ? (b) If the ion has this velocity, what is its kinetic energy in eV, and how does this compare with typical molecular binding energies? Solution ΔxΔp = Δv = (a) h h ⇒ Δp = = m Δv , so 4π 4π ∆x h 6.63 × 10 −34 J ⋅ s = = 9.00 × 10 − m/s 4π mΔ x 4π( 5.86 × 10 − 26 kg )(1.00 × 10 −6 m ) ( )( ) KE = mv = 0.5 × 5.86 ×10 − 26 kg 9.00 × 10 −4 m/s eV = 2.375 ×10 −32 J × = 1.48 × 10 −13 eV −19 1.60 × 10 J (b) 64 Suppose the velocity of an electron in an atom is known to an accuracy of 2.0 × 103 m/s (reasonably accurate compared with orbital velocities) What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom? OpenStax College Physics Solution Instructor Solutions Manual ΔxΔp = Δx( m Δv ) = = Chapter 29 h h ⇒ Δx = 4π 4π m Δv 6.63 ×10 −34 J ⋅ s = 2.90 ×10 −8 m = 29.0 nm 4π 9.11 × 10 −31 kg 2.00 ×10 m/s ( )( ) Δx 2.90 ×10 −8 m = = 290 times larger D 1.00 ×10 −10 m 65 The velocity of a proton in an accelerator is known to an accuracy of 0.250% of the speed of light (This could be small compared with its velocity.) What is the smallest possible uncertainty in its position? Solution ΔxΔp = Δx( m Δv ) ≥ Δx ≥ h ⇒ 4π h 6.63 × 10 −34 J ⋅ s = = 4.21× 10 −14 m 4π mΔv 4π(1.67 × 10 −27 kg )( 0.00250 ) ( 3.00 × 108 m/s ) 66 A relatively long-lived excited state of an atom has a lifetime of 3.00 ms What is the minimum uncertainty in its energy? Solution h 4π , so h 6.63 × 10 −34 J ⋅ s eV −13 ΔE ≥ = = 1.759 × 10 −32 J × = 1.10 × 10 eV −3 −19 4π Δt 4π 3.00 × 10 s 1.60 × 10 J ∆E∆t = ( 67 −20 (a) The lifetime of a highly unstable nucleus is 10 s What is the smallest uncertainty in its decay energy? (b) Compare this with the rest energy of an electron Solution h h 6.63 × 10 −34 J ⋅ s ∆E∆t ≥ ⇒ ΔE ≥ = 4π 4π Δt 4π (1.00 × 10 − 20 s ) MeV −2 = 5.28 ×10 −15 J × = 3.30×10 MeV −13 1.60 × 10 J (a) ΔE ≥ (b) 68 ) h 6.63 ×10 −34 J ⋅ s = 4π Δt 4π(1.00 × 10 − 20 s ) MeV = 5.28 ×10 −15 J × = 3.30 ×10 − MeV −13 1.60 ×10 J The decay energy of a short-lived particle has an uncertainty of 1.0 MeV due to its short lifetime What is the smallest lifetime it can have? OpenStax College Physics Instructor Solutions Manual Chapter 29 Solution 1.60 ×10 −13 J = 1.60 ×10 −13 J; ΔE = (1.0 MeV) MeV h h 6.63 × 10 −34 J ⋅ s ΔEΔt ≥ ⇒ Δt ≥ = = 3.30 ×10 −22 s 4π 4π ΔE 4π (1.60 × 10 −13 J ) 69 The decay energy of a short-lived nuclear excited state has an uncertainty of 2.0 eV due to its short lifetime What is the smallest lifetime it can have? Solution 1.60 × 10 −19 J = 3.2 × 10 −19 J; ΔE = ( 2.0 eV) eV h h 6.63 × 10 −34 J ⋅ s Δ EΔ t ≥ ⇒ Δt ≥ = = 1.6 × 10 −16 s −13 4π 4π ΔE 4π ( 3.2 × 10 J ) 70 What is the approximate uncertainty in the mass of a muon, as determined from its decay lifetime? Solution ΔEΔt = ( Δmc )Δt = Δm = h ⇒ 4π h 6.63 × 10 −34 J ⋅ s = = 2.66 ×10 − 46 kg 2 − 4π c Δt 4π( 3.00 × 10 m/s ) ( 2.20 × 10 s ) 71 Derive the approximate form of Heisenberg’s uncertainty principle for energy and time, ∆E∆t ≈ h , using the following arguments: Since the position of a particle is uncertain by ∆x ≈ λ , where λ is the wavelength of the photon used to examine it, there is an uncertainty in the time the photon takes to traverse ∆x Furthermore, the photon has an energy related to its wavelength, and it can transfer some or all of this energy to the object being examined Thus the uncertainty in the energy of the object is also related to λ Find ∆t and ∆E ; then multiply them to give the approximate uncertainty principle Solution ∆x λ ≈ c c hc hc λ ∆E = ⇒ ∆E∆t ≈ = h λ λ c ∆x = c∆t ⇒ ∆t = 29.8 THE PARTICLE-WAVE DUALITY REVIEWED 72 Integrated Concepts The 54.0-eV electron in Example 29.7 has a 0.167-nm wavelength If such electrons are passed through a double slit and have their first maximum at an angle of 25.0° , what is the slit separation d ? OpenStax College Physics Instructor Solutions Manual Chapter 29 Solution Using the equation d sinθ = mλ , m = 0,1,2 , we can solve for the slit separation (1)( 0.167 nm ) = 0.395 nm mλ d= = sinθ sin 25.0° when m = for the first order maximum: 73 Integrated Concepts An electron microscope produces electrons with a 2.00-pm wavelength If these are passed through a 1.00-nm single slit, at what angle will the first diffraction minimum be found? Solution −12 m mλ −1 (1) 2.00 × 10 D sinθ = mλ ⇒ θ = sin = sin = 0.115 ° −9 D 1.00 × 10 m 74 Integrated Concepts A certain heat lamp emits 200 W of mostly IR radiation averaging 1500 nm in wavelength (a) What is the average photon energy in joules? (b) How many of these photons are required to increase the temperature of a person’s shoulder by 2.0°C , assuming the affected mass is 4.0 kg with a specific heat of 0.83 kcal/kg ⋅ °C Also assume no other significant heat transfer (c) How long does ( −1 ) this take? Solution (a) Eγ = (b) 75 )( ) Q = NEγ = mc∆T ⇒ N= (c) ( hc 6.63 ×10 −34 J ⋅ s 3.00 ×108 m/s = = 1.3 ×10 −19 J λ 1.5 ×10 −6 m/s mc∆T ( 4.00 kg )( 0.83 kcal/kg ⋅ °C )( 4186 J/kcal)( 2.00 °C ) = = 2.1 × 10 23 −19 Eγ 1.326 × 10 J Q = NEγ = pt ⇒ t = NEγ P = ( 2.096 × 10 )(1.326 × 10 23 −19 200 J/s J ) = 139 s = 1.4 × 10 s Integrated Concepts On its high power setting, a microwave oven produces 900 W of 2560 MHz microwaves (a) How many photons per second is this? (b) How many photons are required to increase the temperature of a 0.500-kg mass of pasta by 45.0°C , assuming a specific heat of 0.900 kcal/kg ⋅ °C ? Neglect all other heat transfer (c) How long must the microwave operator wait for their pasta to be ready? Eγ = hf = ( 6.63 × 10 −34 J )( 2.56 ×10 s -1 ) = 1.70 ×10 −24 J Solution (a) P 900 J/s = = 5.30 × 10 26 photon/s Eγ 1.697 ×10 −24 J/photon Q = NEγ = mc∆T ⇒ N= (b) mc∆T ( 0.500 kg )( 0.900 kcal/kg.°C )( 4186 J/kcal)( 45.0 °C ) = = 5.00 × 10 28 − 24 Eγ 1.697 × 10 J OpenStax College Physics (c) 76 Instructor Solutions Manual NEγ = pt ⇒ t = NEγ P = (5.00 ×10 )(1.697 × 10 28 −24 Chapter 29 J) 900 J/s = 94.2 s Integrated Concepts (a) Calculate the amount of microwave energy in joules needed to raise the temperature of 1.00 kg of soup from 20.0°C to 100°C (b) What is the total momentum of all the microwave photons it takes to this? (c) Calculate the velocity of a 1.00-kg mass with the same momentum (d) What is the kinetic energy of this mass? Solution Assume soup has the same specific heat as water (a) Q = mc∆T = (1.00 kg )( 4186 J/kg ⋅ °C )( 80.0 °C ) = 3.35 × 10 J (b) E = pc ⇒ p = E 3.349×10 J = = 1.12 ×10 −3 kg ⋅ m/s c 3.00 ×10 m/s p = 1.12 × 10 −3 m/s m (c) KE = mv = ( 0.5 )(1.00 kg ) 1.116 × 10 −3 m/s (d) v= ( ) = 6.23 × 10 −7 J 77 Integrated Concepts (a) What is γ for an electron emerging from the Stanford Linear Accelerator with a total energy of 50.0 GeV? (b) Find its momentum (c) What is the electron’s wavelength? Solution E 5.00 × 10 MeV = = 9.78 × 10 mc 511 MeV (a) (b) For such a large γ , v ≈ c (within part in 10 ) E = γmc ⇒ γ = p = γmv ≈ γmc = ( 9.78 × 10 )( 9.11 × 10 −31 kg )( 3.00 × 10 m/s ) = 2.67 × 10 −17 kg ⋅ m/s λ= (c) 78 h 6.63 × 10 −34 J ⋅ s = = 2.48 × 10 −17 m = 0.0248 fm p 2.674 × 10 −17 kg ⋅ m/s Integrated Concepts (a) What is γ for a proton having an energy of 1.00 TeV, produced by the Fermilab accelerator? (b) Find its momentum (c) What is the proton’s wavelength? Solution (a) Using E = γmc E 1.00 × 1012 eV 1.60 × 10 −19 J/eV γ= = mc 1.6726 × 10 − 27 kg 3.00 × 108 m/s ( ( )( )( ) ) = 1.063 × 103 = 1.06 × 10 OpenStax College Physics p = γmc = (b) (c) 79 p= Instructor Solutions Manual Chapter 29 E (1.00 × 1012 eV)(1.60 × 10 -19 J/eV) = = 5.33× 10 −16 kg ⋅ m/s c 3.00 × 10 m/s h h 6.63 ×10 −34 kg ⋅ m/s ,λ = = = 1.24 ×10 −18 m λ p 5.33×10 −16 kg ⋅ m/s Integrated Concepts An electron microscope passes 1.00-pm-wavelength electrons through a circular aperture 2.00 μm in diameter What is the angle between two justresolvable point sources for this microscope? Solution 80 θ= 1.22λ (1.22) (1.00 × 10 −12 m ) = = 6.10 × 10 −7 rad = ( 3.50 × 10 −5 )° D 2.00 × 10 −6 m Integrated Concepts (a) Calculate the velocity of electrons that form the same pattern as 450-nm light when passed through a double slit (b) Calculate the kinetic energy of each and compare them (c) Would either be easier to generate than the other? Explain Solution λ= h h = ⇒ p mv v= h 6.63 × 10 -34 kg ⋅ m/s = = 1.617 × 10 m/s = 1.62 × 10 m/s -31 -7 mλ 9.109 × 10 kg 4.50 × 10 m )( ) hc ( 6.63 ×10 J ⋅ s )( 3.00 ×10 E= = λ ( 4.50×10 m) (b) For the photon: (a) ( -34 m/s -7 For the electron: 1 KE = mv = 9.109 × 10 −31 kg 1.617 × 10 m/s 2 = 1.191× 10 −24 J = 1.19 × 10 − 24 J The photon energy is 3.71 ×10 times greater ( )( ) ) = 4.42×10 -19 J (c) The light is probably easier to make, because 450 nm light is blue light and therefore easy to make Creating electrons with 7.43 µeV of energy would not be difficult, but would require a vacuum 81 Solution Integrated Concepts (a) What is the separation between double slits that produces a second-order minimum at 45.0° for 650-nm light? (b) What slit separation is needed to produce the same pattern for 1.00-keV protons d sinθ = m + (a) ( ) ( ) m + λ ( 2.5) 6.50 × 10 −7 m 1 = = 2.30 × 10 −6 m λ ⇒ d = 2 sinθ sin 45.0° OpenStax College Physics Instructor Solutions Manual Chapter 29 1 KE KE = mv ⇒ v = m , so that (b) h h λ= = mv ( 2KE m ) 12 d= 2.5 h ( 2KEm ) sinθ = [(2)(1.60 × 10 ( (2.5) 6.63 × 10 −34 J ⋅ s −16 )( ) J 1.6726 × 10 −27 kg )] (sin 45.0°) = 3.20 × 10 −12 m 82 Integrated Concepts A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.31 eV? (c) Calculate the current of ejected electrons (d) If the photoelectric material is electrically insulated and acts like a 2.00-pF capacitor, how long will current flow before the capacitor voltage stops it? Solution E γ = hf = hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) = = 4.9725 × 10 −19 J −7 λ ( 4.00 ×10 m ) 2.00 × 10 −3 J/s N= = 4.022 × 1015 /s = 4.02 × 1015 /s −19 4.9725 × 10 J (a) KE = hf − BE = E γ − BE 1.60 × 10 −19 J = 1.324 × 10 −19 J = 4.972 × 10 J − ( 2.28 eV) eV 15 −19 P = NKE = 4.022 × 10 /s 1.324 × 10 J = 5.33× 10 − W = 0.533 mW −19 (b) ( )( ) 15 -19 -4 (c) I = Ne = ( 4.022 × 10 /s )(1.60 × 10 C ) = 6.44 × 10 A It Q = CV = It ⇒ V = C is the voltage across the capacitor This V will stop the flow (d) when the energy required to cross the capacitor, eV , equals or exceeds the kinetic energy of the electrons eIt CKE ( 2.00 ×10 −12 F)(1.324×10 −19 J ) eV = = KE ⇒ t = = = 2.57 ×10 −9 s −19 −4 C eI (1.60 ×10 C)( 6.435×10 A ) 83 Integrated Concepts One problem with x rays is that they are not sensed Calculate the temperature increase of a researcher exposed in a few seconds to a nearly fatal accidental dose of x rays under the following conditions The energy of the x-ray 13 photons is 200 keV, and 4.00 × 10 of them are absorbed per kilogram of tissue, the specific heat of which is 0.830 kcal/kg ⋅ °C (Note that medical diagnostic x-ray machines cannot produce an intensity this great.) OpenStax College Physics Instructor Solutions Manual Chapter 29 Solution First, we know the amount of heat absorbed by 1.00 kg of tissue is equal to the number of photons times the energy each one carry, so: 1.60 × 10 −19 J = 1.28 J Q = NEγ = 4.00 × 1013 2.00 × 10 eV eV Next, using the equation Q = mc∆t , we can determine how much 1.00 kg tissue is ( heated: 84 ∆t = )( ) Q 1.282 J = = 3.69 × 10 −4 °C mc (1.00 kg )( 0.830 kcal/kg ⋅ °C )( 4186 J/kcal) Integrated Concepts A 1.00-fm photon has a wavelength short enough to detect some information about nuclei (a) What is the photon momentum? (b) What is its energy in joules and MeV? (c) What is the (relativistic) velocity of an electron with the same momentum? (d) Calculate the electron’s kinetic energy Solution (a) λ= h h 6.63 × 10 −34 kg ⋅ m/s ⇒p= = = 6.63 × 10 −19 kg ⋅ m/s p λ 1.00 × 10 −15 m E = pc = ( 6.63 × 10 −19 kg ⋅ m/s )( 3.00 × 10 m/s ) = 1.99 × 10 −10 J MeV = 1.99 × 10 −10 J × = 1.24 × 10 MeV −13 1.60 × 10 J (b) mu p = γmu = 2 u 1− c (c) , so that c 2 p − u = m 2u ⇒ p − p u = m 2u ⇒ u = 1/ c c mc 1 + p [ ( )] [ ( )] mc Since p is very small, use the binomial expansion: u= c mc 1 + p ( 1/ mc ≈ c1 − p )( 9.11 × 10 −31 kg 3.00 × 10 m/s = c 1 − 6.63 × 10 −19 kg ⋅ m/s ( )] = [(8.50 ×10 ) ] γ = [1 − (u c (d) −1 −1 −8 2 2 Relativistic kinetic energy ( )( ) ) = c − 8.50 × 10 −8 ( ) = 1.18 × 10 )( ) E = ( γ − 1) mc = 1.18 × 10 − 9.11 × 10 −31 kg 3.00 ×10 m/s = 9.67 × 10 −7 J OpenStax College Physics 85 Instructor Solutions Manual Chapter 29 Integrated Concepts The momentum of light is exactly reversed when reflected straight back from a mirror, assuming negligible recoil of the mirror Thus the change in momentum is twice the photon momentum Suppose light of intensity 1.00 kW/m reflects from a mirror of area 2.00 m (a) Calculate the energy reflected in 1.00 s (b) What is the momentum imparted to the mirror? (c) Using the most general form of Newton’s second law, what is the force on the mirror? (d) Does the assumption of no mirror recoil seem reasonable? Solution (a) E = IA = (1.00 × 10 J/m )( 2.00 m ) = 2.00 × 10 J E ( ) ( 2.00 × 10 J ) p= = = 1.33 × 10 -5 kg ⋅ m/s c 3.00 × 10 m/s (b) ∆p 1.33 × 10 -5 kg ⋅ m/s = = 1.33 × 10 -5 N ∆ t 00 s (c) (d) Yes, with such a small force on the mirror, it will move very little during the time it takes for the photons to bounce off the mirror F= 86 Integrated Concepts Sunlight above the Earth’s atmosphere has an intensity of 1.30 kW/m If this is reflected straight back from a mirror that has only a small recoil, the light’s momentum is exactly reversed, giving the mirror twice the incident momentum (a) Calculate the force per square meter of mirror (b) Very low mass mirrors can be constructed in the near weightlessness of space, and attached to a spaceship to sail it Once done, the average mass per square meter of the spaceship is 0.100 kg Find the acceleration of the spaceship if all other forces are balanced (c) How fast is it moving 24 hours later? Solution In one second the incident energy per square meter will be 1.30 kJ E 1.30 × 10 J m/s ∆p = = = 8.67 × 10 −6 kg ⋅ m/s per m c 3.00 × 10 m/s ( (a) (b) F= ) ∆p 8.67 × 10 −6 kg ⋅ m/s per m = = 8.67 × 10 −6 N per m ∆t 1.00 s F = ma ⇒ a = 8.67 ×10 −6 N/m = 8.67 ×10 −5 m/s 2 0.100 kg/m 3600 s v = at = (8.67 ×10 −5 m/s )( 24 h ) = 7.49 m/s h (c) This file is copyright 2016, Rice University All Rights Reserved ... OpenStax College Physics N= Instructor Solutions Manual Chapter 29 5.00 × 10 J/s = 1.16 × 10 32 photon/s = 1.16 × 10 32 photon/s 4.31× 10 −28 J/photon Then, (b) To calculate the flux of photons, we assume... is too large for the given photon energy (c) The electron’s kinetic energy is too large for the given photon energy; it cannot be OpenStax College Physics Instructor Solutions Manual Chapter 29. .. ×1019 photon/s = 4π 7.073 × 10 photon/s ⋅ m ( ) 29. 4 PHOTON MOMENTUM 38 (a) Find the momentum of a 4.00-cm-wavelength microwave photon (b) Discuss why you expect the answer to (a) to be very
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