E differentiate to find kdk m 0 2 1

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Unformatted text preview: 3) (4) (5) (6) Solve for k: k= 2 m* ( 0 ) E (1 + ! E ) ! differentiate to find: kdk = m* ( 0 ) !2 (1 + 2! E ) dE put the two together: k dk = 2 m* ( 0 ) ! 3 2 m* ( 0 ) E (1 + ! E ) (1 + 2! E ) dE Insert in (3) D3 D ( E ) dE = m* ( 0 ) 12 k dk = 2 3 2 m* ( 0 ) E (1 + " E ) (1 + 2" E ) dE !2 !! m* ( 0 ) 12 D3 D ( E ) = 2 k dk = 2 3 2 m* ( 0 ) E (1 + " E ) (1 + 2" E ) ! !! (7) Step 2: From (6) ECE ­656 2 Fall 2013 Mark Lundstrom 9/27/13 dE !2k 1 1 dE !k 1 =* " =# = * dk m ( 0 ) (1 + 2! E ) ! dk m ( 0 ) (1 + 2! E ) Now use (5) ! ( E) = 2 E (1 + " E ) !k 1 1 = * * m ( 0 ) (1 + 2" E ) m ( 0 ) (1 + 2" E ) (8) Step 3: Now use (1), (2), (7), and (8) M 3D ( E ) = M 3D ( E ) = h " ! (E)% h " ! (E)% h E (1 + ( E ) 1 D3 D ( E ) * $ 2 ' D3 D ( E ) = 4 $ 2 ' D3 D ( E ) = 4 & # & 4# 2 m ( 0 ) (1 + 2( E ) # m* ( 0 ) & h E (1 + ! E ) 1 * $ 2 3 2 m ( 0 ) E (1 + ! E ) (1 + 2! E ) ' * 4 2 m ( 0 ) (1 + 2! E ) % " ! ( m* ( 0 ) M 3D ( E ) = E (1 + " E ) 2! ! 2 We have been assuming...
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This document was uploaded on 01/15/2014.

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