ch7 - Chapter 7 ! Atomic Physics 7-1. (Equation 7-4) where...

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151 Chapter 7 ! Atomic Physics 7-1. (Equation 7-4) where and The 1st, 2nd, 3rd, and 5th excited states are degenerate. Energy (×E 0 ) 7-2. (Equation 7-5) is the lowest energy level. where The next nine levels are, increasing order,
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Chapter 7 ! Atomic Physics 152 (Problem 7-2 continued) n 1 n 2 n 3 E (×E 0 ) 112 1.694 121 2.111 113 2.250 122 2.444 123 3.000 114 3.028 131 3.360 132 3.472 124 3.778 7-3. (a) (b) They are identical. The location of the coordinate origin does not affect the energy level structure. 7-4.
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Chapter 7 ! Atomic Physics 153 7-5. (from Equation 7-5) where (a) n 1 n 2 n 3 E (×E 0 ) 111 1 .313 112 1 .500 113 1 .813 121 2 .063 114 2 .250 122 2 .250 123 2 .563 115 2 .813 124 3 .000 116 3 .500 (b) 1,1,4 and 1,2,2 7-6.
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Chapter 7 ! Atomic Physics 154 (Problem 7-6 continued) 7-7. 7-8. (a) Adapting Equation 7-3 to two dimensions (i.e., setting k 3 = 0), we have (b) From Equation 7-5, (c) The lowest energy degenerate states have quantum numbers n 1 = 1, n 2 = 2 and n 1 = 2, n 2 = 1.
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Chapter 7 ! Atomic Physics 155 7-9. (a) For n = 3, R = 0, 1, 2 (b) For R = 0, m = 0 R = 1, m = ! 1, 0, +1 R = 2, m = ! 2, ! 1, 0 +1, +2 (c) There are nine different m -states, each with two spin states, for a total of 18 states for n = 3. 7-10. (a) For n = 2, R = 0, 1 For R = 0, m = 0 with two spin states For R = 1, m = ! 1, 0, +1, each with two spin states The total number of states with n = 2 is eight. (b) For n = 4, R = 0, 1, 2, 3 Adding to those states found in (a), For R = 2, there are 2 R + 1 = 5 m states and 10 total, including spin. For R = 3, there are 2 R + 1 = 7 m states and 14 total, including spin. Thus, for n = 4 there are a total of 8 + 10 + 14 = 32 states, including spin. (c) All n = 2 states have the same energy. All n = 4 states have the same energy. 7-11. (a) (b)
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Chapter 7 ! Atomic Physics 156 7-12. (a) R = 1 | L | = (b) R = 2 | L | =
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Chapter 7 ! Atomic Physics 157 (Problem 7-12 continued) (c) R = 4 | L | = (d) (See diagram above.) 7-13. (a) (b) (c) L x and L y cannot be determined separately. (d) n = 3 7-14. (a) For R = 1, (b) For R = 1, m = ! 1, 0, = 1
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Chapter 7 ! Atomic Physics 158 (Problem 7-14 continued) (c) Z 1 S ! 1 S (d) For R = 3, and m = ! 3. ! 2, ! 1, 0, 1, 2, 3. Z 3 S 2 S 1 S 0 ! 1 S ! 2 S ! 3 S 7-15. and . Since for , i.e., central forces, F is parallel to r , then r × F = 0 and
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Chapter 7 ! Atomic Physics 159 7-16. (a) For R = 3, n = 4, 5, 6, . .. and m = ! 3, ! 2, ! 1, 0, 1, 2, 3 (b) For R = 4, n = 5, 6, 7,. .. and m = ! 4, ! 3, ! 2, ! 1, 0, 1, 2, 3, 4 (c) For R = 0, n = 1 and m = 0 (d) The energy depends only on n . The minimum in each case is: 7-17. (a) 6 f state: n = 6, R = 3 (b) (c) (d) L Z = ! 3 S , ! 2 S , ! 1 S , 0, 1 S , 2 S , 3 S 7-18. Referring to Table 7-2, R 30 = 0 when Letting , this condition becomes Solving for x (quadratic formula or completing the square), x = 1.90, 7.10 ˆ Compare with Figure 7-10(a). 7-19. (a) For the ground state n = 1, R = 0, and m = 0. at
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Chapter 7 ! Atomic Physics 160 (Problem 7-19 continued) (b) at (c) at 7-20. (a) For the ground state, For , at r = a 0 we have (b) For ) r = 0.03a 0 , at r = 2a 0 we have 7-21. For P(r) to be a maximum, This condition is satisfied when or . For r = 0, P(r) = 0 so the maximum P(r) occurs for . 7-22.
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Chapter 7 ! Atomic Physics 161 (Problem 7-22 continued) Letting , we have that and and substituting these above, Integrating on the right side Solving for yields 7-23. (Z = 1 for hydrogen) (a) For ) r = 0.02a 0 , at r = a 0 we have (b) For ) r = 0.02a 0 , at r = 2a 0 we have
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Chapter 7 ! Atomic Physics 162 7-24.
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ch7 - Chapter 7 ! Atomic Physics 7-1. (Equation 7-4) where...

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