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hw26 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 26...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 26. 41.27, 41.32, 41.36, 41.39 41.27. a) The energy of the photon must be equal to, or greater than, the energy gap if it is to excite an electron. The lowest energy photons, those with just enough energy to excite electrons will have the maximum wavelength. Then, m 10 26 . 2 eV . 5 5 m/s) 10 s)( . 3 0 eV 10 . 4 14 ( so , 7 - 8 - 15 max max × = × × = = = g g E hc E hc λ λ λ max = 226 nm ¯¯¯¯¯¯¯¯¯¯¯¯ b) This is shorter than the wavelength of visible light, in the ultraviolet (UV) range. 41.32. a) Since phosphorus has five electrons in its outer shell, adding phosphorus will add electrons. Therefore, the doped material will be n -type . b) Each phosphorus atom will contribute one electron to the conduction band. Therefore, the problem is to calculate how many phosphorus atoms are added per unit volume. One of every 10 7 silicon atoms is replaced by a phosphorus atom, and on page 1050, the solution to sample problem 42-6 shows that there are 5 ×10 28 silicon atoms per cubic meter in crystalline silicon. Then, the added electron density is: (10 -7 )( 5 ×10 28 m -3 ) n added = 5 ×10 21 m -3 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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hw26 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 26...

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