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10hw - Solutions to problem set 10(141A Sp07 1 1(ISSP 8.1(a...

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Solutions to problem set 10 (141A Sp07) 1 1. (ISSP 8.1). (a) ISSP, page 210, equation (51): E d = 13 . 6 2 m e m eV = 13 . 6 18 2 0 . 015 eV = 6 . 3 × 10 - 4 eV (b) ISSP, page 210, equation (52): a = 0 . 53 × m e / m Å = (0 . 53)18 0 . 015 Å = 636Å (c) The wavefunctions of the impurity electrons need to overlap if they are to form a band: then the density needs to be at least one electron per (636Å) 3 , ie n = 1 (636Å) 3 = 1 . 6 × 10 7 m - 3 2. (ISSP 8.2). (a) ISSP, page 213, equation (53), n = n o n d e - E g / (2 k B T ) with n 0 = 2 mk B T 2 π 2 3 / 2 = 2 (0 . 01)(9 . 1 × 10 - 31 )(1 . 38 × 10 - 23 )4 2 π (1 . 05 × 10 - 34 ) 2 3 / 2 = 3 . 9 × 10 19 m - 3 = 3 . 9 × 10 13 cm - 3 and so n 0 n d = 3 . 9 × 10 26 = 1 . 97 × 10 13 cm - 3 . Also, since 4 K corresponds to 0 . 345 × 10 - 3 eV , we have n = (1 . 97 × 10 13 cm - 3 ) e - 1 / (2(0 . 345)) = 4 . 6 × 10 12 cm - 3 (b) In SI units, R H = 1 / ne = 1 (4 . 6 × 10 18 )(1 . 6 × 10 - 19 ) = 1 . 3 m 3 / C 3. (ISSP 8.3). Recall that j x = σ xx E x + σ xy E y j y = σ yx E x + σ yy E y Since the transverse current must be zero, E x = - σ yy E y yx , and so j x = σ xy - σ xx σ yy σ yx E y giving R H = E y j x B = 1 B σ xy - σ xx σ yy σ yx = σ yx B ( σ xy σ yx - σ xx σ yy )
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Solutions to problem set 10 (141A Sp07) 2 Note that σ xy and σ yx are of order B , and so σ xy σ
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