PHOTONS: LIGHT WAVES BEHAVING AS PARTICLES 38.1 38 IDENTIFY: Protons have mass and photons are massless (a) SET UP: For a particle with mass, K = p /2m EXECUTE: p2 = p1 means K = K1 (b) SET UP: For a photon, E = pc p2 = p1 means E2 = E1 EXECUTE: 38.2 EVALUATE: The relation between E and p is different for particles with mass and particles without mass IDENTIFY and SET UP: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a photon c = 3.00 × 108 m/s eV = 1.60 × 10−16 J EXECUTE: (a) f = c λ = 3.00 × 108 m/s 505 × 10 −9 m = 5.94 × 1014 Hz (b) E = hf = (6.626 × 10−34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV 2K 2(3.94 × 10−19 J) = = 9.1 mm/s m 9.5 × 10−15 kg (c) K = 12 mv so v = 38.3 EVALUATE: Compared to kinetic energies of common objects moving at typical speeds, the energy of a visible-light photon is extremely small IDENTIFY and SET UP: Apply c = f λ , p = h /λ and E = pc f = EXECUTE: p= h λ = c λ = 3.00 × 108 m/s 5.20 × 10−7 m 6.63 × 10−34 J ⋅ s 5.20 × 10−7 m = 5.77 × 1014 Hz = 1.28 × 10−27 kg ⋅ m/s E = pc = (1.28 × 10−27 kg ⋅ m/s) (3.00 × 108 m/s) = 3.84 × 10−19 J = 2.40 eV 38.4 EVALUATE: Visible-light photons have energies of a few eV hc energy IDENTIFY and SET UP: Pav = eV = 1.60 × 10−19 J For a photon, E = hf = λ t h = 6.63 × 10−34 J ⋅ s EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10−2 J = 7.5 × 1016 eV (b) E = hc = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = 3.05 × 10−19 J = 1.91 eV λ 652 × 10−9 m (c) The number of photons is the total energy in a pulse divided by the energy of one photon: 1.20 × 10−2 J = 3.93 × 1016 photons 3.05 × 10−19 J/photon EVALUATE: The number of photons in each pulse is very large © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-1 38-2 38.5 Chapter 38 IDENTIFY and SET UP: c = f λ The source emits (0.05)(75 J) = 3.75 J of energy as visible light each second E = hf , with h = 6.63 × 10−34 J ⋅ s EXECUTE: (a) f = c λ 3.00 × 108 m/s = 600 × 10−9 m = 5.00 × 1014 Hz (b) E = hf = (6.63 × 10−34 J ⋅ s)(5.00 × 1014 Hz) = 3.32 × 10−19 J The number of photons emitted per second 3.75 J = 1.13 × 1019 photons 3.32 × 10−19 J/photon EVALUATE: (c) No The frequency of the light depends on the energy of each photon The number of photons emitted per second is proportional to the power output of the source IDENTIFY and SET UP: A photon has zero rest mass, so its energy and momentum are related by Eq (37.40) Eq (38.5) then relates its momentum and wavelength EXECUTE: (a) E = pc = (8.24 × 10−28 kg ⋅ m/s)(2.998 × 108 m/s) = 2.47 × 10−19 J = is 38.6 (2.47 × 10−19 J)(1 eV/1.602 × 10−19 J) = 1.54 eV 6.626 × 10−34 J ⋅ s h = = 8.04 × 10−7 m = 804 nm λ p 8.24 × 10−28 kg ⋅ m/s EVALUATE: This wavelength is longer than visible wavelengths; it is in the infrared region of the electromagnetic spectrum To check our result we could verify that the same E is given by Eq (38.2), using the λ we have calculated h φ IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = f − e e The slope of V0 versus f is h /e The value f th of f when V0 = is related to φ by φ = hf th (b) p = 38.7 h so λ = EXECUTE: (a) From the graph, f th = 1.25 × 1015 Hz Therefore, with the value of h from part (b), φ = hf th = 4.8 eV (b) From the graph, the slope is 3.8 × 10−15 V ⋅ s h = (e)(slope) = (1.60 × 10−16 C)(3.8 × 10−15 V ⋅ s) = 6.1 × 10−34 J ⋅ s (c) No photoelectrons are produced for f < f th (d) For a different metal f th and φ are different The slope is h /e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis EVALUATE: As the frequency f of the light is increased above f th the energy of the photons in the light 38.8 increases and more energetic photons are produced The work function we calculated is similar to that for gold or nickel IDENTIFY and SET UP: λth = 272 nm c = f λ mvmax = hf − φ At the threshold frequency, f th , vmax → h = 4.136 × 10−15 eV ⋅ s EXECUTE: (a) f th = c λth (b) φ = hf th = (4.136 × 10 = −15 3.00 × 108 m/s 272 × 10 −9 m = 1.10 × 1015 Hz eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV mvmax = hf − φ = (4.136 × 10−15 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV EVALUATE: The threshold wavelength depends on the work function for the surface hc IDENTIFY and SET UP: Eq (38.3): mvmax = hf − φ = − φ Take the work function φ from Table 38.1 λ Solve for vmax Note that we wrote f as c /λ (c) 38.9 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles EXECUTE: 38-3 (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) mvmax = − (5.1 eV)(1.602 × 10−19 J/1 eV) 235 × 10−9 m mvmax = 8.453 × 10−19 J − 8.170 × 10−19 J = 2.83 × 10−20 J vmax = 38.10 2(2.83 × 10−20 J) 9.109 × 10−31 kg EVALUATE: The work function in eV was converted to joules for use in Eq (38.3) A photon with λ = 235 nm has energy greater then the work function for the surface hc IDENTIFY and SET UP: φ = hf th = The minimum φ corresponds to the minimum λ λth EXECUTE: φ = 38.11 = 2.49 × 105 m/s hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = 1.77 eV 700 × 10−9 m EVALUATE: A photon of wavelength 700 nm has energy 1.77 eV IDENTIFY: The photoelectric effect occurs The kinetic energy of the photoelectron is the difference between the initial energy of the photon and the work function of the metal SET UP: 12 mvmax = hf − φ , E = hc /λ λth = EXECUTE: Use the data for the 400.0-nm light to calculate φ Solving for φ gives φ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 400.0 × 10−9 m have mv max mv max 38.12 = hf − φ = hc λ −φ = λ − 12 mvmax = − 1.10 eV = 3.10 eV − 1.10 eV = 2.00 eV Then for 300.0 nm, we (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 300.0 × 10−9 m − 2.00 eV, which gives = 4.14 eV − 2.00 eV = 2.14 eV EVALUATE: When the wavelength decreases the energy of the photons increases and the photoelectrons have a larger minimum kinetic energy , where V0 is the stopping potential The stopping potential in IDENTIFY and SET UP: eV0 = mvmax 2 = hf − φ and f = c /λ volts equals eV0 in electron volts mvmax 2 EXECUTE: (a) eV0 = mvmax so eV0 = hf − φ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 250 × 10−9 m potential is 2.7 electron volts (b) mvmax = 2.7 eV (c) vmax = 38.13 hc 2(2.7 eV)(1.60 × 10−19 J/eV) 9.11 × 10−31 kg − 2.3 eV = 4.96 eV − 2.3 eV = 2.7 eV The stopping = 9.7 × 105 m/s EVALUATE: If the wavelength of the light is decreased, the maximum kinetic energy of the photoelectrons increases (a) IDENTIFY: First use Eq (38.4) to find the work function φ hc SET UP: eV0 = hf − φ so φ = hf − eV0 = − eV0 λ EXECUTE: φ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 254 × 10−9 m − (1.602 × 10−19 C)(0.181 V) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-4 Chapter 38 φ = 7.821 × 10−19 J − 2.900 × 10−20 J = 7.531 × 10−19 J(1 eV/1.602 × 10−19 J) = 4.70 eV IDENTIFY and SET UP: The threshold frequency f th is the smallest frequency that still produces photoelectrons It corresponds to K max = in Eq (38.3), so hf th = φ f = EXECUTE: λ says hc λth =φ (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 2.64 × 10−7 m = 264 nm 7.531 × 10−19 J (b) EVALUATE: As calculated in part (a), φ = 4.70 eV This is the value given in Table 38.1 for copper IDENTIFY: The acceleration gives energy to the electrons which is then given to the x ray photons hc SET UP: E = hc /λ , so = eV , where λ is the wavelength of the x ray and V is the accelerating λth = 38.14 hc c φ = λ voltage EXECUTE: λ = 38.15 hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 × 10−11 m = 0.0829 nm eV (1.60 × 10−19 C)(15.0 × 103 V) EVALUATE: This wavelength certainly is in the x ray region of the electromagnetic spectrum IDENTIFY: Apply Eq (38.6) SET UP: For a 4.00-keV electron, eVAC = 4000 eV EXECUTE: eVAC = hf max = 38.16 hc λmin ⇒ λmin = hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 3.11 × 10−10 m eVAC (1.60 × 10−19 C)(4000 V) EVALUATE: This is the same answer as would be obtained if electrons of this energy were used Electron beams are much more easily produced and accelerated than proton beams hc IDENTIFY and SET UP: = eV , where λ is the wavelength of the x ray and V is the accelerating voltage λ hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 kV EXECUTE: (a) V = eλ (1.60 × 10−19 C)(0.150 × 10−9 m) hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 4.14 × 10−11 m = 0.0414 nm eV (1.60 × 10−19 C)(30.0 × 103 V) EVALUATE: Shorter wavelengths require larger potential differences IDENTIFY: Energy is conserved when the x ray collides with the stationary electron hc hc SET UP: E = hc /λ , and energy conservation gives = + Ke λ λ′ ⎛1 ⎞ EXECUTE: Solving for K e gives K e = hc ⎜ − ⎟ = ⎝ λ λ′ ⎠ (b) λ = 38.17 38.18 1 ⎛ ⎞ −16 J = 1.13 keV − (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) ⎜ ⎟ K e = 1.81 × 10 −9 −9 ⎝ 0.100 × 10 m 0.110 × 10 m ⎠ EVALUATE: The electron does not get all the energy of the incident photon hc IDENTIFY and SET UP: The wavelength of the x rays produced by the tube is given by = eV λ h h hc λ′ = λ + (1 − cos φ ) = 2.426 × 10−12 m The energy of the scattered x ray is mc mc λ′ EXECUTE: (a) λ = (b) λ ′ = λ + hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 6.91 × 10−11 m = 0.0691 nm eV (1.60 × 10−19 C)(18.0 × 103 V) h (1 − cos φ ) = 6.91 × 10−11 m + (2.426 × 10−12 m)(1 − cos 45.0°) mc λ ′ = 6.98 × 10−11 m = 0.0698 nm © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles 38-5 hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 17.8 keV λ′ 6.98 × 10−11 m EVALUATE: The incident x ray has energy 18.0 keV In the scattering event, the photon loses energy and its wavelength increases h IDENTIFY: Apply Eq (38.7): λ ′ − λ = (1 − cos φ ) = λC (1 − cos φ ) mc SET UP: Solve for λ ′ : λ ′ = λ + λC (1 − cos φ ) The largest λ ′ corresponds to φ = 180°, so cos φ = −1 (c) E = 38.19 38.20 EXECUTE: λ ′ = λ + 2λC = 0.0665 × 10−9 m + 2(2.426 × 10−12 m) = 7.135 × 10−11 m = 0.0714 nm This wavelength occurs at a scattering angle of φ = 180° EVALUATE: The incident photon transfers some of its energy and momentum to the electron from which it scatters Since the photon loses energy its wavelength increases, λ ′ > λ Δλ IDENTIFY: Apply Eq (38.7): cos φ = − ( h /mc) SET UP: h = 0.002426 nm mc EXECUTE: (a) Δλ = 0.0542 nm − 0.0500 nm, cos φ = − 0.0042 nm = −0.731, and φ = 137° 0.002426 nm 0.0021 nm = 0.134 φ = 82.3° 0.002426 nm (c) Δλ = 0, the photon is undeflected, cos φ = and φ = EVALUATE: The shift in wavelength is larger as φ approaches 180° The photon loses energy in the collision, so the wavelength increases h IDENTIFY and SET UP: The shift in wavelength of the photon is λ ′ − λ = (1 − cos φ ) where λ ′ is the mc h = λC = 2.426 × 10−12 m The energy of a photon of wavelength λ wavelength after the scattering and mc (b) Δλ = 0.0521 nm − 0.0500 nm cosφ = − 38.21 is E = hc λ = 1.24 × 10−6 eV ⋅ m λ Conservation of energy applies to the collision, so the energy lost by the photon equals the energy gained by the electron EXECUTE: (a) λ ′ − λ = λC (1 − cos φ ) = (2.426 × 10−12 m)(1 − cos35.0°) = 4.39 × 10−13 m = 4.39 × 10 −4 nm 38.22 (b) λ ′ = λ + 4.39 × 10−4 nm = 0.04250 nm + 4.39 × 10−4 nm = 0.04294 nm hc hc (c) Eλ = = 2.918 × 104 eV and Eλ ′ = = 2.888 × 104 eV so the photon loses 300 eV of energy λ λ′ (d) Energy conservation says the electron gains 300 eV of energy EVALUATE: The photon transfers energy to the electron Since the photon loses energy, its wavelength increases IDENTIFY: The change in wavelength of the scattered photon is given by Eq 38.7: h h Δλ (1 − cos φ ) ⇒ λ = (1 − cos φ ) = ⎛ Δλ ⎞ λ mcλ mc ⎜ ⎟ ⎝ λ ⎠ SET UP: For backward scattering, φ = 180° Since the photon scatters from a proton, m = 1.67 × 10−27 kg (6.63 × 10−34 J ⋅ s) (1 + 1) = 2.65 × 10−14 m (1.67 × 10 kg)(3.00 × 108 m/s)(0.100) EVALUATE: The maximum change in wavelength, h /mc, is much smaller for scattering from a proton than from an electron EXECUTE: λ = −27 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-6 Chapter 38 38.23 IDENTIFY: During the Compton scattering, the wavelength of the x ray increases by 1.0%, which means that the x ray loses energy to the electron h h SET UP: Δλ = (1 − cos φ ) and = 2.426 × 10−12 m λ ′ = 1.010λ so Δλ = 0.010λ mc mc Δλ (0.010)(0.900 × 10−10 m) =1− = 0.629, so φ = 51.0° h /mc 2.426 × 10−12 m EVALUATE: The scattering angle is less than 90°, so the x ray still has some forward momentum after scattering IDENTIFY: Compton scattering occurs We know speed, and hence the kinetic energy, of the scattered electron Energy is conserved hc hc SET UP: = + Ee where Ee = mv λ λ′ hc hc EXECUTE: Ee = mv = (9.108 × 10−31 kg)(8.90 × 106 m/s) = 3.607 × 10−17 J Using = + Ee , λ λ′ 2 EXECUTE: cos φ = − 38.24 we have 38.25 hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 0.1385 × 10−9 m = 1.434 × 10−15 J Therefore, (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) hc hc = − Ee = 1.398 × 10−15 J, which gives λ ′ = = 0.1421 nm λ′ λ 1.398 × 10−15 J ⎛ h ⎞ λ ′ − λ = ⎜ ⎟ (1 − cos φ ) = 3.573 × 10−12 m, so − cosφ = 1.473, which gives φ = 118° ⎝ mc ⎠ EVALUATE: The photon partly backscatters, but not through 180° (a) IDENTIFY and SET UP: Use Eq (37.36) to calculate the kinetic energy K ⎛ ⎞ EXECUTE: K = mc ⎜ − 1⎟ = 0.1547 mc ⎜ ⎟ 2 ⎝ − v /c ⎠ m = 9.109 × 10−31 kg, so K = 1.27 × 10−14 J (b) IDENTIFY and SET UP: The total energy of the particles equals the sum of the energies of the two photons Linear momentum must also be conserved EXECUTE: The total energy of each electron or positron is E = K + mc = 1.1547 mc = 9.46 × 10−13 J The total energy of the electron and positron is converted into the total energy of the two photons The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero The photons must have equal wavelengths and must be traveling in opposite directions Equal λ means equal energy, so each photon has energy 9.46 × 10−14 J (c) IDENTIFY and SET UP: Use Eq (38.2) to relate the photon energy to the photon wavelength EXECUTE: E = hc/λ so λ = hc/E = hc /(9.46 × 10−14 J) = 2.10 pm 38.26 EVALUATE: When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less IDENTIFY: The uncertainty principle relates the uncertainty in the duration time of the pulse and the uncertainty in its energy, which we know SET UP: E = hc/λ and ΔE Δt = ប/2 EXECUTE: E = hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 625 × 10−9 m = 3.178 × 10−19 J The uncertainty in the energy is 1.0% of this amount, so ΔE = 3.178 × 10−21 J We now use the uncertainty principle Solving ΔE Δt = ប ប 1.055 × 10−34 J ⋅ s = = 1.66 × 10−16 s = 0.166 fs 2ΔE 2(3.178 × 10−19 J) EVALUATE: The uncertainty in the energy limits the duration of the pulse The more precisely we know the energy, the longer the duration must be for the time interval gives Δt = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles 38.27 38-7 IDENTIFY: The wavelength of the pulse tells us the momentum of the photon The uncertainty in the momentum is determined by the uncertainty principle h ប SET UP: p = and ΔxΔpx = λ p= EXECUTE: h λ = 6.626 × 10−34 J ⋅ s 556 × 10−9 m = 1.19 × 10−27 kg ⋅ m/s The spatial length of the pulse is ប Δx = cΔt = (2.998 × 108 m/s)(9.00 × 10−15 s) = 2.698 × 10−6 m The uncertainty principle gives ΔxΔpx = Solving for the uncertainty in the momentum, we have ប 1.055 × 10−34 J ⋅ s Δpx = = = 1.96 × 10−29 kg ⋅ m/s 2Δx 2(2.698 × 10−6 m) 38.28 EVALUATE: This is 1.6% of the average momentum IDENTIFY: We know the beam went through the slit, so the uncertainty in its vertical position is the width of the slit ប h SET UP: ΔyΔp y = and px = Call the x-axis horizontal and the y-axis vertical λ ប EXECUTE: (a) Let Δy = a = 6.20 × 10−5 m Solving ΔyΔp y = for the uncertainty in momentum gives Δp y = ប 1.055 × 10−34 J ⋅ s = = 8.51 × 10−31 kg ⋅ m/s 2Δy 2(6.20 × 10−5 m) (b) px = h λ = 6.626 × 10−34 J ⋅ s 585 × 10−9 m = 1.13 × 10−27 kg ⋅ m/s θ = Δp y px = 8.51 × 10−31 1.13 × 10−27 = 7.53 × 10−4 rad The width is (2.00 m)(7.53 × 10−4 ) = 1.51 × 10−3 m = 1.51 mm 38.29 EVALUATE: We must be especially careful not to confuse the x- and y-components of the momentum IDENTIFY and SET UP: Use c = f λ to relate frequency and wavelength and use E = hf to relate photon energy and frequency EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a single molecule The problem states that it requires 1.00 × 105 J to dissociate one mole of AgBr, and one mole contains Avogadro’s number (6.02 × 1023 ) of molecules, so the energy required to dissociate one AgBr is 1.00 × 105 J/mol 6.02 × 1023 molecules/mol = 1.66 × 10−19 J/molecule The photon is to have this energy, so E = 1.66 × 10−19 J(1 eV/1.602 × 10−19 J) = 1.04 eV (b) E = hc λ so λ = (c) c = f λ so f = hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.20 × 10−6 m = 1200 nm E 1.66 × 10−19 J c λ = 2.998 × 108 m/s 1.20 × 10 −6 m = 2.50 × 1014 Hz (d) E = hf = (6.626 × 10−34 J ⋅ s)(100 × 106 Hz) = 6.63 × 10−26 J E = 6.63 × 10−26 J(1 eV/1.602 × 10−19 J) = 4.14 × 10−7 eV (e) EVALUATE: A photon with frequency f = 100 MHz has too little energy, by a large factor, to 38.30 dissociate a AgBr molecule The photons in the visible light from a firefly individually have enough energy to dissociate AgBr The huge number of 100 MHz photons can’t compensate for the fact that individually they have too little energy IDENTIFY: The number N of visible photons emitted per second is the visible power divided by the energy hf of one photon SET UP: At a distance r from the source, the photons are evenly spread over a sphere of area A = 4π r © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-8 Chapter 38 EXECUTE: (a) N = (b) N P (200 W)(0.10) = = 6.03 × 1019 photons/ sec hf h(5.00 × 1014 Hz) = 1.00 × 1011 photons/ sec ⋅ cm gives 4π r 1/ 38.31 ⎛ ⎞ 6.03 × 1019 photons/ sec r =⎜ = 6930 cm = 69.3 m ⎜ 4π (1.00 × 1011 photons/ sec ⋅ cm ) ⎟⎟ ⎝ ⎠ EVALUATE: The number of photons emitted per second by an ordinary household source is very large c IDENTIFY and SET UP: f = The ( f , V0 ) values are: (8.20 × 1014 Hz,1.48 V), λ 14 (7.41 × 10 Hz,1.15 V), (6.88 × 1014 Hz, 0.93 V), (6.10 × 1014 Hz, 0.62 V), (5.49 × 1014 Hz, 0.36 V), (5.18 × 1014 Hz, 0.24 V) The graph of V0 versus f is given in Figure 38.31 EXECUTE: (a) The threshold frequency, f th , is f where V0 = From the graph this is f th = 4.56 × 1014 Hz (b) λth = c 3.00 × 108 m/s = = 658 nm f th 4.56 × 1014 Hz (c) φ = hf th = (4.136 × 10−15 eV ⋅ s)(4.56 × 1014 Hz) = 1.89 eV φ h ⎛h⎞ (d) eV0 = hf − φ so V0 = ⎜ ⎟ f − The slope of the graph is e e ⎝e⎠ 1.48 V − 0.24 V h ⎛ ⎞ −15 =⎜ V/Hz and ⎟ = 4.11 × 10 e ⎝ 8.20 × 1014 Hz − 5.18 × 1014 Hz ⎠ h = (4.11 × 10−15 V/Hz)(1.60 × 10−19 C) = 6.58 × 10−34 J ⋅ s Figure 38.31 38.32 EVALUATE: The value of h from our calculation is within 1% of the accepted value IDENTIFY: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any excess energy going into kinetic energy of the electron SET UP: Conservation of energy gives hf = hc/λ = K max + φ EXECUTE: (a) Using hc/λ = K max + φ , we solve for the work function: φ = hc /λ − K max = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(124 nm) − 4.16 eV = 5.85 eV (b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal per second N × (energy of a photon) = 2.50 W N (hc /λ ) = 2.50 W N = (2.50 W)(124 nm)/[(6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s (c) N is proportional to the power, so if the power is cut in half, so is N, which gives N = (1.56 × 1018 el/s)/2 = 7.80 ì 1017 el/s â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles 38.33 38-9 (d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ To maintain the same power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s We could also see this from part (b), where N is proportional to λ So if the wavelength is cut in half, so is N EVALUATE: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only reduces the number of ejected electrons In part (d), reducing the wavelength does change the maximum kinetic energy of the photoelectrons because we have increased the energy of each photon IDENTIFY and SET UP: The energy added to mass m of the blood to heat it to Tf = 100°C and to vaporize it is Q = mc(Tf − Ti ) + mLv , with c = 4190 J/kg ⋅ K and Lv = 2.256 × 106 J/kg The energy of one photon is E = hc λ = 1.99 × 10−25 J ⋅ m λ EXECUTE: (a) Q = (2.0 × 10−9 kg)(4190 J/kg ⋅ K)(100°C − 33°C) + (2.0 × 10−9 kg)(2.256 × 106 J/kg) = 5.07 × 10−3 J The pulse must deliver 5.07 mJ of energy (b) P = energy 5.07 × 10−3 J = = 11.3 W t 450 × 10−6 s 38.34 hc 1.99 × 10−25 J ⋅ m = 3.40 × 10−19 J The number N of photons per pulse 585 × 10−9 m is the energy per pulse divided by the energy of one photon: 5.07 × 10−3 J N= = 1.49 × 1016 photons 3.40 × 10−19 J/photon EVALUATE: The power output of the laser is small but it is focused on a small area, so the laser intensity is large hc IDENTIFY: The threshold wavelength λ0 is related to the work function φ by = φ (c) One photon has energy E = λ = λ0 −15 SET UP: For φ in eV, use h = 4.136 × 10 eV ⋅ s hc EXECUTE: (a) λ0 = , and the wavelengths are: cesium: 590 nm, copper: 264 nm, potassium: 539 nm, φ 38.35 zinc: 288 nm EVALUATE: (b) The wavelengths for copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome the threshold energy of these metals Therefore, copper and zinc will not emit photoelectrons when irradiated with visible light h IDENTIFY and SET UP: λ ′ = λ + (1 − cos φ ) mc 2h φ = 180° so λ ′ = λ + = 0.09485 nm Use Eq (38.5) to calculate the momentum of the scattered photon mc Apply conservation of energy to the collision to calculate the kinetic energy of the electron after the scattering The energy of the photon is given by Eq (38.2) EXECUTE: (a) p′ = h /λ ′ = 6.99 × 10−24 kg ⋅ m/s (b) E = E ′ + Ee ; hc/λ = hc/λ ′ + Ee 38.36 λ′ − λ ⎛1 ⎞ Ee = hc ⎜ − ⎟ = (hc ) = 1.129 × 10−16 J = 705 eV λλ ′ ⎝ λ λ′ ⎠ EVALUATE: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the electron This corresponds to a fractional shift in the photon’s wavelength that is also 5% IDENTIFY: Compton scattering occurs For backscattering, the scattering angle of the photon is 180° SET UP: Let +x be in the direction of propagation of the incident photon ⎛ h ⎞ λ′ − λ = ⎜ ⎟ (1 − cos φ ), where φ = 180° ⎝ mc ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-10 Chapter 38 h h h = − + pe Solving = 0.0900 × 10−9 m + 4.852 × 10−12 m = 0.09485 × 10−9 m mc λ λ′ h h 9.000 × 10−11 m + 9.485 × 10−11 m ⎛ λ + λ′ ⎞ = (6.626 × 10−34 J ⋅ s) for pe gives pe = + = h ⎜ ⎟ λ λ′ (9.000 × 10−11 m)(9.485 × 10−11 m) ⎝ λλ ′ ⎠ EXECUTE: λ ′ = λ + 38.37 pe = 1.43 × 10−23 kg ⋅ m/s EVALUATE: The electron gains the most amount of momentum when backscattering occurs IDENTIFY: Compton scattering occurs, and we know the angle of scattering and the initial wavelength (and hence momentum) of the incident photon ⎛ h ⎞ SET UP: λ ′ − λ = ⎜ ⎟ (1 − cos φ ) and p = h/λ Let +x be the direction of propagation of the incident ⎝ mc ⎠ photon and let the scattered photon be moving at 30.0° clockwise from the + y axis ⎛ h ⎞ −9 −12 EXECUTE: λ ′ − λ = ⎜ m)(1 − cos60.0°) = 0.1062 × 10−9 m ⎟ (1 − cos φ ) = 0.1050 × 10 m + (2.426 × 10 ⎝ mc ⎠ h h Pix = Pfx = cos60.0° + pex λ λ′ pex = h λ − h 2λ ′ − λ 2.1243 × 10−10 m − 1.050 × 10−10 m =h = (6.626 × 10−34 J ⋅ s) 2λ ′ (2λ ′)(λ ) (2.1243 × 10−10 m)(1.050 × 10−10 m) pex = 3.191 × 10−24 kg ⋅ m/s Piy = Pfy = pey = − tan θ = 38.38 (6.626 × 10−34 J ⋅ s)sin 60.0° pey pex h λ′ sin 60.0° + pey = −5.403 × 10−24 kg ⋅ m/s pe = 0.1062 × 10−9 m −5.403 = and θ = −59.4° 3.191 pe2x + pe2y = 6.28 × 10−24 kg ⋅ m/s EVALUATE: The electron gets only part of the momentum of the incident photon IDENTIFY and SET UP: Electrical power is VI Q = mcΔT EXECUTE: (a) (0.010)VI = (0.010)(18.0 × 103 V)(60.0 × 10−3 A) = 10.8 W = 10.8 J/s (b) The energy in the electron beam that isn’t converted to x rays stays in the target and appears as thermal energy For t = 1.00 s, Q = (0.990)VI (1.00 s) = 1.07 × 103 J and Q 1.07 × 103 J = = 32.9 K The temperature rises at a rate of 32.9 K/s mc (0.250 kg)(130 J/kg ⋅ K) EVALUATE: The target must be made of a material that has a high melting point IDENTIFY and SET UP: Find the average change in wavelength for one scattering and use that in Δλ in Eq (38.7) to calculate the average scattering angle φ EXECUTE: (a) The wavelength of a MeV photon is hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s) λ= = = × 10−12 m E × 106 eV ΔT = 38.39 The total change in wavelength therefore is 500 × 10−9 m − × 10−12 m = 500 × 10−9 m If this shift is produced in 1026 Compton scattering events, the wavelength shift in each scattering event is Δλ = 500 × 10−9 m × 1026 = × 10−33 m (b) Use this Δλ in Δλ = h (1 − cos φ ) and solve for φ We anticipate that φ will be very small, since mc Δλ is much less than h/mc, so we can use cos φ ≈ − φ /2 h h Δλ = (1 − (1 − φ /2)) = φ mc 2mc © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles 38-11 2Δλ 2(5 × 10−33 m) = = 6.4 × 10−11 rad = (4 × 10−9 )° (h/mc) 2.426 × 10−12 m φ in radians is much less than so the approximation we used is valid (c) IDENTIFY and SET UP: We know the total transit time and the total number of scatterings, so we can calculate the average time between scatterings EXECUTE: The total time to travel from the core to the surface is (106 y)(3.156 × 107 s/y) = 3.2 × 1013 s φ= There are 1026 scatterings during this time, so the average time between scatterings is t= 3.2 × 1013 s 1026 = 3.2 × 10−13 s The distance light travels in this time is d = ct = (3.0 × 108 m/s)(3.2 × 10−13 s) = 0.1 mm 38.40 EVALUATE: The photons are on the average scattered through a very small angle in each scattering event The average distance a photon travels between scatterings is very small IDENTIFY: Apply Eq (38.7) to each scattering + cosθ SET UP: cos(θ /2) = , so cosθ = 2cos (θ /2) − EXECUTE: (a) Δλ1 = (h/mc)(1 − cosθ1 ), Δλ2 = (h/mc)(1 − cosθ ), and so the overall wavelength shift is Δλ = (h/mc)(2 − cosθ1 − cosθ ) (b) For a single scattering through angle θ , Δλs = ( h/mc)(1 − cosθ ) For two successive scatterings through an angle of θ /2 for each scattering, Δλt = 2(h /mc)(1 − cosθ /2) − cosθ = 2(1 − cos (θ /2)) and Δλs = (h/mc)2(1 − cos (θ /2)) cos(θ /2) ≤ so − cos (θ /2) ≥ (1 − cos(θ /2)) and Δλs ≥ Δλt Equality holds only when θ = 180° (c) (h/mc)2(1 − cos30.0°) = 0.268( h/mc) (d) (h/mc)(1 − cos60°) = 0.500(h/mc ), which is indeed greater than the shift found in part (c) EVALUATE: When θ is small, − cosθ ≈ θ and − cos(θ /2) ≈ θ /2 In this limit Δλs and Δλt are 38.41 approximately equal (a) IDENTIFY and SET UP: Conservation of energy applied to the collision gives Eλ = Eλ ′ + Ee , where Ee is the kinetic energy of the electron after the collision and Eλ and Eλ ′ are the energies of the photon before and after the collision The energy of a photon is related to its wavelength according to Eq (38.2) ⎛1 ⎞ ⎛ λ′ − λ ⎞ EXECUTE: Ee = hc ⎜ − ⎟ = hc ⎜ ⎟ ′ ⎝λ λ ⎠ ⎝ λλ ′ ⎠ ⎛ ⎞ 0.0032 × 10−9 m Ee = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) ⎜ ⎜ (0.1100 × 10−9 m)(0.1132 × 10−9 m) ⎟⎟ ⎝ ⎠ Ee = 5.105 × 10−17 J = 319 eV Ee 2(5.105 × 10−17 J) Ee = mv so v = = = 1.06 × 107 m/s m 9.109 × 10−31 kg (b) The wavelength λ of a photon with energy Ee is given by Ee = hc/λ so λ= hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 3.89 nm Ee 5.105 × 10−17 J EVALUATE: Only a small portion of the incident photon’s energy is transferred to the struck electron; this is why the wavelength calculated in part (b) is much larger than the wavelength of the incident photon in the Compton scattering © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 38-12 38.42 Chapter 38 IDENTIFY: Eq (38.7) relates λ and λ ′ to φ Apply conservation of energy to obtain an expression that relates λ and v to λ ′ hc SET UP: The kinetic energy of the electron is K = (γ − 1)mc The energy of a photon is E = λ hc EXECUTE: (a) The final energy of the photon is E ′ = , and E = E ′ + K , where K is the kinetic energy of λ′ the electron after the collision Then, hc hc hc λ′ ( K = mc (γ − 1) since the λ= = = = E ′ + K ( hc/λ ′) + K (hc/λ ′) + (γ − 1)mc ⎤ λ ′mc ⎡ 1+ − 1⎥ ⎢ h ⎣ (1 − v /c )1/2 ⎦ relativistic expression must be used for three-figure accuracy) (b) φ = arccos[1 − Δλ /(h/mc)] (c) γ − = ⎛1 − ⎜ ⎝ 1/2 1.80 ⎞ 3.00 ⎟ ( ) − = 1.25 − = 0.250, h = 2.43 × 10−12 m mc ⎠ 5.10 × 10−3 mm ⇒λ = 1+ (5.10 × 10 −12 m)(9.11 × 10−31 kg)(3.00 × 108 m/s)(0.250) (6.63 × 10−34 J ⋅ s) ⎛ ⎜ ⎝ φ = arccos ⎜ − = 3.34 × 10−3 nm (5.10 × 10−12 m − 3.34 × 10−12 m) ⎞ ⎟⎟ = 74.0° 2.43 × 10−12 m ⎠ EVALUATE: For this final electron speed, v/c = 0.600 and K = 12 mv is not accurate 38.43 IDENTIFY: Apply the Compton scattering formula λ ′ − λ = Δλ = (a) SET UP: Largest Δλ is for φ = 180° EXECUTE: For φ = 180°, Δλ = 2λC = 2(2.426 pm) = 4.85 pm h (1 − cos φ ) = λC (1 − cos φ ) mc (b) SET UP: λ ′ − λ = λC (1 − cos φ ) Wavelength doubles implies λ ′ = 2λ so λ ′ − λ = λ Thus λ = λC (1 − cos φ ) λ is related to E by Eq (38.2) EXECUTE: E = hc/λ , so smallest energy photon means largest wavelength photon, so φ = 180° and λ = 2λC = 4.85 pm Then E= hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 4.85 × 10−12 m = 4.096 × 10−14 J(1 eV/1.602 × 10−19 J) = 0.256 MeV EVALUATE: Any photon Compton scattered at φ = 180° has a wavelength increase of 2λC = 4.85 pm 38.44 4.85 pm is near the short-wavelength end of the range of x-ray wavelengths IDENTIFY: Follow the derivation of Eq (38.7) Apply conservation of energy and conservation of momentum to the collision SET UP: Use the coordinate direction specified in the problem G G G G EXECUTE: Momentum: p + P = p′ + P ′ ⇒ p − P = − p′ − P′ ⇒ p′ = P − ( p + P′) energy: pc + E = p′c + E ′ = p′c + ( P′c ) + (mc ) ⇒ ( pc − p′c + E ) = ( P′c )2 + (mc ) = ( Pc) + (( p + p′)c)2 − P( p + p′)c + ( mc ) ( pc − p′c ) + E = E + ( pc + p′c ) − 2( Pc )( p + p′) + Ec( p − p′) − pp′c + Ec( p − p′) +2( Pc )( p + p′) = ⇒ p′( Pc − pc − Ec) = p (− Ec − Pc ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Photons: Light Waves Behaving as Particles ⇒ p′ = p Ec + Pc 2 pc + Ec − Pc =p 38-13 E + Pc pc + ( E − Pc) 2hc ⎛ 2hc/λ + ( E − Pc) ⎞ ⎛ E − Pc ⎞ ⇒ λ′ = λ ⎜ ⎟ = λ⎜ ⎟+ E + Pc ⎝ ⎠ ⎝ E + Pc ⎠ E + Pc λ ( E − Pc) + 2hc ⇒ λ′ = E + Pc 2 ⎛ ⎞ ⎛ mc ⎞ ⎛ mc ⎞ ⎜ ≈ E − + "⎟ If E mc , Pc = E − (mc ) = E − ⎜ ⎟ ⎜ ⎟ ⎜ E ⎟ ⎜ 2⎜ E ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⇒ E − Pc ≈ λ (mc )2 hc hc ⎛ m 2c 4λ ⎞ (mc )2 ⇒ λ1 ≈ + = ⎜1 + ⎟ E ⎜⎝ E (2 E ) E 4hcE ⎟⎠ E (b) If λ = 10.6 × 10−6 m, E = 1.00 × 1010 eV = 1.60 × 10−9 J ⎛ (9.11 × 10−31 kg)2 c (10.6 × 10−6 m) ⎞ −16 −15 ⎜1 + ⎟⎟ = (1.24 × 10 m)(1 + 56.0) = 7.08 × 10 m 1.60 × 10 J ⎜⎝ 4hc (1.6 × 10−9 J) ⎠ (c) These photons are gamma rays We have taken infrared radiation and converted it into gamma rays! Perhaps useful in nuclear medicine, nuclear spectroscopy, or high energy physics: wherever controlled gamma ray sources might be useful EVALUATE: The photon has gained energy from the initial kinetic energy of the electron Since the photon gains energy, its wavelength decreases ⇒ λ′ ≈ hc −9 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... h = 4 .136 × 10−15 eV ⋅ s EXECUTE: (a) f th = c λth (b) φ = hf th = (4 .136 × 10 = −15 3.00 × 108 m/s 272 × 10 −9 m = 1.10 × 1015 Hz eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV mvmax = hf − φ = (4 .136 ×... (106 y)(3.156 × 107 s/y) = 3.2 × 1 013 s φ= There are 1026 scatterings during this time, so the average time between scatterings is t= 3.2 × 1 013 s 1026 = 3.2 × 10 13 s The distance light travels... EXECUTE: (a) eV0 = mvmax so eV0 = hf − φ = (4 .136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 250 × 10−9 m potential is 2.7 electron volts (b) mvmax = 2.7 eV (c) vmax = 38 .13 hc 2(2.7 eV)(1.60 × 10−19 J/eV) 9.11