1106_PartUniversity Physics Solution

1106_PartUniversity Physics Solution - 40-16 Chapter 40 (c)...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
40-16 Chapter 40 (c) The difference in energy decreases between successive levels. For example: 23 32 1 0 1, 2 1 0.59, 3 2 0.49,. .. −= A sharp step gave ever-increasing level differences 2 (~ ). n A parabola 2 (~ ) gave evenly spaced levels (~ ). xn Now, a linear potential (~ ) gives ever-decreasing level differences (~ ). Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the level differences will increase. If the curvature is less than a parabola, the differences will decrease.
Background image of page 1

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

View Full DocumentRight Arrow Icon
41-1 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 2
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 ⎡⎤ ⎛⎞ == = = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ .
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

1106_PartUniversity Physics Solution - 40-16 Chapter 40 (c)...

This preview shows document pages 1 - 4. Sign up to view the full document.

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