Chapter 3 – Quantization of Charge, Light, and Energy 3-3. B E , u pc 2 u pc , and pc E 2 mc 2 c E 0.561MeV 0.511MeV 2 2 0.2315MeV u 0.2315MeV 0.41 c 0.561MeV 2.0 105V / m B 1.63 103 T 16.3G 0.41c 3-4. F quB and FG mp g 19 6 5 FB quB 1.6 10 C 3.0 10 m / s 3.5 10 T 1.03 109 27 2 FG m p g 1.67 10 kg 9.80m / s 3-5. (a) mu 2 Ek / e e / m R qB e / m B 1/ 2 4 1 2 Ek / e 1 2 4.5 10 eV / e B e/m 0.325T 1.76 1011 kg (b) frequency f u 2 R 1/ 2 2.2 103 m 2.2mm 2 Ek / e e / m 2 R 2 4.5 104 eV / e 1.76 1011 C / kg 3 2 2.2 10 m 1/ 2 9.1109 Hz period T 1/ f 1.11010 s 3-6. (a) 1/ 2mu 2 Ek , so u 2Ek / e e / m u 2 2000eV / e 1.76 1011 C / kg (b) t1 1/ 2 x1 0.05m 1.89 109 s 1.89ns 7 u 2.65 10 m / s 54 2.65 107 m / s Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-6 continued) (c) mu y F t1 eEt1 u y e / m Et1 1.76 1011 C / kg 3.33 103V / m 1.89 109 s 1.11106 m / s 3-7. 3-8. NEk W CV T 3 4.186 J / cal 1.311014 CV T 5 10 cal / C 2C N Ek 2000eV 1.60 1019 J / eV Q1 Q2 25.41 20.64 1019 C 4.47 1019 C n1 n2 e Q2 Q3 20.64 17.47 1019 C 3.17 1019 C n2 n3 e Q4 Q3 19.06 17.47 1019 C 1.59 1019 C n4 n3 e Q4 Q5 19.06 12.70 1019 C 6.36 1019 C n4 n5 e Q6 Q5 14.29 12.70 1019 C 1.59 1019 C n6 n5 e where the ni are integers. Assuming the smallest Δn = 1, then Δn12 = 3.0, Δn23 = 2.0, Δn43 = 1.0, Δn45 = 4.0, and Δn65 = 1.0. The assumption is valid and the fundamental charge implied is 1.59 1019 C. 3-9. For the rise time to equal the field-free fall time, the net upward force must equal the weight. qE mg mg E 2mg / q. 3-10. (See Millikan’s Oil Drop Experiment on the home page at www.whfreeman.com/tiplermodernphysics6e.) The net force in the y-direction is mg bv y ma y . The net force in the x-direction is qE bvx max . At terminal speed ax ay 0 and vx / vt sin . sin vx qE / b qE vt vt bvt 55 Chapter 3 – Quantization of Charge, Light, and Energy 3-23. Equation 3-18: u 8 hc 5 A 5 Letting A hc , B hc / kT , and U ehc / kT 1 eB / 1 5 1 e B / B 2 du d A 5 5 6 B/ A 2 B/ d d eB / 1 e 1 e 1 6 6 B / A B B/ A e B e 5 e B / 1 5 1 e B / 0 2 2 B/ B / e 1 e 1 The maximum corresponds to the vanishing of the quantity in brackets. Thus, 5 1 e B / B . This equation is most efficiently solved by iteration; i.e., guess at a value for B/λ in the expression 5 1 e B / , solve for a better value of B/λ; substitute the new value to get an even better value, and so on. Repeat the process until the calculated value no longer changes. One succession of values is: 5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations repeat the same value (to seven digits), so we have: 6.63 1034 J s 3.00 108 m / s hc hc 4.965114 mT m m kT 4.965114 k 4.965114 1.38 1023 J / K B mT 2.898 103 m K (Equation 3-5) 3-24. Photon energy E hf hc / (a) For λ = 380nm: E 1240eV nm / 380nm 3.26eV For λ = 750nm: E 1240eV nm / 750nm 1.65eV (b) E hf 4.14 1015 eV s 100 106 s 1 4.14 107 eV 3-25. (a) hf hc / 0.47eV . max 4.14 10 hc 4.87eV 15 eV s 3.00 108 m / s 4.87eV 60 2.55 107 m 255nm Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-25 continued) (b) It is the fraction of the total solar power with wavelengths less than 255nm, i.e., the area under the Planck curve (Figure 3-6) up to 255nm divided by the total area. The latter is: R T 4 5.67 108W / m2 K 4 5800K 6.42 107W / m2 . 4 Approximating the former with u with 127nm and 255nm : 8 hc 127 109 m 5 255 109 m 1.23 104 J / m3 u 127nm 255nm 9 ehc / kT 12710 1 R 0 255nm c 1.23 104 J / m 3 4 R 8 4 3 3.00 10 m / s 1.23 10 J / m fraction = 1.4 104 4 6.42 107W / m2 R 0 255nm 3-26. (a) t hc 1240eV nm 653nm, 1.9ev ft h 1.9eV 4.59 104 Hz 15 4.136 10 eV s 1 hc 1 1240eV nm (b) V0 1.9eV 2.23V e e 300nm 1 hc 1 1240eV nm (c) V0 1.9eV 1.20V e e 400nm 3-27. (a) Choose λ =550nm for visible light. nhf E dn dE hf P dt dt 0.05 100W 550 109 m dn P P 1.38 1019 / s 34 8 dt hf hc 6.63 10 J s 3.00 10 m / s (b) flux number radiated / unit time 1.38 1019 / s 2.75 1017 / m2 s 2 area of the sphere 4 2m 61 Chapter 3 – Quantization of Charge, Light, and Energy 3-28. (a) hf ft (b) f c / h 4.22eV 1.02 1015 Hz 15 4.14 10 eV s 3.00 108 m / s 5.36 1014 Hz 9 560 10 m No. Available energy/photon hf 4.14 1015 eV s 5.36 1014 Hz 2.22eV . This is less than . 3-29. (a) E hf hc / hc / E For E 4.26eV : 1240eV nm / 4.26eV 291nm and since f c / , f 3.00 108 m / s / 291nm 1.03 1015 s 1 (b) This photon is in the ultraviolet region of the electromagnetic spectrum. 3-30. (a) First, add a row f 1014 Hz to the table in the problem, then plot a graph Ek ,max versus f . The slope of the graph is h/e; the intercept on the Ek ,max axis is work function. The graph below is a least squares fit to the data. λ nm 544 Ek ,max eV 0.360 0.199 0.156 0.117 0.062 f 1014 Hz 5.51 594 5.05 62 604 4.97 612 4.90 633 4.74 Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-30 continued) Slope h / e 3.90 1015 eV s (b) (3.90 1015 4.14 1015 ) eV s 5.6 percent 4.14 1015 eV s (c) The work function is the magnitude of the intercept on the Ek ,max axis, 1.78 eV. (d) cesium En hc 3-32. (a) hc 3-31. (b) Ek hc 60 6.63 1034 J s 3.00 108 m / s 9 550 10 m 1240eV nm 1.90eV 653nm 1240eV nm 1.90eV 2.23eV 300nm 63 2.17 1017 J Chapter 3 – Quantization of Charge, Light, and Energy 3-33. Equation 3-25: 2 1 6.63 10 9.1110 34 31 1 100 J s 1 cos135 kg 3.00 10 m / s 3-36. p h 8 4.14 1012 m 4.14 103 nm 4.14 103 nm 100 5.8% 0.0711nm 3-34. Equation 3-24: m 3-35. h 1 cos mc 1.24 103 1.24 103 nm 0.016nm V 80 103V hc c (a) p 1240eV nm 6.63 1034 J s 3.10eV / c 1.66 1027 kg m / s c 400nm 400 109 m (b) p 1240eV nm 6.63 1034 J s 1.24 104 eV / c 6.63 1024 kg m / s c 0.1nm 0.1109 m (c) p 1240eV nm 6.63 1034 J s 5 4.14 10 eV / c 2.211032 kg m / s 3 102 m c 3 107 nm (d) p 1240eV nm 6.63 1034 J s 620eV / c 3.32 1025 kg m / s 9 c 2nm 2 10 m 2 1 6.63 1034 J s 1 cos110 h 1 cos 3.26 1012 m 31 8 mc 9.1110 kg 3.00 10 m / s 6.63 1034 J s 3 108 m / s hc 1 2.43 1012 m E1 0.511106 eV 1.60 1019 J / eV 2 1 3.26 1012 m 2.43 3.26 1012 m 5.69 1012 m 64 Chapter 3 – Quantization of Charge, Light, and Energy 3-33. Equation 3-25: 2 1 6.63 10 9.1110 34 31 1 100 J s 1 cos135 kg 3.00 10 m / s 3-36. p h 8 4.14 1012 m 4.14 103 nm 4.14 103 nm 100 5.8% 0.0711nm 3-34. Equation 3-24: m 3-35. h 1 cos mc 1.24 103 1.24 103 nm 0.016nm V 80 103V hc c (a) p 1240eV nm 6.63 1034 J s 3.10eV / c 1.66 1027 kg m / s c 400nm 400 109 m (b) p 1240eV nm 6.63 1034 J s 1.24 104 eV / c 6.63 1024 kg m / s c 0.1nm 0.1109 m (c) p 1240eV nm 6.63 1034 J s 5 4.14 10 eV / c 2.211032 kg m / s 3 102 m c 3 107 nm (d) p 1240eV nm 6.63 1034 J s 620eV / c 3.32 1025 kg m / s 9 c 2nm 2 10 m 2 1 6.63 1034 J s 1 cos110 h 1 cos 3.26 1012 m 31 8 mc 9.1110 kg 3.00 10 m / s 6.63 1034 J s 3 108 m / s hc 1 2.43 1012 m E1 0.511106 eV 1.60 1019 J / eV 2 1 3.26 1012 m 2.43 3.26 1012 m 5.69 1012 m 64 Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-36 continued) E2 hc 2 1240eV nm 2.18 105 eV 0.218MeV 3 5.69 10 nm Electron recoil energy Ee E1 E2 (Conservation of energy) Ee 0.511MeV 0.218MeV 0.293MeV . The recoil electron momentum makes an angle θ with the direction of the initial photon. PE θ h/λ1 110° h/λ2 20° h 2 cos 20 pe sin 1/ c E 2 mc 2 sin 2 3.00 10 m / s 6.63 10 8 sin 5.69 10 12 (Conservation of momentum) 34 J s cos 20 2 2 m 0.804MeV 0.511MeV 1/ 2 1.60 10 13 J / MeV 0.330 or 19.3 3-37. (a) First, add a row ( 1 cos ) to the table in the problem, then plot a graph of versus ( 1 cos ). The slope of the graph is the Compton wavelength of the electron. pm degrees 1 cos 0.647 1.67 2.45 3.98 4.80 45 75 90 135 170 0.293 0.741 1.000 1.707 1.985 65 Chapter 3 – Quantization of Charge, Light, and Energy h (4.5 1.0) nm slope 2.43nm mc (1.88 0.44) (b) The graph above is a least squares fit to the data. The percent difference is (2.43 2.426) nm 0.004 nm 100 100 0.15percent 2.426 nm 2.426 nm 3-38. 2 1 1 100 3-39. (a) E1 h 1 cos 100 0.00243nm 1 cos90 0.243nm mc hc 1 (b) 2 1 (c) E2 h 1 cos 0.011 Equation 3-25 mc hc 2 1240eV nm 1.747 104 eV 0.0711nm h 1 cos 0.0711nm 0.00243nm 1 cos180 0.0760nm mc 1240eV nm 1.634 104 eV 0.0760nm (d) Ee E1 E2 1.128 103 eV 66 Chapter 3 – Quantization of Charge, Light, and Energy 3-42. (a) Compton wavelength = electron: h mc h 6.63 1034 J s 2.43 1012 m 0.00243nm 31 8 mc 9.1110 kg 3.00 10 m / s h 6.63 1034 J s proton: 1.32 1015 m 1.32 fm 27 8 mc 1.67 10 kg 3.00 10 m / s (b) E hc (i) electron: E (ii) proton: E 1240eV nm 5.10 105 eV 0.510MeV 0.00243nm 1240eV nm 9.39 108 eV 939MeV 6 1.32 10 nm 3-43. Photon energy E hf hc / allows us to rewrite Equation 3-25 as hc hc h (1 cos ) E2 E1 mc Rearranging the above, hc h hc (1 cos ) E2 mc E1 Dividing both sides of the equation by hc yields 1 1 1 ( E1 / mc 2 )(1 cos ) 1 (1 cos ) E2 mc 2 E1 E1 Or E2 E1 ( E1 / mc )(1 cos ) 1 2 3-44. (a) eV0 hf hc / e 0.52V hc / 450nm (i) e 1.90V hc / 300nm (ii) 68