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# hw3.sol - Physics 240B Spring 2008 Homework 3 Solutions...

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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 γ = σT 4 = 7 . 36 × 10 7 W/m 2 . This can also be done using the Planck formula to derive σ = 2 π 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 γ 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 sun’s outgoing power is very, very small. This solid angle is d Ω = πR 2 (1 A.u. ) 2 = π parenleftbigg . 696 149 . 6 parenrightbigg 2 = 6 . 8 × 10- 5 steradians (1) The fraction of the intensity reaching the earth is f Ω = d Ω /π , giving a total incident power of f Ω I γ = 1600 W/m 2 . c) P out = hν g N ( ν g ) A (2) where E g = hν g is the energy gap, N ( ν g ) is the number of photons incident with energy greater than the bandgap and...
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hw3.sol - Physics 240B Spring 2008 Homework 3 Solutions...

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