Weeks10-11HWSolutions

M m e f 2 m m e f 2 accordingly

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Unformatted text preview: time is: !m = !0 " ( s + 5 2) " (5 2 ) Note also that 2 2 ! m = ! 0 ( E k BTL ) 2s is in power law form with a characteristic exponent of 2s instead of s, so 2 !m 2 = !0 " ( 2s + 5 2 ) " (5 2 ) and rH = ! ( 2s + 5 2 ) ! (5 2 ) " ! ( s + 5 2)$ # % 2 = 1.93 Now using (i), we find rH = ! ( 2s + 5 2 ) ! (5 2 ) " ! ( s + 5 2)$ # % 2 Assuming s = 3 / 2 for II scattering, we find rH = ! (11 2 ) ! (5 2 ) " ! ( 4)$ # % 2 ECE ­656 13 Fall 2013 Mark Lundstrom 10/27/13 Recall some properties of the Gamma function: ! ( n) = ( n " 1)! (when n is an integer) ! (1 / 2) = " ! ( p + 1) = p! ( p ) Accordingly, we find: ! ( 4) = ( 3)! = 6 9 97 975 ! (11 / 2) = ! (9 2) = ! (7 2) = ! (5 2) 2 22 222 3 31 3 ! (5 / 2) = ! (3 2) = ! (1 2) = " 2 22 4 Putting it all together: rH = ! (11 2 ) ! (5 2 ) " ! ( 4)$ # % 2 9753 3 & & 22& 2224 4 = = = 1.9828 6'6 36 rH = 1.9828 7b) Develop an expression for the Hall factor in 2D. Is it the same for parabolic energy bands and for graphene? Solution: Recall that for 2D, parabolic bands, non ­degenerate conditions !m = !0 " ( s + 2) " ( 2) As in 7a), we find 2 !m 2 = !0 " ( 2s + 2 ) " ( 2) So the Hall factor is rH ! ECE...
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This document was uploaded on 01/15/2014.

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