chap41 - 41.1: 2 34 34 2 s J 10 054 . 1 s J 10 4.716 ) 1 (...

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Unformatted text preview: 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 . 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 , . 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 ( ), , 4 ( ), 3 , 3 ( ), 2 , 3 ( ), 1 , 3 ( a total of 25. b) Each state has the same energy ( n is the same), eV. 544 . 25 eV 60 . 13- =- 41.5: J 10 3 . 2 m 10 . 1 C) 10 60 . 1 ( 4 1 . 4 1 18 10 2 19 2 1--- - = - = = r q q 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 / 2 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 r P . 0803 . 2 5 1 1 =- =- e b) The difference in the probabilities is . 243 . ) 2 )( 2 5 ( ) ) 2 5 ( 1 ( ) 5 1 ( 2 1 1 2 =- =------- e e e e 41.7: a) ) )( ( | ) ( | | ) ( | | | 2 2 2 l l im im * Ae Ae r R +- = = , | ) ( | | ) ( | 2 2 2 r R A = which is independent of b) = = = = A A d A d 2 2 2 2 2 . 2 1 1 2 | ) ( | 41.8: . ) 75 . ( 2 2 ) 4 ( 1 1 1 2 1 1 2 12 2 2 4 2 E E E E E E n e m 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 4 C Nm 10 988 . 8 s) J 10 055 . 1 ( 2 C) 10 kg)(1.602 10 9.109 ) 4 ( =--- 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 4 = = = m m 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 4 = = m m e m r l and the 1 2 transition gives (10.19 eV)(185.8)=1893 eV....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap41 - 41.1: 2 34 34 2 s J 10 054 . 1 s J 10 4.716 ) 1 (...

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