39 PARTICLES BEHAVING AS WAVES 39.1 IDENTIFY and SET UP: λ = h h = For an electron, m = 9.11 × 10−31 kg For a proton, p mv m = 1.67 × 10−27 kg EXECUTE: (a) λ = 6.63 × 10−34 J ⋅ s (9.11 × 10 −31 kg)(4.70 × 10 m/s) = 1.55 × 10−10 m = 0.155 nm ⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ = 8.46 × 10−14 m , so λp = λe ⎜ e ⎟ = (1.55 × 10−10 m) ⎜ ⎜ 1.67 × 10−27 kg ⎟⎟ ⎜ mp ⎟ m ⎝ ⎠ ⎝ ⎠ EVALUATE: For the same speed the proton has a smaller de Broglie wavelength h p2 hc IDENTIFY and SET UP: For a photon, E = For an electron or proton, p = and E = , so λ λ 2m (b) λ is proportional to 39.2 E= h2 2mλ EXECUTE: (a) E = hc λ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = 6.2 keV ⎛ 6.63 × 10−34 J ⋅ s ⎞ = = 6.03 × 10−18 J = 38 eV (b) E = ⎜⎜ ⎟⎟ −9 −31 2mλ ⎝ 0.20 × 10 m ⎠ 2(9.11 × 10 kg) ⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ = 0.021 eV (c) Ep = Ee ⎜ e ⎟ = (38 eV) ⎜ ⎜ 1.67 × 10−27 kg ⎟⎟ ⎜ mp ⎟ ⎝ ⎠ ⎝ ⎠ EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has more energy than a proton h p2 IDENTIFY: For a particle with mass, λ = and K = 2m p h2 39.3 0.20 × 10−9 m SET UP: eV = 1.60 × 10−19 J EXECUTE: (a) λ = h h (6.63 × 10−34 J ⋅ s) ⇒ p= = = 2.37 × 10−24 kg ⋅ m/s p λ (2.80 × 10−10 m) p (2.37 × 10−24 kg ⋅ m/s) = = 3.08 × 10−18 J = 19.3 eV 2m 2(9.11 × 10−31 kg) EVALUATE: This wavelength is on the order of the size of an atom This energy is on the order of the energy of an electron in an atom (b) K = © 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 39-1 39-2 39.4 Chapter 39 IDENTIFY: For a particle with mass, λ = p2 h and E = 2m p SET UP: eV = 1.60 × 10−19 J EXECUTE: λ = 39.5 h h (6.63 × 10−34 J ⋅ s) = = = 7.02 × 10−15 m −27 −19 p 2mE 2(6.64 × 10 kg) (4.20 × 10 eV) (1.60 × 10 J/eV) EVALUATE: This wavelength is on the order of the size of a nucleus h h IDENTIFY and SET UP: The de Broglie wavelength is λ = = In the Bohr model, mvrn = n(h /2π ), p mv so mv = nh /(2π rn ) Combine these two expressions and obtain an equation for λ in terms of n Then ⎛ 2π rn ⎞ 2π rn ⎟= n ⎝ nh ⎠ λ = h⎜ EXECUTE: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10 −10 m, so λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit (b) For n = 4, λ = 2π r4 /4 rn = n a0 so r4 = 16a0 λ = 2π (16a0 )/4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10−9 m 1 λ = 2π r4 /4; the de Broglie wavelength is = times the circumference of the orbit n EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength decreases For any n, the circumference of the orbits equals an integer number of de Broglie wavelengths 39.6 IDENTIFY: λ = h p SET UP: eV = 1.60 × 10−19 J An electron has mass 9.11 × 10−31 kg EXECUTE: (a) For a nonrelativistic particle, K = p2 h , so λ = = 2m p h Km (b) (6.63 × 10−34 J ⋅ s) / 2(800 eV)(1.60 × 10−19 J/eV)(9.11 × 10−31 kg) = 4.34 × 10−11 m 39.7 EVALUATE: The de Broglie wavelength decreases when the kinetic energy of the particle increases IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave properties SET UP: The de Broglie wavelength of the person is λ = h /mv EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s λ = h /mv = (6.626 × 10−34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10−36 m EVALUATE: (b) A typical doorway is about m wide, so the person’s de Broglie wavelength is much too small to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength Hence ordinary objects not show wave behavior in everyday life 39.8 IDENTIFY and SET UP: Combining Eqs 37.38 and 37.39 gives p = mc γ − EXECUTE: (a) λ = h = ( h /mc)/ γ − = 4.43 × 10−12 m (The incorrect nonrelativistic calculation gives p 5.05 × 10−12 m ) (b) (h /mc)/ γ − = 7.07 × 10−13 m EVALUATE: The de Broglie wavelength decreases when the speed increases © 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 Particles Behaving as Waves 39.9 39-3 IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq (38.2) An electron has mass Its energy is related to its momentum by E = p /2m and its wavelength is related to its momentum by Eq (39.1) hc hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) EXECUTE: (a) photon: E = so λ = = = 62.0 nm λ E (20.0 eV)(1.602 × 10−19 J/eV) electron: E = p /(2m) so p = 2mE = 2(9.109 × 10−31 kg)(20.0 eV)(1.602 × 10−19 J/eV) = 2.416 × 10−24 kg ⋅ m/s λ = h /p = 0.274 nm (b) photon: E = hc/R = 7.946 × 10−19 J = 4.96 eV electron: λ = h /p so p = h /λ = 2.650 × 10−27 kg ⋅ m/s E = p /(2m) = 3.856 × 10−24 J = 2.41 × 10−5 eV (c) EVALUATE: You should use a probe of wavelength approximately 250 nm An electron with λ = 250 nm has much less energy than a photon with λ = 250 nm, so is less likely to damage the molecule Note that λ = h /p applies to all particles, those with mass and those with zero mass E = hf = hc /λ applies only to photons and E = p /2m applies only to particles with mass 39.10 IDENTIFY: Knowing the de Broglie wavelength for an electron, we want to find its speed h h = 1.00 mm, m = 9.11 × 10−31 kg SET UP: λ = = p mv h 6.63 × 10−34 J ⋅ s = = 0.728 m/s mλ (9.11 × 10−31 kg)(1.00 × 10−3 m) EVALUATE: Electrons normally move much faster than this, so their de Broglie wavelengths are much much smaller than a millimeter IDENTIFY and SET UP: Use Eq (39.1) h h 6.626 × 10−34 J ⋅ s EXECUTE: λ = = = = 3.90 × 10−34 m p mv (5.00 × 10−3 kg)(340 m/s) EXECUTE: v = 39.11 39.12 EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties IDENTIFY: The kinetic energy of the electron is equal to the energy of the photon We want to find the wavelengths of each of them SET UP: Both for particles with mass (electrons) and for massless particles (photons) the wavelength is h related to the momentum p by λ = But for each type of particle there is a different expression that p relates the energy E and momentum p For an electron E = 12 mv = EXECUTE: photon: p = p2 but for a photon E = pc 2m E h h E hc 1.24 × 10−6 eV ⋅ m and p = so = and λ = = = 49.6 nm λ λ c c E 25 eV electron: Solving for p gives p = 2mE This gives p = 2(9.11 × 10−31 kg)(25 eV)(1.60 × 10−19 J/eV) = 2.70 × 10−24 kg ⋅ m/s The wavelength is therefore λ= h 6.63 × 10−34 J ⋅ s = = 0.245 nm p 2.70 × 10−24 kg ⋅ m/s EVALUATE: The wavelengths are quite different For the electron λ = h hc and for the photon λ = , E 2mE so for an electron λ is proportional to E −1/2 and for a photon λ is proportional to E −1 It is incorrect to say p = E for a particle such as an electron that has mass; the correct relation is p = c E − ( mc ) c © 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 39-4 Chapter 39 39.13 IDENTIFY: The acceleration gives momentum to the electrons We can use this momentum to calculate their de Broglie wavelength SET UP: The kinetic energy K of the electron is related to the accelerating voltage V by K = eV For an electron E = 12 mv = h p2 hc and λ = For a photon E = λ 2m p EXECUTE: (a) For an electron p = h λ = 6.63 × 10−34 J ⋅ s 5.00 × 10−9 m = 1.33 × 10−25 kg ⋅ m/s and p (1.33 × 10−25 kg ⋅ m/s) K 9.71 × 10−21 J = = 9.71 × 10−21 J V = = = 0.0607 V The electrons − 31 2m e 1.60 × 10−19 C 2(9.11 × 10 kg) would have kinetic energy 0.0607 eV E= (b) E = hc λ = 1.24 × 10−6 eV ⋅ m 5.00 × 10−9 m = 248 eV (c) E = 9.71 × 10−21 J so λ = 39.14 hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 20.5 μ m E 9.71 × 10−21 J EVALUATE: If they have the same wavelength, the photon has vastly more energy than the electron h IDENTIFY: λ = Apply conservation of energy to relate the potential difference to the speed of the p electrons hc SET UP: The mass of an electron is m = 9.11 × 10−31 kg The kinetic energy of a photon is E = λ EXECUTE: (a) λ = h /mv → v = h /mλ Energy conservation: eΔV = mv 2 ⎛ h ⎞ m⎜ ⎟ mv h2 (6.626 × 10−34 J ⋅ s) mλ ⎠ ΔV = = ⎝ = = = 66.9 V − 19 2e 2e 2emλ 2(1.60 × 10 C)(9.11 × 10−31 kg) (0.15 × 10−9 m) (b) Ephoton = hf = ΔV = Ephoton e = hc λ = (6.626 × 10−34 J ⋅ s) (3.0 × 108 m/s) 1.33 × 10−15 J 1.6 × 10−19 C 0.15 × 10−9 m = 8310 V EVALUATE: The electron in part (b) has wavelength λ = 39.15 = 1.33 × 10−15 J eΔV = K = Ephoton and h = p h = 0.0135 nm, much shorter than the 2mE wavelength of a photon of the same energy h hc IDENTIFY: For an electron, λ = and K = 12 mv For a photon, E = The wavelength should be 0.10 nm p λ SET UP: For an electron, m = 9.11 × 10−31 kg EXECUTE: (a) λ = 0.10 nm p = mv = h /λ so v = h /(mλ ) = 7.3 × 106 m/s (b) K = mv = 150 eV (c) E = hc /λ = 12 keV EVALUATE: (d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure being probed © 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 Particles Behaving as Waves 39.16 39-5 IDENTIFY: The electrons behave like waves and are diffracted by the slit SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find their de Broglie wavelength, which is λ = h /mv Finally we know that the first dark fringe for single-slit diffraction occurs when a sin θ = λ EXECUTE: (a) Use energy conservation to find the speed of the electron: mv = eV 2eV 2(1.60 × 10−19 C)(100 V) = = 5.93 × 106 m/s m 9.11 × 10−31 kg v= which is about 2% the speed of light, so we can ignore relativity (b) First find the de Broglie wavelength: λ= h 6.626 × 10−34 J ⋅ s = = 1.23 × 10−10 m = 0.123 nm mv (9.11 × 10−31 kg)(5.93 × 106 m/s) For the first single-slit dark fringe, we have a sin θ = λ , which gives a= 39.17 λ sin θ = 1.23 × 10−10 m = 6.16 × 10−10 m = 0.616 nm sin(11.5°) EVALUATE: The slit width is around times the de Broglie wavelength of the electron, and both are much smaller than the wavelength of visible light h IDENTIFY: The intensity maxima are located by Eq (39.4) Use λ = for the wavelength of the p neutrons For a particle, p = 2mE SET UP: For a neutron, m = 1.67 × 10−27 kg EXECUTE: For m = 1, λ = d sin θ = E= h2 2md sin θ = h 2mE (6.63 × 10−34 J ⋅ s)2 2(1.675 × 10−27 kg) (9.10 × 10−11 m) sin (28.6°) = 6.91 × 10−20 J = 0.432 eV EVALUATE: The neutrons have λ = 0.0436 nm, comparable to the atomic spacing 39.18 IDENTIFY: Intensity maxima occur when d sin θ = mλ λ = h h mh = so d sin θ = p 2ME 2ME SET UP: Here m is the order of the maxima, whereas M is the mass of the incoming particle EXECUTE: (a) d = (2)(6.63 × 10−34 J ⋅ s) mh = = 2ME sin θ 2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV) sin(60.6°) 2.06 × 10−10 m = 0.206 nm (b) m = also gives a maximum ⎛ ⎞ (1)(6.63 × 10−34 J ⋅ s) ⎟ = 25.8° This is the only other θ = arcsin ⎜ ⎜ 2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV)(2.06 × 10−10 m) ⎟ ⎝ ⎠ one If we let m ≥ 3, then there are no more maxima (c) E = m2h2 Md sin θ = (1) (6.63 × 10−34 J ⋅ s) 2(9.11 × 10−31 kg) (2.60 × 10−10 m) sin (60.6°) = 7.49 × 10−18 J = 46.8 eV Using this energy, if we let m = 2, then sin θ > Thus, there is no m = maximum in this case EVALUATE: As the energy of the electrons is lowered their wavelength increases and a given intensity maximum occurs at a larger angle © 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 39-6 Chapter 39 39.19 IDENTIFY: The condition for a maximum is d sin θ = mλ λ = h h ⎛ mh ⎞ = , so θ = arcsin ⎜ ⎟ p Mv ⎝ dMv ⎠ SET UP: Here m is the order of the maximum, whereas M is the incoming particle mass ⎛ h ⎞ EXECUTE: (a) m = ⇒ θ1 = arcsin ⎜ ⎟ ⎝ dMv ⎠ ⎛ ⎞ 6.63 × 10−34 J ⋅ s = arcsin ⎜ = 2.07° ⎜ (1.60 × 10−6 m)(9.11 × 10−31 kg)(1.26 × 104 m/s) ⎟⎟ ⎝ ⎠ ⎛ ⎞ (2)(6.63 × 10−34 J ⋅ s) = 4.14° m = ⇒ θ = arcsin ⎜ ⎜ (1.60 × 10−6 m)(9.11 × 10−31 kg)(1.26 × 104 m/s) ⎟⎟ ⎝ ⎠ ⎛ π radians ⎞ (b) For small angles (in radians!) y ≅ Dθ , so y1 ≈ (50.0 cm) (2.07°) ⎜ ⎟ = 1.81 cm, ⎝ 180° ⎠ ⎛ π radians ⎞ y2 ≈ (50.0 cm) (4.14°) ⎜ ⎟ = 3.61 cm and y2 − y1 = 3.61 cm − 1.81 cm = 1.80 cm ⎝ 180° ⎠ EVALUATE: For these electrons, λ = h = 0.0577 μm λ is much less than d and the intensity maxima mv occur at small angles 39.20 IDENTIFY: λ = h p2 Conservation of energy gives eV = K = , where V is the accelerating voltage p 2m SET UP: The electron mass is 9.11 × 10−31 kg and the proton mass is 1.67 × 10−27 kg EXECUTE: (a) eV = K = p ( h /λ ) (h /λ ) = = 419 V , so V = 2m 2m 2me (b) The voltage is reduced by the ratio of the particle masses, (419 V) 9.11 × 10−31 kg 1.67 × 10−27 kg = 0.229 V h h = For the same λ , particles of greater mass have smaller E, so a smaller p 2mE accelerating voltage is needed for protons EVALUATE: λ = 39.21 IDENTIFY and SET UP: For a photon Eph = hc λ = 1.99 × 10−25 J ⋅ m λ For an electron Ee = p2 ⎛h⎞ h2 = = ⎜ ⎟ 2m 2m ⎝ λ ⎠ 2mλ EXECUTE: (a) photon Eph = electron Ee = Eph 1.99 × 10−25 J ⋅ m 10.0 × 10−9 m = 1.99 × 10−17 J (6.63 × 10−34 J ⋅ s) 2(9.11 × 10−31 kg)(10.0 × 10−9 m) = 2.41 × 10−21 J 1.99 × 10−17 J = 8.26 × 103 2.41 × 10−21 J (b) The electron has much less energy so would be less damaging Ee = EVALUATE: For a particle with mass, such as an electron, E ~ λ −2 For a massless photon E ~ λ −1 © 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 Particles Behaving as Waves 39.22 39-7 IDENTIFY: The kinetic energy of the alpha particle is all converted to electrical potential energy at closest approach The force on the alpha particle is the electrical repulsion of the nucleus q1q2 SET UP: The electrical potential energy of the system is U = and the kinetic energy is 4πε r K = 12 mv The electrical force is R = 2.5 m (at closest approach) (a) Equating the initial kinetic energy and the final potential energy and solving for the separation radius r gives (92e) (2e) (184) (1.60 × 10−19 C) = = 5.54 × 10−14 m 4πε 4πε (4.78 × 106 eV)(1.60 × 10−19 J/eV) K (b) The above result may be substituted into Coulomb’s law Alternatively, the relation between the U magnitude of the force and the magnitude of the potential energy in a Coulomb field is F = U = K, r r= so F = 39.23 K (4.78 × 106 eV) (1.6 × 10−19 J/ev) = = 13.8 N r (5.54 × 10−14 m) EVALUATE: The result in part (a) is comparable to the radius of a large nucleus, so it is reasonable The force in part (b) is around pounds, which is large enough to be easily felt by a person q1q2 (a) IDENTIFY: If the particles are treated as point charges, U = 4π ⑀0 r SET UP: q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U EXECUTE: U = (8.987 × 109 N ⋅ m /C ) (2)(82)(1.602 × 10−19 C) 6.50 × 10−14 m = 5.82 × 10−13 J U = 5.82 × 10−13 J (1 eV/1.602 × 10−19 J) = 3.63 × 106 eV = 3.63 MeV (b) IDENTIFY: Apply conservation of energy: K1 + U1 = K + U SET UP: Let point be the initial position of the alpha particle and point be where the alpha particle momentarily comes to rest Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = Alpha particle stops implies K = EXECUTE: Conservation of energy thus says K1 = U = 5.82 × 10−13 J = 3.63 MeV (c) K = 39.24 2K 2(5.82 × 10−13 J) mv so v = = = 1.32 × 107 m/s m 6.64 × 10−27 kg EVALUATE: v /c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v When the alpha particle stops, all its initial kinetic energy has been converted to electrostatic potential energy IDENTIFY: The minimum energy the photon would need is the 3.84 eV bond strength hc SET UP: The photon energy E = hf = must equal the bond strength λ hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 327 nm λ 3.80 eV 3.80 eV EVALUATE: Any photon having a shorter wavelength would also spell doom for the Horta! IDENTIFY and SET UP: Use the energy to calculate n for this state Then use the Bohr equation, Eq (39.6), to calculate L EXECUTE: En = −(13.6 eV)/n , so this state has n = 13.6/1.51 = In the Bohr model, L = n so for EXECUTE: 39.25 hc = 3.80 eV, so λ = this state L = 3ћ = 3.16 × 10−34 kg ⋅ m /s EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different result © 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 39-8 39.26 Chapter 39 13.6 eV hc ΔE = , where ΔE is the magnitude of λ n2 the energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted IDENTIFY and SET UP: For a hydrogen atom En = − ⎛ 1⎞ EXECUTE: ΔE = E4 − E1 = −(13.6 eV) ⎜ − ⎟ = +12.75 eV ⎝4 ⎠ λ= hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) c = = 97.3 nm f = = 3.08 × 1015 Hz λ ΔE 12.75 eV EVALUATE: This photon is in the ultraviolet region of the electromagnetic spectrum 39.27 IDENTIFY: The force between the electron and the nucleus in Be3+ is F = Ze2 4π ⑀0 r , where Z = is the nuclear charge All the equations for the hydrogen atom apply to Be3+ if we replace e by Ze (a) SET UP: Modify Eq (39.14) EXECUTE: En = − En = − m( Ze ) ⑀02 8n 2h me ⑀02 8n 2h (hydrogen) becomes ⎛ me ⎞ ⎛ 13.60 eV ⎞ 3+ = Z2⎜− 2 ⎟ = Z2⎜− ⎟ (for Be ) ⎜ ⑀ 8n h ⎟ n ⎝ ⎠ ⎝ ⎠ ⎛ 13.60 eV ⎞ The ground-level energy of Be3+ is E1 = 16 ⎜ − ⎟ = −218 eV 12 ⎝ ⎠ EVALUATE: The ground-level energy of Be3+ is Z = 16 times the ground-level energy of H (b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the n = level energy EXECUTE: The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV EVALUATE: This is 16 times the ionization energy of hydrogen (c) SET UP: ⎛ 1 ⎞ = R ⎜ − ⎟ just as for hydrogen but now R has a different value ⎜ ⎟ λ ⎝ n1 n2 ⎠ EXECUTE: RH = RBe = Z me4 8⑀02 h3c me4 8⑀02 h3c = 1.097 × 107 m −1 for hydrogen becomes = 16(1.097 × 107 m −1 ) = 1.755 × 108 m −1 for Be3+ For n = to n = 1, ⎛1 ⎞ = RBe ⎜ − ⎟ = 3RBe /4 λ ⎝1 ⎠ λ = 4/(3RBe ) = 4/(3(1.755 × 108 m −1)) = 7.60 × 10−9 m = 7.60 nm EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding transition in the hydrogen atom (d) SET UP: Modify Eq (39.8): rn = ⑀ EXECUTE: rn = ⑀ n2h2 π m( Ze2 ) n2h2 π me2 (hydrogen) (Be3+ ) EVALUATE: For a given n the orbit radius for Be3+ is smaller by a factor of Z = compared to the corresponding radius for hydrogen © 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 Particles Behaving as Waves 39.28 IDENTIFY and SET UP: En = − EXECUTE: (a) En = − 13.6 eV n2 13.6 eV n2 39-9 and En +1 = − 13.6 eV (n + 1) ⎡ 1⎤ n − ( n + 1) 2n + ΔE = En +1 − En = ( −13.6 eV) ⎢ − ⎥ = − (13.6 eV) ΔE = (13.6 eV) 2 + + ( 1) ( )( 1) n n n n ( n )( n + 1) ⎣ ⎦ 2n As n becomes large, ΔE → (13.6 eV) = (13.6 eV) Thus ΔE becomes small as n becomes large n n (b) rn = n r1 so the orbits get farther apart in space as n increases EVALUATE: There are an infinite number of bound levels for the hydrogen atom As n increases the energies of the bound levels converge to the ionization threshold 39.29 IDENTIFY: Apply Eqs (39.8) and (39.9) SET UP: The orbital period for state n is the circumference of the orbit divided by the orbital speed EXECUTE: (a) = n = ⇒ v2 = v1 v = 1.09 × 106 m/s n = ⇒ v3 = = 7.27 × 105 m/s (b) Orbital period = n = ⇒ T1 = e2 (1.60 × 10−19 C) : n = ⇒ v1 = = 2.18 × 106 m/s ε 2nh ε (6.63 × 10−34 J ⋅ s) 2π rn 2ε n h /me2 4ε 02 n3h3 = = me4 1/ε ⋅ e2 /2nh 4ε 02 (6.63 × 10−34 J ⋅ s)3 (9.11 × 10−31 kg)(1.60 × 10−19 C) = 1.53 × 10−16 s n = 2: T2 = T1 (2)3 = 1.22 × 10−15 s n = 3: T3 = T1 (3)3 = 4.13 × 10−15 s (c) number of orbits = 1.0 × 10−8 s 1.22 × 10−15 s = 8.2 × 106 EVALUATE: The orbital speed is proportional to1/n, the orbital radius is proportional to n , and the orbital period is proportional to n3 39.30 IDENTIFY and SET UP: The ionization threshold is at E = The energy of an absorbed photon equals the energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom EXECUTE: (a) ΔE = − (−20 eV) = 20 eV (b) When the atom in the n = level absorbs an 18-eV photon, the final level of the atom is n = The possible transitions from n = and corresponding photon energies are n = → n = 3, eV; n = → n = 2, eV; n = → n = 1, 18 eV Once the atom has gone to the n = level, the following transitions can occur: n = → n = 2, eV; n = → n = 1,15 eV Once the atom has gone to the n = level, the following transition can occur: n = → n = 1, 10 eV The possible energies of emitted photons are: eV, eV, eV, 10 eV, 15 eV and 18 eV (c) There is no energy level eV higher in energy than the ground state, so the photon cannot be absorbed (d) The photon energies for n = → n = and for n = → n = are eV and 15 eV The photon energy for n = → n = is eV The work function must have a value between eV and eV EVALUATE: The atom has discrete energy levels, so the energies of emitted or absorbed photons have only certain discrete energies © 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 39-10 39.31 Chapter 39 IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy Ei − Ef of the atom by Ei − Ef = hc λ where hc = 1.240 × 10−6 eV ⋅ m EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0, so E1 = −17.50 eV For n = → n = 1, λ = 73.86 nm and E5 − E1 = 1.240 × 10−6 eV ⋅ m 73.86 × 10−9 m = 16.79 eV E5 = −17.50 eV + 16.79 eV = −0.71 eV For n = → n = 1, λ = 75.63 nm and E4 = −1.10 eV For n = → n = 1, λ = 79.76 nm and E3 = −1.95 eV For n = → n = 1, λ = 94.54 nm and E2 = −4.38 eV (b) Ei − Ef = E4 − E2 = −1.10 eV − (−4.38 eV) = 3.28 eV and λ = 39.32 hc 1.240 × 10−6 eV ⋅ m = = 378 nm Ei − Ef 3.28 eV EVALUATE: The n = → n = transition energy is smaller than the n = → n = transition energy so the wavelength is longer In fact, this wavelength is longer than for any transition that ends in the n = state IDENTIFY and SET UP: For the Lyman series the final state is n = and the wavelengths are given by ⎛1 ⎞ = R ⎜ − ⎟ , n = 2, 3,… For the Paschen series the final state is n = and the wavelengths are given λ ⎝1 n ⎠ ⎞ ⎛ = R ⎜ − ⎟ , n = 4, 5,… R = 1.097 × 107 m −1 The longest wavelength is for the smallest n and λ n ⎠ ⎝3 the shortest wavelength is for n → ∞ ⎛ 1 ⎞ 3R EXECUTE: Lyman: Longest: = R ⎜ − ⎟ = = 121.5 nm λ= λ 3(1.097 × 107 m −1) ⎝1 ⎠ by Shortest: λ 1 ⎞ ⎛1 = R ⎜ − ⎟ = R λ = = 91.16 nm 1.097 × 107 m −1 ⎝1 ∞ ⎠ Paschen: Longest: ⎞ R ⎛ = R⎜ − ⎟ = ∞ ⎠ ⎝3 EVALUATE: The Lyman series is in the ultraviolet The Paschen series is in the infrared IDENTIFY: Apply conservation of energy to the system of atom and photon hc SET UP: The energy of a photon is Eγ = Shortest: 39.33 144 ⎞ 7R ⎛ = R⎜ − ⎟ = = 1875 nm .λ= λ 7(1.097 × 107 m −1 ) ⎠ 144 ⎝3 1 λ λ hc (6.63 × 10 J ⋅ s)(3.00 × 108 m/s) EXECUTE: (a) Eγ = = = 2.31 × 10−19 J = 1.44 eV So the internal λ 8.60 × 10−7 m energy of the atom increases by 1.44 eV to E = −6.52 eV + 1.44 eV = −5.08 eV −34 (b) Eγ = hc λ = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) −7 = 4.74 × 10−19 J = 2.96 eV So the final internal energy of 4.20 × 10 m the atom decreases to E = −2.68 eV − 2.96 eV = −5.64 eV 39.34 EVALUATE: When an atom absorbs a photon the energy of the atom increases When an atom emits a photon the energy of the atom decreases 1 ⎞ ⎛ IDENTIFY and SET UP: Balmer’s formula is = R ⎜ − ⎟ For the Hγ spectral line n = Once we λ n ⎠ ⎝2 have λ , calculate f from f = c /λ and E from Eq (38.2) EXECUTE: (a) ⎞ ⎛ ⎛ 25 − ⎞ ⎛ 21 ⎞ = R⎜ − ⎟ = R⎜ ⎟ = R⎜ ⎟ λ ⎠ ⎝2 ⎝ 100 ⎠ ⎝ 100 ⎠ © 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 39-16 39.55 Chapter 39 (a) IDENTIFY and SET UP: Apply Eq (39.17): mr = EXECUTE: mr = 207 me mp m1m2 = m1 + m2 207 me + mp 207(9.109 × 10−31 kg)(1.673 × 10−27 kg) 207(9.109 × 10−31 kg) + 1.673 × 10−27 kg = 1.69 × 10−28 kg We have used me to denote the electron mass (b) IDENTIFY: In Eq (39.14) replace m = me by mr : En = − mr e ⑀ 02 8n 2h ⎛ m ⎞ ⎛ m e4 ⎞ m e4 SET UP: Write as En = ⎜ r ⎟ ⎜ − H2 ⎟ , since we know that H = 13.60 eV Here mH denotes ⎜ ⎟ ⑀ 8h ⎝ mH ⎠ ⎝ ⑀ 8n h ⎠ the reduced mass for the hydrogen atom; mH = 0.99946(9.109 × 10−31 kg) = 9.104 × 10−31 kg ⎛ m ⎞ ⎛ 13.60 eV ⎞ EXECUTE: En = ⎜ r ⎟ ⎜ − ⎟ n2 ⎠ ⎝ mH ⎠ ⎝ E1 = 1.69 × 10−28 kg 9.109 × 10−31 kg (−13.60 eV) = 186(−13.60 eV) = −2.53 keV ⎛ m ⎞ ⎛ R ch ⎞ (c) SET UP: From part (b), En = ⎜ r ⎟ ⎜ − H2 ⎟ , where RH = 1.097 × 107 m −1 is the Rydberg constant ⎝ mH ⎠ ⎝ n ⎠ hc = Ei − Ef to find an expression for 1/λ The initial level for for the hydrogen atom Use this result in λ the transition is the ni = level and the final level is the nf = level EXECUTE: λ = = hc λ = mr ⎛ RH ch ⎛ RH ch ⎞ ⎞ − ⎜ − ⎟⎟ ⎜− ⎜ n ⎟⎟ mH ⎜⎝ ni2 ⎝ f ⎠⎠ ⎛ mr ⎞ RH ⎜ − ⎟ ⎜ ⎟ mH ⎝ nf ni ⎠ 1.69 × 10−28 kg λ 9.109 × 10 λ = 0.655 nm −31 ⎛1 ⎞ (1.097 × 107 m −1) ⎜ − ⎟ = 1.527 × 109 m −1 kg ⎝1 ⎠ EVALUATE: From Example 39.6 the wavelength of the radiation emitted in this transition in hydrogen is m 122 nm The wavelength for muonium is H = 5.39 × 10−3 times this The reduced mass for hydrogen is mr very close to the electron mass because the electron mass is much less then the proton mass: mp /me = 1836 The muon mass is 207 me = 1.886 × 10−28 kg The proton is only about 10 times more massive than the muon, so the reduced mass is somewhat smaller than the muon mass The muon-proton atom has much more strongly bound energy levels and much shorter wavelengths in its spectrum than for hydrogen 39.56 IDENTIFY: Apply conservation of momentum to the system of atom and emitted photon SET UP: Assume the atom is initially at rest For a photon E = hc λ and p = h λ EXECUTE: (a) Assume a non-relativistic velocity and conserve momentum ⇒ mv = h λ ⇒v= h mλ 1 ⎛ h ⎞ h (b) K = mv = m ⎜ ⎟ = 2 ⎝ mλ ⎠ 2mλ © 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 Particles Behaving as Waves 39-17 K h2 λ h = ⋅ = Recoil becomes an important concern for small m and small λ since this E 2mλ hc 2mcλ ratio becomes large in those limits hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) (d) E = 10.2 eV ⇒ λ = = = 1.22 × 10−7 m = 122 nm E (10.2 eV)(1.60 × 10−19 J/eV) (c) K= 39.57 (6.63 × 10−34 J ⋅ s) 2(1.67 × 10−27 kg)(1.22 × 10−7 m) = 8.84 × 10−27 J = 5.53 × 10−8 eV K 5.53 × 10−8 eV = = 5.42 × 10−9 This is quite small so recoil can be neglected E 10.2 eV EVALUATE: For emission of photons with ultraviolet or longer wavelengths the recoil kinetic energy of the atom is much less than the energy of the emitted photon IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the n = to n = transition En = − 13.6 eV n2 hc λ = ΔE ⎛ 1⎞ EXECUTE: (a) The atom must be given an amount of energy E3 − E1 = −(13.6 eV) ⎜ − ⎟ = 12.1 eV ⎝3 ⎠ hc (b) There are three possible transitions n = → n = 1: ΔE = 12.1 eV and λ = = 103 nm; ΔE ⎞ ⎛ n = → n = : ΔE = −(13.6 eV) ⎜ − ⎟ = 1.89 eV and λ = 657 nm; n = → n = 1: ⎠ ⎝3 39.58 ⎛ 1⎞ ΔE = −(13.6 eV) ⎜ − ⎟ = 10.2 eV and λ = 122 nm ⎝2 ⎠ EVALUATE: The larger the transition energy for the atom, the shorter the wavelength n − ( E − E )/kT IDENTIFY: Apply = e ex g n1 −13.6 eV = −3.4 eV Eg = −13.6 eV Eex − Eg = 10.2 eV = 1.63 × 10−18 J −( Eex − Eg ) n2 − (1.63 × 10−18 J) = 10−12 T = = 4275 K EXECUTE: (a) T = k ln( n2 /n1) n1 (1.38 × 10−23 J/K) ln(10−12 ) SET UP: Eex = E2 = (b) − (1.63 × 10−18 J) n2 = 6412 K = 10−8 T = n1 (1.38 × 10−23 J/K) ln(10−8 ) n2 − (1.63 × 10−18 J) = 10−4 T = = 12824 K n1 (1.38 × 10−23 J/K) ln(10−4 ) EVALUATE: (d) For absorption to take place in the Balmer series, hydrogen must start in the n = state From part (a), colder stars have fewer atoms in this state leading to weaker absorption lines (a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom Some of this energy overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron Apply hf = Ef − Ei , the energy given to the electron in the atom when a photon is absorbed (c) 39.59 EXECUTE: The energy of one photon is hc λ hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 85.5 × 10−9 m = 2.323 × 10−18 J(1 eV/1.602 × 10−19 J) = 14.50 eV The final energy of the electron is Ef = Ei + hf In the ground state of the hydrogen atom the energy of the electron is Ei = −13.60 eV Thus Ef = −13.60 eV + 14.50 eV = 0.90 eV © 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 39-18 Chapter 39 (b) EVALUATE: At thermal equilibrium a few atoms will be in the n = excited levels, which have an energy of −13.6 eV/4 = −3.40 eV,10.2 eV greater than the energy of the ground state If an electron with E = −3.40 eV gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV − 3.4 eV = 11.1 eV of kinetic energy 39.60 IDENTIFY: For circular motion, L = mvr and a = v2 mM Newton’s law of gravitation is Fg = G , r r with G = 6.67 × 10−11 N ⋅ m /kg SET UP: The period T is 2.00 h = 7200 s EXECUTE: (a) mvr = n n= h 2π mvr 2π r n = v = So h T 2π (2π r )2 m (2π) (8.06 × 106 m) (20.0 kg) = = 1.08 × 1046 hT (6.63 × 10−34 J s)(7200 s) (b) F = ma gives G mmE r2 =m v GmE nh GmE n2h2 = v The Bohr postulate says v = so = 2 r r 2πmr r 4π m r ⎛ ⎞ h2 r =⎜ n This is in the form r = kn 2, with ⎜ 4π Gm m ⎟⎟ E ⎝ ⎠ h2 k= 4π GmE m = (6.63 × 10−34 J.s)2 4π (6.67 × 10 −11 N ⋅ m /kg )(5.97 × 1024 kg) = 7.0 × 10−86 m (c) Δr = rn +1 − rn = k ([n + 1]2 − n ) = (2n + 1)k = (2[1.08 × 1046 ] + 1)(7.0 × 10−86 m) = 1.5 × 10−39 m EVALUATE: (d) Δr is exceedingly small, so the separation of adjacent orbits is not observable (e) There is no measurable difference between quantized and classical orbits for this satellite; either method of calculation is totally acceptable 39.61 IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan-Boltzmann law apply to its radiation SET UP: Wien’s displacement law is λ peak = 2.90 × 10−3 m ⋅ K , and the Stefan-Boltzmann law says that T the intensity of the radiation is I = σ T , so the total radiated power is P = σ AT EXECUTE: (a) First use Wien’s law to find the peak wavelength: λ m = (2.90 × 10−3 m ⋅ K)/(3000 K) = 9.667 × 10−7 m Call N the number of photons/second radiated N × (energy per photon) = IA = σ AT N (hc/λm ) = σ AT N = N= λmσ AT hc (9.667 × 10−7 m)(5.67 × 10−8 W/m ⋅ K )(4π )(600 × 6.96 × 108 m) (3000 K) (6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s) N = × 1049 photons/s (b) I B AB σ ABTB4 4π RB2TB4 ⎛ 600 RS ⎞ ⎛ 3000 K ⎞ = = =⎜ ⎟ ⎜ ⎟ = × 10 IS AS σ ASTS4 4π RS2TS4 ⎝ RS ⎠ ⎝ 5800 K ⎠ EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun! © 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 Particles Behaving as Waves 39.62 39-19 IDENTIFY: The diffraction grating allows us to determine the peak-intensity wavelength of the light Then Wien’s displacement law allows us to calculate the temperature of the blackbody, and the StefanBoltzmann law allows us to calculate the rate at which it radiates energy SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ Wien’s displacement law is λ peak = 2.90 × 10−3 m ⋅ K , and the Stefan-Boltzmann law says that the intensity of the radiation is T I = σ T , so the total radiated power is P = σ AT EXECUTE: (a) First find the wavelength of the light: λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10−7 m Now use Wien’s law to find the temperature: T = (2.90 × 10−3 m ⋅ K)/(5.22 × 10−7 m) = 5550 K (b) The energy radiated by the blackbody is equal to the power times the time, giving U = Pt = IAt = σ AT 4t , which gives t = U /(σ AT ) = (12.0 × 106 J)/[(5.67 × 10−8 W/m ⋅ K )(4π )(0.0750 m)2 (5550 K)4 ] = 3.16 s 39.63 EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of energy IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the n = and the n = states Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength 13.6 eV SET UP: In the Bohr model, the energy of an electron in shell n is En = − , and Wien’s n2 2.90 × 10−3 m ⋅ K The energy of a photon is E = hf = hc /λ T EXECUTE: First find the energy (ΔE) that a photon would need to excite the atom The ground state of the atom is n = and the third excited state is n = This energy is the difference between the two energy ⎛ 1⎞ levels Therefore ΔE = (−13.6 eV) ⎜ − ⎟ = 12.8 eV Now find the wavelength of the photon having ⎝4 ⎠ this amount of energy hc /λ = 12.8 eV and displacement law is λ m = λ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(12.8 eV) = 9.73 × 10−8 m Now use Wien’s law to find the temperature T = (0.00290 m ⋅ K)/(9.73 × 10−8 m) = 2.98 × 104 K 39.64 EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in excited states during everyday conditions IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody The net heat entering the water causes evaporation Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is dQ 4 = σ A Tsphere − Twater Since this heat evaporates water, the rate at which water evaporates is dt ( ) dQ dm 2.90 × 10−3 m ⋅ K = Lv Wien’s displacement law is λ m = , and the wavelength in the water is dt dt T λ w =λ0 /n EXECUTE: (a) The net radiated heat is ( ) dQ 4 = σ A Tsphere − Twater and the evaporation rate is dt dQ dm = Lv , where dm is the mass of water that evaporates in time dt Equating these two rates gives dt dt © 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 39-20 Chapter 39 ( ) 4 dm dm σ (4π R ) Tsphere − Twater 4 Lv = σ A Tsphere − Twater = dt dt Lv ( dm (5.67 × 10 = dt ) −8 W/m ⋅ K )(4π )(0.120 m) ⎡ (498 K) − (373 K) ⎤ ⎣ ⎦ = 1.92 × 10−4 kg/s = 0.193 g/s 2256 × 103 J/kg (b) (i) Wien’s law gives λ m = (0.00290 m ⋅ K)/(498 K) = 5.82 × 10−6 m But this would be the wavelength in vacuum In the water the thermophile organism would measure λ w = λ0 /n = (5.82 × 10−6 m)/1.333 = 4.37 ì 106 m = 4.37 àm (ii) The frequency is the same as if the wave were in air, so f = c /λ0 = (3.00 × 108 m/s)/(5.82 × 10−6 m) = 5.15 × 1013 Hz EVALUATE: An alternative way is to use the quantities in the water: f = 39.65 c /n λ0 /n = c /λ0 , which gives the same answer for the frequency An organism in the water would measure the light coming to it through the water, so the wavelength it would measure would be reduced by a factor of 1/n IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom plus photon (a) SET UP: Let Etr be the transition energy, Eph be the energy of the photon with wavelength λ ′, and Er be the kinetic energy of the recoiling atom Conservation of energy gives Eph + Er = Etr Eph = hc hc hc so = Etr − Er and λ ′ = λ′ λ′ Etr − Er EXECUTE: If the recoil energy is neglected then the photon wavelength is λ = hc /Etr ⎛ ⎞ 1 ⎞ ⎛ hc ⎞⎛ Δλ = λ ′ − λ = hc ⎜ − − 1⎟ ⎟=⎜ ⎟⎜ − − E E E E E / E r tr ⎠ ⎝ tr ⎠⎝ r tr ⎝ tr ⎠ −1 ⎛ E E ⎞ E = ⎜1 − r ⎟ ≈ + r since r Etr Etr − Er /Etr ⎝ Etr ⎠ (We have used the binomial theorem, Appendix B.) Thus Δλ = hc ⎛ Er ⎞ ⎛ Er ⎞ ⎜ ⎟ , or since Etr = hc /λ , Δλ = ⎜ ⎟ λ Etr ⎝ Etr ⎠ ⎝ hc ⎠ SET UP: Use conservation of linear momentum to find Er : Assuming that the atom is initially at rest, the momentum pr of the recoiling atom must be equal in magnitude and opposite in direction to the momentum pph = h /λ of the emitted photon: h /λ = pr EXECUTE: Er = h2 pr2 , where m is the mass of the atom, so Er = 2m 2mλ ⎛ h ⎞⎛ λ ⎞ h ⎛E ⎞ Use this result in the above equation: Δλ = ⎜ r ⎟ λ = ⎜ ; ⎟⎟ = ⎟⎜ ⎜ ⎟⎜ hc hc mc m λ ⎝ ⎠ ⎝ ⎠⎝ ⎠ note that this result for Δλ is independent of the atomic transition energy (b) For a hydrogen atom m = mp and Δλ = 6.626 × 10−34 J ⋅ s h = = 6.61 × 10−16 m 2mpc 2(1.673 × 10−27 kg)(2.998 × 108 m/s) EVALUATE: The correction is independent of n The wavelengths of photons emitted in hydrogen atom transitions are on the order of 100 nm = 10−7 m, so the recoil correction is exceedingly small © 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 Particles Behaving as Waves 39.66 39-21 IDENTIFY: Combine I = σT , P = IA, and ΔE = Pt SET UP: In the Stefan-Boltzmann law the temperature must be in kelvins 200°C = 473 K ΔE (100 J) = 8.81 × 103 s = 2.45 h (4.00 × 10 m )(5.67 × 10−8 W/m ⋅ K )(473 K)4 Aσ T EVALUATE: P = 0.0114 W Since the area of the hole is small, the rate at which the cavity radiates EXECUTE: t = 39.67 = −6 energy through the hole is very small IDENTIFY and SET UP: Follow the procedures specified in the problem 2π hc c 2π hc 2π hf EXECUTE: (a) I (λ ) = hc /λ kT = but λ = ⇒ I ( f ) = f − 1) λ (e (c /f )5 (ehf /kT − 1) c3 (ehf /kT − 1) (b) ∫0 ⎛ −c ⎞ I (λ ) d λ = ∫ I ( f ) df ⎜⎜ ⎟⎟ ∞ ⎝ f ⎠ =∫ ∞ 2π hf df ∞ c (ehf /kT − 1) (c) The expression = 2π (kT ) c h3 2π 5k 15h3c ∞ ∫0 x3 ex − dx = 2π ( kT ) (2π )5 (kT ) 2π 5k 4T 4 = = (2 π ) 240h3c 15c 2h3 c 2h3 240 = σ as shown in Eq (39.28) Plugging in the values for the constants we get σ = 5.67 × 10−8 W/m ⋅ K 39.68 EVALUATE: The Planck radiation law, Eq (39.24), predicts the Stefan-Boltzmann law, Eq (39.19) h h IDENTIFY: λ = = From Chapter 36, if λ a then the width w of the central maximum is p 2mE Rλ w=2 , where R = 2.5 m and a is the width of the slit a SET UP: vx = 2E , since the beam is traveling in the x-direction and Δv y m EXECUTE: (a) λ = (b) vx (6.63 × 10−34 J ⋅ s) h = = 1.94 × 10−10 m −31 −19 2mK 2(9.11 × 10 kg)(40 eV)(1.60 × 10 J/eV) (2.5 m)(9.11 × 10−31 kg)1/2 R R = = = 6.67 × 10−7 s −19 v E /m 2(40 eV)(1.6 × 10 J/eV) λ (c) The width w is w = R ' and w = Δv yt = Δp yt /m, where t is the time found in part (b) and a is the slit a 2mλ R width Combining the expressions for w, Δp y = = 2.65 × 10−28 kg ⋅ m/s at (d) Δy = Δp y = 0.20 μm, which is the same order of magnitude of the width of the slit EVALUATE: For these electrons λ = 1.94 × 10−10 m This is much smaller than a and the approximate Δp y 2Rλ 2E is very accurate Also, vx = = 2.9 × 102 m/s, so it = 3.75 × 106 m/s Δv y = a m m is the case that vx Δv y expression w = 39.69 p2 = qΔV , where ΔV is the 2m λ λ accelerating voltage To exhibit wave nature when passing through an opening, the de Broglie wavelength of the particle must be comparable with the width of the opening SET UP: An electron has mass 9.109 × 10−31 kg A proton has mass 1.673 × 10−27 kg EXECUTE: (a) E = hc /λ = 12 eV IDENTIFY: For a photon E = hc For a particle with mass, p = h and E = © 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 39-22 Chapter 39 (b) Find E for an electron with λ = 0.10 × 10−6 m λ = h /p so p = h /λ = 6.626 × 10−27 kg ⋅ m/s E = p /(2m) = 1.5 × 10−4 eV E = qΔV so ΔV = 1.5 × 10−4 V v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(9.109 × 10−31 kg) = 7.3 × 103 m/s (c) Same λ so same p E = p /(2m) but now m = 1.673 × 10−27 kg so E = 8.2 × 10−8 eV and ΔV = 8.2 × 10−8 V v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(1.673 × 10−27 kg) = 4.0 m/s 39.70 EVALUATE: A proton must be traveling much slower than an electron in order to have the same de Broglie wavelength IDENTIFY: The de Broglie wavelength of the electrons must be such that the first diffraction minimum occurs at θ = 20.0° h SET UP: The single-slit diffraction minima occur at angles θ given by a sin θ = mλ p = λ EXECUTE: (a) λ = a sin θ = (150 × 10 v= 6.626 × 10 −34 J⋅s (9.11 × 10−31 kg)(5.13 × 10−8 m) −9 m)sin 20° = 5.13 × 10 −8 m λ = h /mv → v = h /mλ = 1.42 × 104 m/s (b) No electrons strike the screen at the location of the second diffraction minimum a sin θ = 2λ ⎛ 5.13 × 10−8 m ⎞ = ±2 ⎜ = ±0.684 θ = ±43.2° ⎜ 150 × 10−9 m ⎟⎟ a ⎝ ⎠ EVALUATE: The intensity distribution in the diffraction pattern depends on the wavelength λ and is the same for light of wavelength λ as for electrons with de Broglie wavelength λ IDENTIFY: The electrons behave like waves and produce a double-slit interference pattern after passing through the slits SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ /2 The de Broglie wavelength of each of the electrons is λ = h /mv EXECUTE: (a) First find the wavelength of the electrons For the first dark fringe, we have d sin θ = λ /2, which gives (1.25 nm)(sin 18.0°) = λ /2, and λ = 0.7725 nm Now solve the de Broglie wavelength sin θ = ±2 39.71 λ equation for the speed of the electron: v= 6.626 × 10−34 J ⋅ s h = = 9.42 × 105 m/s mλ (9.11 × 10−31 kg)(0.7725 × 10−9 m) which is about 0.3% the speed of light, so they are nonrelativistic (b) Energy conservation gives eV = 12 mv and V = mv /2e = (9.11 × 10−31 kg)(9.42 × 105 m/s) /[2(1.60 × 10−19 C)] = 2.52 V EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction 39.72 IDENTIFY: The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after passing through the hole SET UP: For a round hole, the first dark ring occurs at the angle θ for which sin θ = 1.22λ /D, where D is the diameter of the hole The de Broglie wavelength for a particle is λ = h /p = h /mv EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives sin θα 1.22λα λα = = sin θ p 1.22λ p λ p The de Broglie wavelength gives λ p = h /pp and λα = h /pα , so sin θα h /pα pp = = Using K = p /2m, sin θ p h /pp pα we have p = 2mK Since the alpha particle has twice the charge of the proton and both are accelerated © 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 Particles Behaving as Waves 39-23 through the same potential difference, Kα = K p Therefore pp = 2mp K p and pα = 2mα Kα = 2mα (2 K p ) = 4mα K p Substituting these quantities into the ratio of the sines gives 2mp K p mp sin θα pp = = = sin θ p pα mα 4mα K p Solving for sin θα gives sin θα = 1.67 × 10−27 kg 2(6.64 × 10−27 kg) sin15.0° and θα = 5.3° EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha particles form their first dark ring at a smaller angle than the ring for the lighter protons 39.73 IDENTIFY: Both the electrons and photons behave like waves and exhibit single-slit diffraction after passing through their respective slits SET UP: The energy of the photon is E = hc /λ and the de Broglie wavelength of the electron is λ = h /mv = h /p Destructive interference for a single slit first occurs when a sin θ = λ EXECUTE: (a) For the photon: λ = hc /E and a sin θ = λ Since the a and θ are the same for the photons and electrons, they must both have the same wavelength Equating these two expressions for λ gives h a sinθ = hc /E For the electron, λ = h /p = and a sinθ = λ Equating these two expressions for λ 2mK h h Equating the two expressions for a sin θ gives hc /E = , which gives a sin θ = 2mK 2mK gives E = c 2mK = (4.05 × 10−7 J1/ ) K E c 2mK 2mc = = Since v c, mc > K , so the square root is > Therefore E /K > 1, K K K meaning that the photon has more energy than the electron EVALUATE: When a photon and a particle have the same wavelength, the photon has more energy than the particle IDENTIFY: The de Broglie wavelength of the electrons must equal the wavelength of the light SET UP: The maxima in the two-slit interference pattern are located by d sin θ = mλ For an electron, h h λ= = p mv (b) 39.74 EXECUTE: λ = d sin θ (40.0 × 10−6 m)sin(0.0300 rad) = = 600 nm The velocity of an electron with this m (6.63 × 10−34 J ⋅ s) p h = = = 1.21 × 103 m/s Since m mλ (9.11 × 10−31 kg)(600 × 10−9 m) this velocity is much smaller than c we can calculate the energy of the electron classically 1 K = mv = (9.11 × 10−31 kg)(1.21 × 103 m/s)2 = 6.70 × 10−25 J = 4.19 μeV 2 hc EVALUATE: The energy of the photons of this wavelength is E = = 2.07 eV The photons and wavelength is given by Eq (39.1) v = λ electrons have the same wavelength but very different energies 39.75 IDENTIFY and SET UP: The de Broglie wavelength of the blood cell is λ = h mv 6.63 × 10−34 J ⋅ s = 1.66 × 10−17 m (1.00 × 10−14 kg)(4.00 × 10−3 m/s) EVALUATE: We need not be concerned about wave behavior EXECUTE: λ = © 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 39-24 39.76 Chapter 39 IDENTIFY: An electron and a photon both have the same wavelength We want to use this fact to calculate the energy of each of them h SET UP: The de Broglie wavelength is λ = The energy of the electron is its kinetic energy, p K = 12 mv = p /2m The energy of the photon is E = hf = hc /λ EXECUTE: (a) p = E= λ = 6.626 × 10−34 J ⋅ s 400 × 10−9 m = 1.656 × 10−27 kg ⋅ m/s p (1.656 × 10−27 kg ⋅ m/s) = = 1.506 × 10−24 J = 9.40 × 10−6 eV 2m 2(9.109 × 10−31 kg) hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 4.966 × 10−19 J = 3.10 eV 400 × 10−9 m EVALUATE: The photon has around 300,000 times as much energy as the electron IDENTIFY and SET UP: Follow the procedures specified in the problem (b) E = 39.77 h λ = 1/ ⎛ v2 ⎞ h ⎜1 − ⎟ ⎜ c ⎟ h ⎠ EXECUTE: (a) λ = = ⎝ p mv ⇒ v2 = (b) v = 39.78 h2 ⎛ 2 h ⎞ ⎜⎜ λ m + ⎟⎟ c ⎠ ⎝ c 1/2 ⎛ ⎛ λ ⎞2 ⎞ ⎜1 + ⎜ ⎟ ⎜ ⎝ (h /mc ) ⎟⎠ ⎟ ⎝ ⎠ (c) λ = 1.00 × 10−15 m ⎛ v2 ⎞ h 2v v2 ⇒ λ m 2v = h ⎜ − ⎟ = h − ⇒ λ m 2v + h 2 = h ⎜ c ⎟ c c ⎝ ⎠ = c2 ⎛λ m c ⎞ + 1⎟ ⎜⎜ ⎟ h ⎝ ⎠ 2 ⇒v= c 1/ ⎛ ⎛ mcλ ⎞2 ⎞ ⎜1 + ⎜ ⎟ ⎜ ⎝ h ⎟⎠ ⎟ ⎝ ⎠ ⎛ ⎛ mcλ ⎞2 ⎞ m 2c 2λ ⎟ = (1 − Δ)c Δ = ≈ c ⎜1 − ⎜ ⎟ ⎜ 2⎝ h ⎠ ⎟ 2h ⎝ ⎠ h (9.11 × 10−31 kg) (3.00 × 108 m/s) (1.00 × 10−15 m) Δ= = 8.50 × 10−8 mc 2(6.63 × 10−34 J ⋅ s) ⇒ v = (1 − Δ )c = (1 − 8.50 × 10−8 )c EVALUATE: As Δ → 0, v → c and λ → IDENTIFY and SET UP: The minimum uncertainty product is Δ xΔp x = /2 Δ x = r1, where r1 is the radius of the n = Bohr orbit In the n = Bohr orbit, mv1r1 = h h and p1 = mv1 = 2π 2π r1 1.055 × 10−34 J ⋅ s = 1.0 × 10−24 kg ⋅ m/s This is the same as the 2(0.529 × 10−10 m) magnitude of the momentum of the electron in the n = Bohr orbit EVALUATE: Since the momentum is the same order of magnitude as the uncertainty in the momentum, the uncertainty principle plays a large role in the structure of atoms EXECUTE: Δpx = 39.79 2Δ x = 2r1 = IDENTIFY and SET UP: Combining the two equations in the hint gives pc = K ( K + 2mc ) and λ= hc K ( K + 2mc ) EXECUTE: (a) With K = 3mc this becomes λ = (b) (i) K = 3mc = 3(9.109 × 10 −31 hc 2 3mc (3mc + 2mc ) = h 15mc kg)(2.998 × 10 m/s) = 2.456 × 1013 J = 1.53 MeV © 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 Particles Behaving as Waves λ= 39-25 h 6.626 × 10−34 J ⋅ s = = 6.26 × 10−13 m 15mc 15(9.109 × 10−31 kg)(2.998 × 108 m/s) (ii) K is proportional to m, so for a proton K = (mp /me )(1.53 MeV) = 1836(1.53 MeV) = 2810 MeV λ is proportional to 1/m, so for a proton λ = (me /mp )(6.26 × 10−13 m) = (1/1836)(6.26 × 10−13 m) = 3.41 × 10−16 m 39.80 EVALUATE: The proton has a larger rest mass energy so its kinetic energy is larger when K = 3mc The proton also has larger momentum so has a smaller λ IDENTIFY: Apply the Heisenberg Uncertainty Principle Consider only one component of position and momentum SET UP: Δ xΔp x ≥ /2 Take Δx ≈ 5.0 × 10−15 m K = E − mc For a proton, m = 1.67 × 10−27 kg EXECUTE: (a) Δpx = Δx = (1.055 × 10−34 J ⋅ s) 2(5.0 × 10−15 m) = 1.1 × 10−20 kg ⋅ m/s (b) K = ( pc )2 + ( mc ) − mc = 3.3 × 10−14 J = 0.21 MeV EVALUATE: (c) The result of part (b), about × 105 eV, is many orders of magnitude larger than the 39.81 potential energy of an electron in a hydrogen atom (a) IDENTIFY and SET UP: Δ xΔp x ≥ /2 Estimate Δ x as Δ x ≈ 5.0 × 10−15 m EXECUTE: Then the minimum allowed Δpx is Δpx ≈ 2Δx = 1.055 × 10−34 J ⋅ s 2(5.0 × 10−15 m) = 1.1 × 10−20 kg ⋅ m/s (b) IDENTIFY and SET UP: Assume p ≈ 1.1 × 10−20 kg ⋅ m/s Use Eq (37.39) to calculate E, and then K = E − mc EXECUTE: E = (mc ) + ( pc) mc = (9.109 × 10−31 kg)(2.998 × 108 m/s)2 = 8.187 × 10−14 J pc = (1.1 × 10−20 kg ⋅ m/s)(2.998 × 108 m/s) = 3.165 × 10−12 J E = (8.187 × 10−14 J)2 + (3.165 × 10−12 J)2 = 3.166 × 10−12 J K = E − mc = 3.166 × 10−12 J − 8.187 × 10−14 J = 3.084 × 10−12 J × (1 eV/1.602 × 10−19 J) = 19 MeV (c) IDENTIFY and SET UP: The Coulomb potential energy for a pair of point charges is given by Eq (23.9) The proton has charge + e and the electron has charge – e ke2 (8.988 × 109 N ⋅ m /C2 )(1.602 × 10−19 C) =− = −4.6 × 10−14 J = −0.29 MeV r 5.0 × 10−15 m EVALUATE: The kinetic energy of the electron required by the uncertainty principle would be much larger than the magnitude of the negative Coulomb potential energy The total energy of the electron would be large and positive and the electron could not be bound within the nucleus IDENTIFY: Apply the Heisenberg Uncertainty Principle Let the uncertainty product have its minimum possible value, so ΔxΔp x = /2 SET UP: Take the direction of the electron beam to be the x -direction and the direction of motion perpendicular to the beam to be the y -direction EXECUTE: U = − 39.82 EXECUTE: (a) Δv y = Δp y 2mΔy = 1.055 × 10−34 J ⋅ s = 0.12 m/s 2(9.11 × 10−31 kg)(0.50 × 10−3 m) (b) The uncertainty Δr in the position of the point where the electrons strike the screen is Δp y x x Δr = Δv yt = = = 4.78 × 10−10 m m vx 2mΔy K /m EVALUATE: (c) This is far too small to affect the clarity of the picture m = © 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 39-26 39.83 Chapter 39 IDENTIFY and SET UP: ΔE Δt ≥ /2 Take the minimum uncertainty product, so ΔE = Δt = 8.4 × 10−17 s m = 264me Δm = EXECUTE: ΔE = 39.84 1.055 × 10−34 J ⋅ s 2(8.4 × 10−17 s) ΔE c2 2Δt , with = 6.28 × 10−19 J Δm = 6.28 × 10−19 J (3.00 × 108 m/s) = 7.0 × 10−36 kg Δm 7.0 × 10−36 kg = = 2.9 × 10−8 m (264)(9.11 × 10−31 kg) EVALUATE: The fractional uncertainty in the mass is very small IDENTIFY: The insect behaves like a wave as it passes through the hole in the screen SET UP: (a) For wave behavior to show up, the wavelength of the insect must be of the order of the diameter of the hole The de Broglie wavelength is λ = h /mv EXECUTE: The de Broglie wavelength of the insect must be of the order of the diameter of the hole in the screen, so λ ≈ 4.00 mm The de Broglie wavelength gives v= h 6.626 × 10−34 J ⋅ s = = 1.33 × 10−25 m/s mλ (1.25 × 10−6 kg)(0.00400 m) (b) t = x /v = (0.000500 m)/(1.33 × 10−25 m/s) = 3.77 × 10 21 s = 1.4 × 1010 yr The universe is about 14 billion years old (1.4 × 1010 yr) so this time would be about 85,000 times the age 39.85 of the universe EVALUATE: Don’t expect to see a diffracting insect! Wave behavior of particles occurs only at the very small scale IDENTIFY and SET UP: Use Eq (39.1) to relate your wavelength and speed EXECUTE: (a) λ = h h 6.626 × 10−34 J ⋅ s , so v = = = 1.1 × 10−35 m/s mv mλ (60.0 kg)(1.0 m) distance 0.80 m = = 7.3 × 1034 s(1 y/3.156 × 107 s) = 2.3 × 1027 y velocity 1.1 × 10−35 m/s Since you walk through doorways much more quickly than this, you will not experience diffraction effects EVALUATE: A 1-kg object moving at m/s has a de Broglie wavelength λ = 6.6 × 10−34 m, which is exceedingly small An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelike properties IDENTIFY: The transition energy E for the atom and the wavelength λ of the emitted photon are related by hc E = Apply the Heisenberg Uncertainty Principle in the form ΔE Δt ≥ λ (b) t = 39.86 SET UP: Assume the minimum possible value for the uncertainty product, so that ΔE Δt = EXECUTE: (a) E = 2.58 eV = 4.13 × 10−19 J, with a wavelength of λ = (b) ΔE = 2Δt = (1.055 × 10−34 J ⋅ s) 2(1.64 × 10−7 s) hc = 4.82 × 10−7 m = 482 nm E = 3.22 × 10−28 J = 2.01 × 10−9 eV (c) λ E = hc, so (Δλ ) E + λΔE = 0, and ΔE /E = Δλ /λ , so ⎛ 3.22 × 10−28 J ⎞ Δλ = λ ΔE /E = (4.82 × 10−7 m) ⎜ = 3.75 × 10−16 m = 3.75 × 10−7 nm ⎜ 4.13 × 10−19 J ⎟⎟ ⎝ ⎠ EVALUATE: The finite lifetime of the excited state gives rise to a small spread in the wavelength of the emitted light © 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 Particles Behaving as Waves 39.87 39-27 IDENTIFY: The electrons behave as waves whose wavelength is equal to the de Broglie wavelength SET UP: The de Broglie wavelength is λ = h /mv, and the energy of a photon is E = hf = hc /λ EXECUTE: (a) Use the de Broglie wavelength to find the speed of the electron v= h 6.626 × 10−34 J ⋅ s = = 7.27 × 105 m/s mλ (9.11 × 10−31 kg)(1.00 × 10−9 m) which is much less than the speed of light, so it is nonrelativistic (b) Energy conservation gives eV = 12 mv V = mv /2e = (9.11 × 10−31 kg)(7.27 × 105 m/s) /[2(1.60 × 10−19 C)] = 1.51 V (c) K = eV = e(1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl molecule, so the electron would not be too damaging (d) E = hc /λ = (4.136 × 10−15 eV s)(3.00 × 108 m/s)/(1.00 × 10−9 m) = 1240 eV which would certainly destroy the molecules under study EVALUATE: As we have seen in Problems 39.73 and 39.76, when a particle and a photon have the same wavelength, the photon has much more energy 39.88 IDENTIFY: Assume both the x rays and electrons are at normal incidence and scatter from the surface plane of the crystal, so the maxima are located by d sin θ = mλ , where d is the separation between adjacent atoms in the surface plane SET UP: Let primed variables refer to the electrons λ ′ = EXECUTE: sin θ ′ = h h = p′ 2mE ′ h λ′ ⎛ ⎞ sin θ , and λ ′ = (h /p′) = (h / 2mE′ ), and so θ ′ = arcsin ⎜ sin θ ⎟ λ ⎝ λ 2mE′ ⎠ ⎛ (6.63 × 10−34 J ⋅ s)sin 35.8° θ ′ = arcsin ⎜ ⎜ (3.00 × 10 −11 m) 2(9.11 × 10−31 kg)(4.50 × 10+3 eV)(1.60 × 10−19 ⎝ 39.89 ⎞ ⎟ = 20.9° J/eV) ⎟⎠ EVALUATE: The x rays and electrons have different wavelengths and the m = maxima occur at different angles IDENTIFY: The interference pattern for electrons with de Broglie wavelength λ is the same as for light with wavelength λ h h SET UP: For an electron, λ = = p 2mE EXECUTE: (a) The maxima occur when 2d sin θ = mλ (b) λ = (6.63 × 10−34 J ⋅ s) 2(9.11 × 10 −37 kg)(71.0 eV)(1.60 × 10 −19 ⎛ mλ ⎞ = 1.46 × 10−10 m = 0.146 nm θ = sin −1 ⎜ ⎟ ⎝ 2d ⎠ J/eV) ⎛ (1)(1.46 × 10−10 m) ⎞ = 53.3° (Note: This m is the order of the maximum, not the mass.) θ = sin −1 ⎜ ⎜ 2(9.10 × 10−11 m) ⎟⎟ ⎝ ⎠ EVALUATE: (c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming electrons by eφ An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength A smaller wavelength gives a smaller angle θ (see part (b)) 39.90 IDENTIFY: The photon is emitted as the atom returns to the lower energy state The duration of the excited state limits the energy of that state due to the uncertainty principle hc SET UP: The wavelength λ of the photon is related to the transition energy E of the atom by E = λ ΔE Δt ≥ /2 The minimum uncertainty in energy is ΔE ≥ 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 39-28 Chapter 39 EXECUTE: (a) The photon energy equals the transition energy of the atom, 3.50 eV hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) λ= = = 355 nm E 3.50 eV (b) ΔE = 39.91 1.055 × 10−34 J ⋅ s −6 = 1.32 × 10−29 J = 8.2 × 10−11 eV 2(4.0 × 10 s) EVALUATE: The uncertainty in the energy could be larger than that found in (b), but never smaller IDENTIFY: The wave (light or electron matter wave) having less energy will cause less damage to the virus SET UP: For a photon Eph = EXECUTE: (a) E = λ = λ = 1.24 × 10 1.24 × 10−6 eV ⋅ m −6 5.00 × 10 λ eV ⋅ m −9 m For an electron Ee = p2 h2 = 2m 2mλ = 248 eV (6.63 × 10−34 J ⋅ s) = 9.65 × 10−21 J = 0.0603 eV 2mλ 2(9.11 × 10−31 kg)(5.00 × 10−9 m) EVALUATE: The electron has much less energy than a photon of the same wavelength and therefore would cause much less damage to the virus IDENTIFY and SET UP: Assume px ≈ h and use this to express E as a function of x E is a minimum for (b) Ee = 39.92 h2 hc hc = that x that satisfies dE = dx EXECUTE: (a) Using the given approximation, E = ((h /x) /m + kx ), (dE /dx) = kx − ( h /mx3 ), and the h The minimum energy is then h k /m minimum energy occurs when kx = (h /mx ), or x = mk EVALUATE: (b) U = 12 kx = 39.93 h k p2 h2 h k K= = = At this x the kinetic and potential energies m 2m 2mx 2 m are the same (a) IDENTIFY and SET UP: U = A x Eq (7.17) relates force and potential The slope of the function A x is not continuous at x = so we must consider the regions x > and x < separately EXECUTE: For x > 0, x = x so U = Ax and F = − d ( Ax ) = − A For x < 0, x = − x so U = − Ax and dx d ( − Ax) = + A We can write this result as F = − A x /x, valid for all x except for x = dx (b) IDENTIFY and SET UP: Use the uncertainty principle, expressed as ΔpΔ x ≈ h, and as in Problem 39.80 estimate Δp by p and Δ x by x Use this to write the energy E of the particle as a function of x F =− Find the value of x that gives the minimum E and then find the minimum E EXECUTE: E = K + U = p2 +Ax 2m px ≈ h, so p ≈ h /x Then E ≈ h2 2mx For x > 0, E = +Ax h2 2mx + Ax To find the value of x that gives minimum E set dE = dx © 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 Particles Behaving as Waves 0= −2 h 2mx3 39-29 +A 1/3 x3 = ⎛ h2 ⎞ h2 and x = ⎜ ⎜ mA ⎟⎟ mA ⎝ ⎠ With this x the minimum E is E= h ⎛ mA ⎞ ⎜ ⎟ 2m ⎝ h ⎠ 2/3 1/3 ⎛ h2 ⎞ + A⎜ ⎜ mA ⎟⎟ ⎝ ⎠ = h 2/3m −1/3 A2/3 + h 2/3m −1/3 A2/3 1/3 ⎛ h A2 ⎞ E= ⎜ ⎟ ⎜⎝ m ⎟⎠ 39.94 EVALUATE: The potential well is shaped like a V The larger A is, the steeper the slope of U and the smaller the region to which the particle is confined and the greater is its energy Note that for the x that minimizes E, K = U (a) IDENTIFY and SET UP: Let the y-direction be from the thrower to the catcher, and let the x-direction be horizontal and perpendicular to the y-direction A cube with volume V = 125 cm3 = 0.125 × 10−3 m3 has side length l = V 1/3 = (0.125 × 10−3 m3 )1/3 = 0.050 m Thus estimate Δ x as Δ x ≈ 0.050 m Use the uncertainty principle to estimate Δp x EXECUTE: Δ xΔp x ≥ /2 then gives Δpx ≈ 2Δ x = 0.01055 J ⋅ s = 0.11 kg ⋅ m/s (The value of 2(0.050 m) in this other universe has been used.) (b) IDENTIFY and SET UP: Δ x = ( Δvx )t is the uncertainty in the x-coordinate of the ball when it reaches the catcher, where t is the time it takes the ball to reach the second student Obtain Δvx from Δ p x EXECUTE: The uncertainty in the ball’s horizontal velocity is Δvx = Δp x 0.11 kg ⋅ m/s = = 0.42 m/s m 0.25 kg 12 m = 2.0 s The uncertainty in the 6.0 m/s x-coordinate of the ball when it reaches the second student that is introduced by Δvx is Δ x = ( Δvx )t = (0.42 m/s)(2.0 s) = 0.84 m The ball could miss the second student by about 0.84 m The time it takes the ball to travel to the second student is t = EVALUATE: A game of catch would be very different in this universe We don’t notice the effects of the uncertainty principle in everyday life because h is so small 39.95 IDENTIFY and SET UP: The period was found in Exercise 39.29b: T = 4ε 02 n3h3 me Eq (39.14) gives the energy of state n of a hydrogen atom EXECUTE: (a) The frequency is f = (b) Eq (39.5) tells us that f = n2 = n and n1 = n + 1, then n22 Therefore, for large n, f ≈ − me = 3 T 4ε n h me ⎛ 1 ⎞ ( E2 − E1 ) So f = ⎜ − ⎟ (from Eq (39.14)) If ⎜ h 8ε h ⎝ n2 n1 ⎟⎠ n12 = n − (n + 1) = ⎞ ⎛ ⎛ ⎛ ⎞⎞ ≈ ⎜1 − ⎜1 − + … ⎟ ⎟ = 1− 2⎜ 2⎟ ⎜ ⎟ n ⎝ (1 + 1/n) ⎠ n ⎝ ⎝ n ⎠⎠ n me 4ε 02 n3h3 EVALUATE: We have shown that for large n we obtain the classical result that the frequency of revolution of the electron is equal to the frequency of the radiation it emits © 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 39-30 39.96 Chapter 39 IDENTIFY: Follow the steps specified in the hint SET UP: The value of Δxi that minimizes Δx f satisfies d ( Δx f ) = d ( Δxi ) EXECUTE: Time of flight of the marble, from a free-fall kinematic equation is just 2y 2(25.0 m) t ⎛ Δp ⎞ + Δ xi To minimize Δ x f t= = = 2.26 s Δ x f = Δ xi + (Δvx )t = Δ xi + ⎜ x ⎟ t = Δ m x g 9.80 m/s ⎝ ⎠ im with respect to Δ xi , ⇒ Δx f (min) = d (Δ x f ) d (Δ xi ) =0= t t + = 2m 2m − t ⎛ t ⎞ + ⇒ Δxi (min) = ⎜ ⎟ 2m( Δ xi ) ⎝ 2m ⎠ t = m 2(1.055 × 10−34 J ⋅ s)(2.26 s) = 1.54 × 10−16 m = 1.54 × 10−7 nm 0.0200 kg EVALUATE: The uncertainty introduced by the uncertainty principle is completely negligible in this situation © 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 ... En = − EXECUTE: (a) En = − 13. 6 eV n2 13. 6 eV n2 39-9 and En +1 = − 13. 6 eV (n + 1) ⎡ 1⎤ n − ( n + 1) 2n + ΔE = En +1 − En = ( 13. 6 eV) ⎢ − ⎥ = − (13. 6 eV) ΔE = (13. 6 eV) 2 + + ( 1) ( )(... kg)(5 .13 × 10−8 m) −9 m)sin 20° = 5 .13 × 10 −8 m λ = h /mv → v = h /mλ = 1.42 × 104 m/s (b) No electrons strike the screen at the location of the second diffraction minimum a sin θ = 2λ ⎛ 5 .13. .. this state Then use the Bohr equation, Eq (39.6), to calculate L EXECUTE: En = − (13. 6 eV)/n , so this state has n = 13. 6/1.51 = In the Bohr model, L = n so for EXECUTE: 39.25 hc = 3.80 eV, so λ