PC Chapter 09

PC Chapter 09 - 9 Atomic Structure 9-1 E = 2 B B = hf 2 (...

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139 9 Atomic Structure 9-1 μ == 2 B E B hf ( 29 ( 29 ( 29 -- × 24 34 2 9.27 10 J T 0.35 T 6.63 10 Js f so 9 9.79 10 Hz f 9-2 Since L rp , and = md dt r p is tangent to the path at every point, the orbit is confined to the plane perpendicular to L . The figure below shows the positions of the particle in this orbital plane at consecutive times t and + t dt . The area (of the triangle) swept out in this time is (29 π θθ = - = 1 11 sin sin 2 22 d A d dd rr . Then, = × = ×= 111 2 d Ad dt dt m m r r r pL . d r π θ r ( t + dt ) r ( t ) 9-3 (a) = 1 n ; for = 1 n , = 0 l , = 0 l m , 1 2 s m 2 sets n l l m s m 1 0 0 –1/2 1 0 0 +1/2 2 2 2 2 12 n
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140 CHAPTER 9 ATOMIC STRUCTURE (b) For = 2 n we have n l l m s m 2 0 0 ± 1/2 2 1 –1 ± 1/2 1 0 ± 1/2 1 ± 1/2 Yields 8 sets; (29 == 2 2 2 2 28 n . Note that the number is twice the number of l m values. Also that for each l there are + 21 l l m values. Finally, l can take on values ranging from 0 to - 1 n , so the general expression is ( 29 - =+ 1 0 2 n sl . The series is an arithmetic progression: +++ K 261 01 4 , the sum of which is [ ] [ ] = + - -= 2 2 1 where 2, 4 2 4 1 42 2 n s a n d ad n s nn (c) = 3 n : + + = + + = = 2 2 21 23 25 2 6 1 0 1 8 2 23 1 8 n (d) = 4 n : ( 29( +++= = 2 2 2 1 2 3 2 5 2 7 32 2 2 4 32 n (e) = 5 n : + =+== = = 2 2 32 2 9 32 18 50 2 2 5 50 n 9-4 (a) 3 d subshell ⇒=⇒ = -- 2 2 , 1 ,0 ,1 ,2 l lm and 1 2 s m for each l m l l m s m 3 2 –2 –1/2 3 2 –2 +1/2 3 2 –1 –1/2 3 2 –1 +1/2 3 2 0 –1/2 3 2 0 +1/2 3 2 1 –1/2 3 2 1 +1/2 3 2 2 –1/2 3 2 2 +1/2 ( b ) 3 p subshell: for a p state, = 1 l . Thus l m can take on values - l to l , or –1, 0, 1. For each l m , s m can be ± 1 2 .
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MODERN PHYSICS 141 n l l m s m 3 1 –1 –1/2 3 1 –1 +1/2 3 1 0 –1/2 3 1 0 +1/2 3 1 1 –1/2 3 1 1 +1/2 9-5 The time of passage is == 1 m 0.01 s 100 m s t . Since the field gradient is assumed uniform, so is the force, and hence the acceleration. Thus the deflection is = 2 1 2 d at , or = 2 2 d a t for the acceleration. The required force is then (29 ( 29 -- - - × = = 2 73 24 22 2 2 108 u 1.66 10 kg u 10 m 2 3.59 10 N 1 0s z Md F t . The magnetic moment of the silver atom is due to a single unpaired electron spin, so μμ -  = = =   h 24 2 2 9.27 10 J T 2 z zB ee S mm . Thus, μ - - × = × 24 24 3.59 10 N 0.387 T m 9.27 10 N zz z d BF dz . 9-6 The exiting beams differ in the spin orientation of the outermost atomic electron. The energy difference derives from the magnetic energy of this spin in the applied field B : μ - =- ⋅ = =- 2 s z Bs e U g S B g Bm m B μ . With = 2 g for electrons, the energy difference between the up spin = 1 2 s m and down spin 1 2 s m orientations is ( 29 ( 29 ( 29 μ - = = × = × 2 4 2 45 2 9.273 10 J T 0.5 T 9.273 10 J 5.80 10 eV B UgB . 9-7 The angular momentum L of a spinning ball is related to the angular velocity of rotation ϖ as = I L ϖ . I , the moment of inertia, is given in terms of the mass m and radius R of the ball as = 2 2 5 I mR . For the electron this gives
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142 CHAPTER 9 ATOMIC STRUCTURE ( 29( 29 -- = × × 2 3 2 6 6 22 2 511 10 eV 3 10 nm 1.840 10 eV nm 5 I cc .
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This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.

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PC Chapter 09 - 9 Atomic Structure 9-1 E = 2 B B = hf 2 (...

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