Problem%20Set%2010%20Solutions

# Problem%20Set%2010%20Solutions - 13 The rate at which...

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abs abs emit 2 (0.80) . 4 A RR r π = Given that 62 abs 2.00 10 m A and 3.00 m, r = with abs 4.000 photons/s, R = we find the rate at which photon is emitted to be () 22 8 emit abs abs 44 ( 3 . 0 0 m ) 4.000 photons/s 2.83 10 photons/s (0.80) (0.80)(2.00 10 m ) r A ππ == = × × . Since the energy of each emitted photon is ph 1240 eV nm 2.48 eV 500nm hc E λ = , the power output of source is 88 1 0 emit emit ph 2.83 10 photons/s (2.48 eV) 7.0 10 eV/s 1.1 10 W. PR E × = × = × 13. The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source via

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22. To find the longest possible wavelength λ max (corresponding to the lowest possible energy) of a photon which can produce a photoelectric effect in platinum, we set K max = 0 in Eq. 38-5 and use hf = hc / λ . Thus hc / λ max = Φ . We solve for λ max : λ max . == = hc Φ 1240 532 233 eV nm nm nm.
Consequently, the new photon energy is E hc ' '. .. == = λ 1240 0 0757 164 10 164 4 eV nm nm eV keV . By energy conservation, then, the kinetic energy of the electron must equal E' – E = 17.5 keV – 16.4 keV = 1.1 keV. 36. The initial wavelength of the photon is (using hc = 1240 eV·nm) λ= = = hc E 1240 17500 0 07086 eV nm eV . or 70.86 pm. The maximum Compton shift occurs for φ = 180°, in which case Eq. 38-11 (applied to an electron) yields ∆λ = F H G I K J −° = × F H G I K J −− = hc mc e 23 1 180 1240 511 10 1 1 0 00485 (c o s ) (( ) ) . eV nm eV where Table 37-3 is used. Therefore, the new photon wavelength is λ ' = 0.07086 nm + 0.00485 nm = 0.0757 nm.

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48. (a) Since Km c == 7 5 4 932 2 ., MeV < <M e V α bg we may use the non-relativistic formula pm K = 2 . Using Eq. 38-43 (and noting that 1240 eV·nm = 1240 MeV·fm), we obtain λ= = = = h p
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Problem%20Set%2010%20Solutions - 13 The rate at which...

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