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Unformatted text preview: Physics 240B, Spring 2008 Homework 3 Solutions Jesse Noffsinger April 7, 2008 Solutions are in general not the original work of the author. Problem 1) The Ultimate efficiency of a single-gap solar cell. a) The total power is the intensity times the area. Assuming all the power from the blackbody is concentrated on an area of 1 m 2 , the incident power is I = T 4 = 7 . 36 10 7 W/m 2 . This can also be done using the Planck formula to derive = 2 5 15 k 4 B c 2 h 3 , arriving at the same numerical answer. So the incident power is P in = 7 . 36 10 7 W/m 2 . b) I computed in (a) is much, much greater than 1000 W/m 2 . The single largest contributing factor to this is that the solid angle subtended by the earth of the suns outgoing power is very, very small. This solid angle is d = R 2 (1 A.u. ) 2 = parenleftbigg . 696 149 . 6 parenrightbigg 2 = 6 . 8 10- 5 steradians (1) The fraction of the intensity reaching the earth is f = d / , giving a total incident power of f I = 1600 W/m 2 . c) P out = h g N ( g ) A (2) where E g = h g is the energy gap, N ( g ) is the number of photons incident with energy greater than the bandgap and...
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