ESc314-Answers_to_Homework_Problems-Fall

ESc314-Answers_to_Homework_Problems-Fall - Answers to the...

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Answers to the Homework Problems Chapter 3 Prob. 3.1: (a) N ph = 2.16×10 30 photons per second; (b) Γ ph = 4.02×10 21 photons m -2 s -1 ; (c) E = 868 V m -1 . Prob. 3.3: Φ υ (650 nm) = 0.35 lm; Φ υ (532 nm) = 1.09 lm; Φ 650nm = 1.63×10 16 photons s -1 ; Φ 532nm = 5.35×10 15 photons s -1 . Prob. 3.4: (a) E ph (rod) = 3.92×10 -19 J or 2.45 eV; E ph (cones) =3.58×10 -19 J or 2.24 eV; (b) Φ = 900 photons s -1 ; E total = 3.53×10 -17 J or 221 eV; P = 3.53×10 -16 W; (c) I th = 1.41×10 -7 J s -1 m -2 or 1.41×10 -7 W m -2 ; Total # of rods in test spot = 375 rods; (d) P st = 7.17×10 -15 J s -1 or 7.17×10 -15 W. Prob. 3.5: E ph = 9.945×10 -15 J or 62.15 keV; Φ = 1×10 7 photons cm -2 . Prob. 3.7: (a) Φ = 4.733×10 -19 J or 2.96 eV; (b) KE = 1.18 eV; (c) J = 48.4 A m -2 = 4.84 mA cm -2 . Prob. 3.8: (a) λ max = 653.9×10 -7 m = 653.9 nm; (b) E ph = 4.417×10 -19 J = 2.76 eV; KE = 0.86 eV; The stopping voltage for this wavelength will be -0.86 V; (c) I = 7.691×10 -4 A = 0.769 mA. Prob. 3.9: (a) J = 2.54×10 -3 A cm -2 ; KE = 1.36 eV; (b) The voltage to extinguish the current is -1.36 V; (c) I = 0.105 J s -1 cm -2 or 0.105 W cm -2 . Prob. 3.14: ∆λ = 9.20×10 -15 m or 9.20 fm; l = 6.0 m. Chapter 1 Prob. 1.19: (a) n A = 6.34×10 -9 m -3 ; (b) n A = 1.61×10 16 m -3 . (If using the method of numerical integration, (a) n A = 6.16×10 -8 m -3 ; (b) n A = 8.21×10 16 m -3 .) Prob. 1.22: (a) a = 3.147×10 -10 m = 0.3147 nm; n at = 6.415×10 22
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This note was uploaded on 11/11/2008 for the course E SC 314 taught by Professor Lenahan,patrickm during the Fall '07 term at Pennsylvania State University, University Park.

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ESc314-Answers_to_Homework_Problems-Fall - Answers to the...

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