P32_011 - 11. (a) Since ml = 0, Lorb,z = ml h/2 = 0. (b)...

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11. (a) Since m l =0, L orb ,z = m l h/ 2 π =0. (b) Since m l =0, µ orb ,z = m l µ B =0. (c) Since m l = 0, then from Eq. 32-12, U = µ orb ,z B ext = m l µ B B ext =0. (d) Regardless of the value of m l , we Fnd for the spin part U = µ s,z B = ± µ B B = ± ( 9 . 27 × 10 24 J / T ) (35 mT) = ± 3 . 2 × 10 25 J . (e) Now m l = 3, so L orb ,z = m l h 2 π = ( 3) ( 6 . 63 × 10 27 J · s ) 2 π = 3 . 16 × 10 34 J · s and µ orb ,z = m l µ B = ( 3) ( 9 . 27 × 10 24 J / T ) =2 . 78 × 10 23 J / T . The potential energy associated with the electron’s orbital magnetic moment is now
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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