P32_044 - z component of the net angular momentum of the...

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44. (a) For a given value of l , m l varies from l to + l . Thus, in our case l = 3, and the number of di±erent m l ’s is 2 l + 1 = 2(3) + 1 = 7. Thus, since L orb ,z m l , there are a total of seven di±erent values of L orb ,z . (b) Similarly, since µ orb ,z m l , there are also a total of seven di±erent values of µ orb ,z . (c) Since L orb ,z = m l h/ 2 π , the greatest allowed value of L orb ,z is given by | m l | max h/ 2 π =3 h/ 2 π ; while the least allowed value is given by | m l | min h/ 2 π =0. (d) Similar to part (c), since µ orb ,z = m l µ B , the greatest allowed value of µ orb ,z is given by | m l | max µ B = 3 eh/ 4 πm e ; while the least allowed value is given by | m l | min µ B =0. (e) From Eqs. 32-3 and 32-9 the
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Unformatted text preview: z component of the net angular momentum of the electron is given by L net ,z = L orb ,z + L s,z = m l h 2 π + m s h 2 π . For the maximum value of L net ,z let m l = [ m l ] max = 3 and m s = 1 2 . Thus [ L net ,z ] max = µ 3 + 1 2 ¶ h 2 π = 3 . 5 h 2 π . (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 z component of L net ,z is given by 2[ m J ] max + 1 = 2(3 . 5) + 1 = 8....
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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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