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Unformatted text preview: 1455 Chapter 38 1. (a) With E = hc / min = 1240 eV·nm/ min = 0.6 eV, we obtain = 2.1 10 3 nm = 2.1 m. (b) It is in the infrared region. 2. Let 1 2 2 m v E hc e photon and solve for v : v 2 hc m e 2 hc m e c 2 c 2 c 2 hc m e c 2 2.998 10 8 m/s 2 1240eV nm 588nm 511 10 3 eV 8.6 10 5 m/s. Since v c , the nonrelativistic formula K mv 1 2 2 may be used. The m e c 2 value of Table 373 and 1240eV nm hc are used in our calculation. 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 = 6.626 10 –34 J·s is the Planck constant, f is the frequency of the light emitted, and is the wavelength. Thus P = Rhc / and 26 45 34 8 550nm 3.9 10 W 1.0 10 photons/s. 6.63 10 J s 2.998 10 m/s P R hc 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 383, the rate is given by CHAPTER 38 1456 R A P hc d 2 / 4 4 633nm 5.0 10 3 W 6.63 10 34 J s 2.998 10 8 m/s 3.0 10 3 m 2 2.3 10 21 photons/m 2 s . 5. 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 = (1, 650, 763.73) –1 m = 6.0578021 10 –7 m = 605.78021 nm, we find the energy to be E hc 1240 60578021 2 047 eV nm nm eV. . . 6. 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 7. The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source via abs abs emit 2 (0.80) ....
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This document was uploaded on 04/22/2011.
 Spring '11
 Photon

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