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Chapter 41 - 41.1 L l(l 1 l(l 1 L 2 4.716 1 054 10 10 34 34...

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41.1: 2 34 34 2 s J 10 054 . 1 s J 10 4.716 ) 1 ( ) 1 ( × × = = + + = - - L l l l l L . 4 0 . 20 ) 1 ( = = + l l l 41.2: a) . 2 so , 2 max max 2 = = z l L m b) . 45 . 2 6 ) 1 ( 2 2 2 = = + l l c) The angle is arcco , 6 arccos = l z m L L and the angles are, for , 0 . 90 , 1 . 114 , 7 . 144 , 2 to 2 ° ° ° = - = l l m m . 3 . 35 , 9 . 65 ° ° The angle corresponding to l m l = will always be larger for larger . l 41.3: . ) 1 ( 2 + = l l L The maximum orbital quantum number : if So . 1 - = n l 2 2 2 2 2 2 2 5 . 199 ) 200 ( 199 199 200 49 . 19 ) 20 ( 19 19 20 41 . 1 2 ) 1 ( 1 2 = = = = = = = = = = + = = = L l n L l n l l L l n The maximum angular momentum value gets closer to the Bohr model value the larger the value of . n 41.4: The ) , ( l m l combinations are (0, 0), (1, 0), ) 1 , 1 ( ± , (2, 0), ), 1 , 2 ( ± ), 2 , 2 ( ± (3, 0), 4), (4, and ), 3 , 4 ( ), 2 , 4 ( ), 1 , 4 ( ), 0 , 4 ( ), 3 , 3 ( ), 2 , 3 ( ), 1 , 3 ( ± ± ± ± ± ± ± a total of 25. b) Each state has the same energy ( n is the same), eV. 544 . 0 25 eV 60 . 13 - = - 41.5: J 10 3 . 2 m 10 0 . 1 C) 10 60 . 1 ( 4 1 . 4 1 18 10 2 19 0 2 1 0 - - - × - = × × - = = πε r q q πε U eV. 4 . 14 eV J 10 60 . 1 J 10 3 . 2 19 18 - = × × - = - - 41.6: a) As in Example 41.3, the probability is - - - = = - 2 / 0 2 0 2 3 2 2 3 2 2 1 4 2 2 4 4 | | a a a r s e a r a ar a dr πr ψ P . 0803 . 0 2 5 1 1 = - = - e b) The difference in the probabilities is . 243 . 0 ) 2 )( 2 5 ( ) ) 2 5 ( 1 ( ) 5 1 ( 2 1 1 2 = - = - - - - - - - e e e e
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41.7: a) ) )( ( | ) ( | | ) ( | | | 2 2 2 φ φ l l im im * Ae Ae θ r R ψ ψ ψ + - Θ = = , | ) ( | | ) ( | 2 2 2 θ r R A Θ = which is independent of φ b) = = = = Φ π π π A πA d A d 2 0 2 2 0 2 2 . 2 1 1 2 | ) ( | φ φ θ 41.8: . ) 75 . 0 ( 2 2 ) 4 ( 1 1 1 2 1 1 2 12 2 2 4 2 0 E E E E E E n e m πε E r n - = - = - = - = a) kg 10 11 . 9 If 31 - × = = m m r ( 29 2 2 9 2 34 4 9 1 31 2 2 0 4 C Nm 10 988 . 8 s) J 10 055 . 1 ( 2 C) 10 kg)(1.602 10 9.109 ) 4 ( × × × × = - - - πε e m r eV. 59 . 13 J 10 177 . 2 18 = × = - For 1 2 transition, the coefficient is (0.75)(13.59 eV)=10.19 eV. b) If , 2 m m r = using the result from part (a), eV. 795 . 6 2 eV 59 . 13 2 eV) 59 . 13 ( ) 4 ( 2 2 0 4 = = = m m πε e m r l Similarly, the 1 2 transition, eV. 095 . 5 2 eV 19 . 10 = c) If , 8 . 185 m m r = using the result from part (a), eV, 2525 185.8 eV) 59 . 13 ( ) 4 ( 2 2 0 4 = = m m πε e m r l and the 1 2 transition gives (10.19 eV)(185.8)=1893 eV. 41.9: a) 2 19 31 2 34 0 2 2 0 1 C) 10 kg)(1.602 10 109 . 9 ( ) s J 10 626 . 6 ( : - - - × × × = = = π ε πme h ε a m m r m 10 293 . 5 11 1 - × = a b) . m 10 059 . 1 2 2 10 1 2 - × = = = a a m m r c) . m 10 849 . 2 185.8 1 m 8 . 185 13 1 3 - × = = = a a m r 41.10: ), sin( ) cos( φ φ φ l l im m i m e l + = and to be periodic with period π m π l 2 , 2 must be an integer multiple of l m π so , 2 must be an integer.
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41.11: - = = a a a r s dr πr e πa dV ψ a P 0 0 2 2 3 1 ) 4 ( 1 ) ( . 5 1 ) ( 4 4 2 2 4 4 2 2 4 4 ) ( 2 0 3 2 3 3 3 3 0 2 2 2 2 3 2 2 3 - - - - - = + - - - = - - - = = e a P e a e a a a a e a r a ar a dr e r a a P a a r a r a o 41.12: a) , 2 b) , eV 10 2.32 T) 400 . 0 )( T V e 10 79 . 5 ( 5 5 B - = × = × = = - - l m B μ E the lowest possible value of .
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