ch32-p061 - 61(a For a given value of m varies from to Thus...

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(c) Since L orb, z = m A h /2 π , the greatest allowed value of L orb, z is given by | m A | max h /2 π = 3 h /2 π . (d) Similar to part (c), since µ orb, z = – m A B , the greatest allowed value of orb, z is given by | m A | max B = 3 eh /4 π m e . (e) From Eqs. 32-23 and 32-29 the z component of the net angular momentum of the electron is given by net, orb, , . 22 s zz s z mh LLL =+ = + π π A For the maximum value of L net, z let m A = [ m A ] max = 3 and m s = 1 2 . Thus L hh z net, max . . F H G I K J = 3 1 35 2 ππ (f) Since the maximum value of L net, z is given by [ m J ] max h /2 π with [ m J ] max = 3.5 (see the last part above), the number of allowed values for the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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