Daily_006.SOLN - they are the same Q3S.8 The calculation...

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Phys 274 – c ± Eric B. Szarmes 31 January 2011 Daily Homework 6 SOLUTIONS Q3S.7 The maximum potential V max developed between the plates is related to the maximum kinetic energy of the ejected electrons by the relation K max = eV max . Thus, if ultraviolet light of wavelength 250 nm develops a maximum potential of 0.82 V, the associated kinetic energy is K max = 0 . 82 eV, and we have K max = 0 . 82 eV = hc λ - W = 1240 eV · nm 250 nm - W , (1) from which we calculate W = 4 . 14 eV for lead. If the wavelength of the incident light is reduced to 215 nm, the resulting maximum kinetic energy of the electrons is given by K max = hc λ - W = 1240 eV · nm 215 nm - 4 . 14 eV = 1 . 63 eV , (2) and, consequently, the maximum potential which develops between the plates is V max = 1 . 63 V. (Note the different usage of electron-volts (eV) and volts (V): the potential between the plates is measured in volts. The kinetic energy of an electron falling through this potential is measured in electron-volts, though numerically
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Unformatted text preview: they are the same.) Q3S.8 The calculation proceeds as follows: Photon rate = # photons sec = Energy/sec Energy/photon = I · A eye hf ave , (3) where the intensity I is I = P bulb 4 πR 2 . (4) for uniform illumination at distance R = 100 m, and A eye = πr 2 eye with r eye = 1 2 (2 mm) = 1 mm. We also compute the average photon energy to be hf ave = hc λ ave = 1240 eV · nm 550 nm = 2 . 255 eV = 3 . 6(10)-19 J . (5) Then, combining eqs. (3,4,5): Photon rate = I · A eye hf ave = P bulb 4 πR 2 πr 2 eye hf ave = 60 W 4 π (100 m) 2 π (0 . 001 m) 2 3 . 6(10)-19 J = 4 . 2(10) 9 photons/sec . (6) The problem did not specify visible photons, but if you specified this fraction (for example, ∼ 15%, which would yield 6.3(10) 8 photons/sec for the final answer) then this was acceptable as well. 1...
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