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Unformatted text preview: Chapter 3 Solutions 3.1 a. Given: power P = 1000 W and frequency = 700 10 3 s1 . Then the photon energy is E ph = h = (6.626 1034 J s)(700 10 3 s1 ) E ph = 4.638 1028 J/photon or 2.895 109 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 1034 J s)(3.0 10 8 m s1 )/(800 109 m) E ph = 2.485 1019 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 m2 s1 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 m1 c. If each photon gives rise to one electron then the current density J is: J = e ph = (1.602 1019 C)(4.02 10 21 m2 s1 ) = 644 A m2 3.5 Photon energy, E ph = hc / = (6.63 1034 J s)(3.0 10 8 m/s)/(0.02 109 m) E ph = 9.945 1015 J or 62.15 keV Photons per unit, = P /hc = (0.1106 J cm2 ) (0.02109 m)/(6.631034 J s)(310 8 ms1 ) = 1 10 7 photons cm2 3.7 Solution a. We are given max = 420 nm. The work function is then: = h o = hc / max = (6.626 1034 J s)(3.0 10 8 m s1 ) /(420 109 m) = 4.733 1019 J or 2.96 eV b. Given = 300 nm, the photon energy is then: E ph = h = hc / = (6.626 1034 J s)(3.0 10 8 m s1 )/(300 109 m) E ph = 6.626 1019 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.
 Spring '08
 Morelli

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