ch38 - 1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1...

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1. (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.
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2. 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 .. . Js 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
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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 / λ , where h 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 Js m /s photons/ s. b gc h c hc h . .. .
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4. 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 3, the rate is given by () ( ) 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 . ms RP A hc d −− × λ == π π× × ×
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5. Since λ = (1, 650, 763.73) –1 m = 6.0578021 × 10 –7 m = 605.78021 nm, the energy is (using the fact that 1240eV nm hc =⋅ ), E hc == = λ 1240 60578021 2 047 eV nm nm eV. . .
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6. Let 1 2 2 mv E hc e == photon λ and solve for v : v hc m hc mc cc hc ee e = × 22 2 2 998 10 2 1240 590 511 10 86 10 2 2 2 8 3 5 λλ λ c h c h b g b gc h .. m/s eV nm nm eV m/s. Since vc << , the non-relativistic formula Km v = 1 2 2 may be used. The m e c 2 value of Table 38-3 and 1240eV nm hc =⋅ are used in our calculation.
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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 / λ . Therefore, the number of photons emitted is N E E Pt hc == = × ×⋅ × × −− ph Ws Js m /s m 093 0 93 60 2 628 10 6 63 10 2 998 10 630 10 47 10 6 34 8 9 26 . / .. / λ b gb gc h c hc h c h
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8. Following Sample Problem 38-1, we have P Rhc == ×⋅ × × λ 100 6 63 10 2 998 10 550 10 36 10 34 8 9 17 / .. . sJ sm / s m W. b gc hc h
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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 / λ , where h is the Planck constant, f is the frequency of the light emitted, and λ is the wavelength. Thus P = Rhc / λ and R = λ P / hc . The lamp emitting light with the longer wavelength (the 700 nm lamp) emits more photons per unit time. The energy of each photon is less, so it must emit photons at a greater rate. (b) Let R be the rate of photon production for the 700 nm lamp. Then, R P hc == ×⋅ λ 700 400 160 10 1240 141 10 19 21 nm J / s J / eV eV nm photon / s. b gb g c hb g . .
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10. (a) The rate at which solar energy strikes the panel is P == 139 260 361 ..
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This homework help was uploaded on 10/01/2007 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell University (Engineering School).

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ch38 - 1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1...

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