CH3 - 3 The Quantum Theory of Light 3-2 Assume that your...

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3 The Quantum Theory of Light 3-2 Assume that your skin can be considered a blackbody. One can then use Wien’s displacement law, max T 0.289 8 10 2 m K with T 35 0 C 308 K to find max 0.289 8 10 2 m K 308 K 9.41 10 6 m 9 410 nm . 3-4 (a) From Stefan’s law, one has P A T 4 . Therefore, P A 5.7 10 8 Wm 2 K 4 3000 K  4 4.62 10 6 2 . (b) A P 4.62 10 6 2 75 W 4.62 10 6 2 16.2 mm 2 . 3-5 (a) Planck’s radiation energy density law as a function of wavelength and temperature is given by u , T 8 hc 5 e hc B T 1 . Using u 0 and setting x hc max k B T , yields an extremum in u T with respect to . The result is 0  5 hc max k B T e hc max k B T e hc max k B T 1 1 or x 51 e x . (b) Solving for x by successive approximations, gives x 4.965 or max T hc k B 4.965 2.90 10 3 m K . 3-10 The energy per photon, E hf and the total energy E transmitted in a time t is Pt where power P 100 kW . Since E nhf where n is the total number of photons transmitted in the time t , and f 94 MHz , there results nhf 100 kW t 10 5 W t , or n t 10 5 W hf 10 5 Js 6.63 10 34 J s 94 10 6 s 1 1.60 10 30 photons s . 3-14 (a) K hf hc 1 240 eV nm 350 nm 2.24 eV 1.30 eV (b) At c , K 0 and hc 1 240 eV nm 2.24 eV 554 nm
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3-16 (a) hc K , 1 240 eV nm 300 nm 2.23 eV 1.90 eV (b) V s 1 240 eV nm 400 nm e 1.90 eV e 1.20 V 3-17 The energy of one photon of light of wavelength 300 nm is E hc 1 240 eV nm 300 nm 4.13 eV . (a) As lithium and beryllium have work functions that are less than 4.13 eV, they will exhibit the photoelectric effect for incident light with this energy. However, mercury will not because its work function is greater than 4.13 eV. (b) The maximum kinetic energy is given by K hc , so K Li  1 240 eV nm 300 nm 2.3 eV 1.83 eV , and K Be 1 240 eV nm 300 nm 3.9 eV 0.23 eV .
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This note was uploaded on 04/12/2011 for the course PHYS 2D taught by Professor Hirsch during the Winter '08 term at UCSD.

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CH3 - 3 The Quantum Theory of Light 3-2 Assume that your...

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