Week6HWSolutions

# 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 + &quot; E ) (1 + 2&quot; E ) dE !2 !! m* ( 0 ) 12 D3 D ( E ) = 2 k dk = 2 3 2 m* ( 0 ) E (1 + &quot; E ) (1 + 2&quot; E ) ! !! (7) Step 2: From (6) ECE ­656 2 Fall 2013 Mark Lundstrom 9/27/13 dE !2k 1 1 dE !k 1 =* &quot; =# = * dk m ( 0 ) (1 + 2! E ) ! dk m ( 0 ) (1 + 2! E ) Now use (5) ! ( E) = 2 E (1 + &quot; E ) !k 1 1 = * * m ( 0 ) (1 + 2&quot; E ) m ( 0 ) (1 + 2&quot; E ) (8) Step 3: Now use (1), (2), (7), and (8) M 3D ( E ) = M 3D ( E ) = h &quot; ! (E)% h &quot; ! (E)% h E (1 + ( E ) 1 D3 D ( E ) * \$ 2 ' D3 D ( E ) = 4 \$ 2 ' D3 D ( E ) = 4 &amp; # &amp; 4# 2 m ( 0 ) (1 + 2( E ) # m* ( 0 ) &amp; h E (1 + ! E ) 1 * \$ 2 3 2 m ( 0 ) E (1 + ! E ) (1 + 2! E ) ' * 4 2 m ( 0 ) (1 + 2! E ) % &quot; ! ( m* ( 0 ) M 3D ( E ) = E (1 + &quot; E ) 2! ! 2 We have been assuming...
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## This document was uploaded on 01/15/2014.

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