KaSap Electronic Materials_Chapter_3_problem_solutions

KaSap Electronic Materials_Chapter_3_problem_solutions -...

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Unformatted text preview: Chapter 3 Solutions 3.1 a. Given: power P = 1000 W and frequency = 700 10 3 s-1 . Then the photon energy is E ph = h = (6.626 10-34 J s)(700 10 3 s-1 ) E ph = 4.638 10-28 J/photon or 2.895 10-9 eV/photon The number of photons emitted from the antenna per unit time ( N ph ) is therefore: J 10 638 . 4 W 1000 28- = = ph E P ph N = 2.16 10 30 photons per second b. Average intensity I average = 1000 W/m 2 and maximum wavelength max = 800 nm. The photon energy at max is, E ph = hc / max = (6.626 10-34 J s)(3.0 10 8 m s-1 )/(800 10-9 m) E ph = 2.485 10-19 J or 1.551 eV The photon flux ph is the number of photons arriving per unit time per unit area, J 10 485 . 2 m W 1000 19 2 average = = ph E I ph = 4.02 10 21 photons m-2 s-1 For the electric field we use classical physics. I average = (1/2) c o E 2 , so that ( ) ( )( ) 1 12 1 8 2 average m F 10 854 . 8 s m 10 . 3 m W 1000 2 2 = = o c I E = 868 V m-1 c. If each photon gives rise to one electron then the current density J is: J = e ph = (1.602 10-19 C)(4.02 10 21 m-2 s-1 ) = 644 A m-2 3.5 Photon energy, E ph = hc / = (6.63 10-34 J s)(3.0 10 8 m/s)/(0.02 10-9 m) E ph = 9.945 10-15 J or 62.15 keV Photons per unit, = P /hc = (0.110-6 J cm-2 ) (0.0210-9 m)/(6.6310-34 J s)(310 8 ms-1 ) = 1 10 7 photons cm-2 3.7 Solution a. We are given max = 420 nm. The work function is then: = h o = hc / max = (6.626 10-34 J s)(3.0 10 8 m s-1 ) /(420 10-9 m) = 4.733 10-19 J or 2.96 eV b. Given = 300 nm, the photon energy is then: E ph = h = hc / = (6.626 10-34 J s)(3.0 10 8 m s-1 )/(300 10-9 m) E ph = 6.626 10-19 J = 4.14 eV The kinetic energy KE of the emitted electron can then be found: KE = - E ph = 4.14 eV - 2.96 eV = 1.18 eV= 1....
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This note was uploaded on 04/24/2008 for the course MSE MSE 350 taught by Professor Morelli during the Spring '08 term at Michigan State University.

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KaSap Electronic Materials_Chapter_3_problem_solutions -...

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