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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 r p , 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 ( 29 π θ θ = - = = × 1 1 1 sin sin 2 2 2 dA d d d r r r r r r . Then, = × = × = 1 1 1 2 2 2 dA d dt dt m m r r r p L . 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 ( 29 = = 2 2 2 2 1 2 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 2 1 0 ± 1/2 2 1 1 ± 1/2 Yields 8 sets; ( 29 = = 2 2 2 2 2 8 n . Note that the number is twice the number of l m values. Also that for each l there are + 2 1 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 2 1 n s l . The series is an arithmetic progression: + + + K 2 6 10 14 , the sum of which is ( 29 [ ] ( 29 [ ] = + - = = = + - = 2 2 1 where 2, 4 2 4 1 4 2 2 n s a n d a d n s n n (c) = 3 n : ( 29 ( 29 ( 29 ( 29 + + = + + = = = = 2 2 2 1 2 3 2 5 2 6 10 18 2 2 3 18 n (d) = 4 n : ( 29 ( 29 ( 29 ( 29 ( 29 + + + = = = = 2 2 2 1 2 3 2 5 2 7 32 2 2 4 32 n (e) = 5 n : ( 29 ( 29 + = + = = = = 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 l m and = ± 1 2 s m for each l m n 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 .
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( 29 ( 29 - - - - × = = = × 27 3 24 2 2 2 2 108 u 1.66 10 kg u 10 m 2 3.59 10 N 10 s z M d F t . The magnetic moment of the silver atom is due to a single unpaired electron spin, so ( 29 μ μ - = = = = × h 24 2 2 9.27 10 J T 2 2 2 z z B e e e e S m m . Thus, μ - - × = = = × 24 24 3.59 10 N 0.387 T m 9.27 10 N z z z dB F 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 : ( 29 μ - = - = = - 2 s z B s e U g S B g Bm m B μ . With = 2 g for electrons, the energy difference between the up spin ( 29 = 1 2 s m and down spin ( 29 = - 1 2 s m orientations is ( 29 ( 29 ( 29 μ - - - = = × = × = × 24 24 5 2 9.273 10 J T 0.5 T 9.273 10 J 5.80 10 eV B U g B . 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 2 2 2 511 10 eV 3 10 nm 1.840 10 eV nm 5 I c c .
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PC Chapter 09 - 9 Atomic Structure 9-1 E = 2 B B = hf 2...

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