YF_ISM_41

YF_ISM_41 - ATOMIC STRUCTURE 41 L = l (l + 1) . Lz = ml . l...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
41-1 A TOMIC S TRUCTURE 41.1. IDENTIFY and SET UP: (1 ) Ll l =+ = . zl Lm = = . 0, 1, 2,. .., 1 ln =− . 0, 1, 2,. l ml =±± ± . cos / z LL θ = . EXECUTE: (a) 0 l = : 0 L = , 0 z L = . 1 l = : 2 L = = , ,0, z L == . 2 l = : 6 L = = , 2 , ,0, , 2 z L = = . (b) In each case cos / z = . 0 L = : not defined. 2 L = = : 45.0 , 90.0 , 135.0 °° ° . 6 L = = : 35.3 , 65.9 , 90.0 , 114.1 , 144.7 ° ° °°° . EVALUATE: There is no state where G L is totally aligned along the z axis. 41.2. IDENTIFY and SET UP: ) l = . = = . 0,1,2,. .., 1 . 0, 1, 2,. .., l ± . cos / z = . EXECUTE: (a) 0 l = : 0 L = , 0 z L = . 1 l = : 2 L = = , ,0, z L . 2 l = : 6 L = = , 2 , ,0, , 2 z L = = . 3 l = : 23 L = = , 3,2,,0 , ,2,3 z L = = = = . 4 l = : 25 L = = , 4,3,2,,0 , ,2,3,4 z L = = = = ==== . (b) 0 L = : not defined. 2 L = = : 45.0 ,90.0 ,135.0 ° . 6 L = = : 35.3 ,65.9 ,90.0 ,114.1 ,144.7 ° ° . L = = : 54.7 ,73.2 ,90.0 ,106.8 ,125.3 ,150.0 ° ° °°°° . L = = : 26.6 ,47.9 ,63.4 ,77.1 ,90.0 ,102.9 ,116.6 ,132.1 ,153.4 ° ° ° ° °°°°° . (c) The minimum angle is 26.6 ° and occurs for 4 l = , 4 l m . The maximum angle is 153.4 ° and occurs for 4 l = , 4 l m . 41.3. IDENTIFY and SET UP: The magnitude of the orbital angular momentum L is related to the quantum number l by Eq.(41.4): ) , 10 , 1 , 2 , l = = EXECUTE: 2 34 2 34 4.716 10 kg m /s ) 2 0 1.055 10 J s L ll ⎛⎞ ×⋅ += = = ⎜⎟ ⎝⎠ = And then ( 1) 20 gives that 4. l = EVALUATE: l must be integer. 41.4. (a) max max () 2 , s o 2 . lz mL = (b) ( 6 2.45 . = = (c) The angle is arccos arccos , 6 L = and the angles are, for 2 to 2, 144.7 , mm = ° 114.1 , 90.0 , 65.9 , 35.3 . The angle corresponding to l = will always be larger for larger . l 41.5. IDENTIFY and SET UP: The angular momentum L is related to the quantum number l by Eq.(41.4), ) . l = The maximum l , max , l for a given n is max 1. EXECUTE: For max 2, 1 and 2 1.414 . nl L = = For max 20, 19 and (19)(20) 19.49 . L = = For max 200, 199 and (199)(200) 199.5 . L = = EVALUATE: As n increases, the maximum L gets closer to the value n = postulated in the Bohr model. 41.6. The ( , ) l lm combinations are (0, 0), (1, 0), (1, ± , (2, 0), (2, 1), ± (2, 2), ± (3, 0), (3, 1), (3, 2), (3, 3), (4, 0), (4, 1), (4, 2), (4, 3), and (4, 4), ±± ± ± ± a total of 25. (b) Each state has the same energy ( n is the same), 13.60 eV 0.544 eV. 25 −= 41.7. 19 2 18 12 10 00 11 ( 1 . 6 0 1 0 C ) 2.3 10 J 44 1 . 0 1 0 m qq U π r π −× = × × PP 18 19 2.3 10 J 14.4 eV. 1.60 10 J eV U × 41
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
41-2 Chapter 41 41.8. (a) As in Example 41.3, the probability is 2 22 3 1 /2 2 1 3 0 0 45 | | 4 1 0.0803 4 2 a a ra s ar a r a e P ψπ rd r e a ⎡⎤ ⎛⎞ == = = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ . (b) The difference in the probabilities is 21 1 2 (1 5 ) (1 (5 2) ) (5 2)( 2 ) 0.243. ee e e −− = − = 41.9. (a) 2 || |( ) | ) | ( ) ( ) ll im im * ψψ ψ Rr θ Ae Ae φφ −+ Θ 222 | ( )| | ( )| , AR r θ which is independent of φ . (b) 2 00 1 ) | 2 1 . 2 ππ dA d π AA π Φ= = = = ∫∫ 41.10. 4 1 12 2 1 1 1 2 2 0 1 (0.75) . (4 ) 2 2 r n me E EE E E E E π n = Δ = −= −= = P (a) 31 r If 9.11 10 kg mm × () 43 1 1 9 4 2 92 1 8 r 3 4 2 0 (9.109 10 kg)(1.602 10 C) 8.988 10 N m C 2.177 10 J 13.59 eV (4 ) 2(1.055 10 J s) π ×× = × = ×⋅ = P For 2 1 transition, the coefficient is (0.75)(13.59 eV) = 10.19 eV. (b) If r , 2 m m = using the result from part (a), 4 r 0 3 . 5 9 e V (13.59 eV) 6.795 eV. (4 ) 2 m π m = = P Similarly, the 2 1 transition, 10.19 eV 5.095 eV. 2 = (c) If r 185.8 , = using the result from part (a), 4 0 185.8 (13.59 eV) 2525 eV, (4 ) r m π m = P and the 2 1 transition gives (10.19 eV)(185.8) = 1893 eV. 41.11. IDENTIFY and SET UP: Eq.(41.8) gives rr 4 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

YF_ISM_41 - ATOMIC STRUCTURE 41 L = l (l + 1) . Lz = ml . l...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online