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Unformatted text preview: Chapter 38 1. (a) Let E = 1240 eV·nm/ min = 0.6 eV to get = 2.1 10 3 nm = 2.1 m. (b) It is in the infrared region. 2. The energy of a photon is given by E = hf , where h is the Planck constant and f is the frequency. The wavelength is related to the frequency by f = c , so E = hc / . Since h = 6.626 10 –34 J·s and c = 2.998 10 8 m/s, hc 6 626 10 2 998 10 1602 10 10 1240 34 8 19 9 . . . J s m / s J / eV m / nm eV nm. c hc h c hc h Thus, E 1240eV nm . With 589 nm , we obtain 1240eV nm 2.11eV. 589nm hc E ￗ 3. Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon. Then the power output of the Sun is given by P = RE . Now E = hf = hc / , where h is the Planck constant, f is the frequency of the light emitted, and is the wavelength. Thus P = Rhc / and R P hc 550 39 10 6 63 10 2 998 10 10 10 26 34 8 45 nm W J s m / s photons/ s. b gc h c hc h . . . . 4. We denote the diameter of the laser beam as d . The crosssectional area of the beam is A = d 2 /4. From the formula obtained in problem 3, the rate is given by 1529 CHAPTER 38 ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 34 8 3 21 2 4 633nm 5.0 10 W / 4 6.63 10 J s 2.998 10 m/s 3.5 10 m photons 1.7 10 . m s R P A hc d ￗ ״ ￗ ￗ 5. Since = (1, 650, 763.73) –1 m = 6.0578021 10 –7 m = 605.78021 nm, the energy is (using the fact that 1240eV nm hc ￗ ), E hc 1240 60578021 2 047 eV nm nm eV. . . 6. Let 1 2 2 m v E hc e photon and solve for v : v hc m hc m c c c hc m c e e e 2 2 2 2 998 10 2 1240 590 511 10 8 6 10 2 2 2 8 3 5 c h c h b g b gc h . . m / s eV nm nm eV m / s. Since v c , the nonrelativistic formula K mv 1 2 2 may be used. The m e c 2 value of Table 383 and 1240eV nm hc ￗ are used in our calculation. 7. The total energy emitted by the bulb is E = 0.93 Pt , where P = 60 W and t = 730 h = (730 h)(3600 s/h) = 2.628 10 6 s. The energy of each photon emitted is E ph = hc / . Therefore, the number of photons emitted is N E E Pt hc ph W s J s m / s m 0 93 0 93 60 2 628 10 6 63 10 2 998 10 630 10 4 7 10 6 34 8 9 26 . / . . . . / . . b gb gc h c hc hc h 8. Following Sample Problem 381, we have 250 P Rhc 100 6 63 10 2 998 10 550 10 36 10 34 8 9 17 / . . . s J s m / s m W. b gc hc h 9. (a) Let R be the rate of photon emission (number of photons emitted per unit time) and let E be the energy of a single photon. Then, the power output of a lamp is given by P = RE if all the power goes into photon production. Now, E = hf = hc / , where h is the Planck constant, f is the frequency of the light emitted, and is the wavelength. Thus is the wavelength....
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 Fall '08
 Many
 Physics, Electron, Energy, Kinetic Energy, Photon, ev

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