Chapter-38 - 1489 Chapter 38 1. Let R be the rate of photon...

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Unformatted text preview: 1489 Chapter 38 1. 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 / λ , where h = 6.626 × 10 –34 J·s is the Planck constant, f is the frequency of the light emitted, and λ is the wavelength. Thus P = Rhc / λ and R P hc = = × × ⋅ × = ×- λ 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 . . . . 2. We denote the diameter of the laser beam as d . The cross-sectional area of the beam is A = π d 2 /4. From the formula obtained in Problem 38-1, the rate is given by ( 29 ( 29 ( 29 ( 29( 29( 29 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 1.7 10 photons/m s . R P A hc d--- × λ = = π π × ⋅ × × = × ⋅ 3. The energy of a photon is given by E = hf , where h is the Planck constant and f is the frequency. The wavelength λ is related to the frequency by λ f = c , so E = hc / λ . 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 λ . With λ = (1, 650, 763.73) –1 m = 6.0578021 × 10 –7 m = 605.78021 nm, we find the energy to be E hc = = ⋅ = λ 1240 60578021 2 047 eV nm nm eV. . . CHAPTER 38 1490 4. The energy of a photon is given by E = hf , where h is the Planck constant and f is the frequency. The wavelength λ is related to the frequency by λ f = c , so E = hc / λ . 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 λ . With 589 nm λ = , we obtain 1240eV nm 2.11eV. 589nm hc E ⋅ = = = λ 5. (a) Let E = 1240 eV·nm/ λ min = 0.6 eV to get λ = 2.1 × 10 3 nm = 2.1 μ m. (b) It is in the infrared region. 6. Let 1 2 2 m v E hc e = = photon λ 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 λ λ λ c h c h b g b gc h . . m / s eV nm nm eV m / s. Since v c << , the non-relativistic formula K mv = 1 2 2 may be used. The m e c 2 value of Table 37-3 and 1240eV nm hc = ⋅ are used in our calculation. 7. (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 / λ , where h is the Planck constant, f is the frequency of the light emitted, and λ is the wavelength. Thus Rhc P P R hc λ λ = ⇒ = ....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

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Chapter-38 - 1489 Chapter 38 1. Let R be the rate of photon...

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