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# ssm_ch38 - Chapter 38 7(a Let R be the rate of photon...

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Chapter 38 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=¸ ,whe re h is the Planck constant, f is the frequency of the light emitted, and ¸ is the wavelength. Thus P = Rhc=¸ and R = ¸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 = (700 £ 10 ¡ 9 m)(400 J = s) (6 : 626 £ 10 ¡ 34 J ¢ s)(2 : 9979 £ 10 8 m = s) =1 : 41 £ 10 21 photon = s : 17 The energy of an incident photon is E = hf = hc=¸ ,whe re h is the Planck constant, f is the frequency of the electromagnetic radiation, and ¸ is its wavelength. The kinetic energy of the most energetic electron emitted is K m = E ¡ © =( hc=¸ ) ¡ © ,whe re © is the work function for sodium. The stopping potential V 0 is related to the maximum kinetic energy by eV 0 = K m ,so eV 0 =( hc=¸ ) ¡ © and ¸ = hc eV 0 + © = (6 : 626 £ 19 ¡ 34 J ¢ s)(2 : 9979 £ 10 8 m = s) (5 : 0eV+2 : 2eV)(1 : 602 £ 10 ¡ 19 J = eV) =1 : 7 £ 10 ¡ 7 m : Here eV 0 =5 : 0eV was used. 21 (a) The kinetic energy K m of the fastest electron emitted is given by K m = hf ¡ © =( hc=¸ ) ¡ © , where © is the work function of aluminum, f is the frequency of the incident radiation, and ¸ is its wavelength. The relationship f

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ssm_ch38 - Chapter 38 7(a Let R be the rate of photon...

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