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

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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 ) 1 1 2 1.694 1 2 1 2.111 1 1 3 2.250 1 2 2 2.444 1 2 3 3.000 1 1 4 3.028 1 3 1 3.360 1 3 2 3.472 1 2 4 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 ) 1 1 1 1.313 1 1 2 1.500 1 1 3 1.813 1 2 1 2.063 1 1 4 2.250 1 2 2 2.250 1 2 3 2.563 1 1 5 2.813 1 2 4 3.000 1 1 6 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, = 0, 1, 2 (b) For = 0, m = 0 = 1, m = 1, 0, +1 = 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, = 0, 1 For = 0, m = 0 with two spin states For = 1, m = 1, 0, +1, each with two spin states The total number of states with n = 2 is eight. (b) For n = 4, = 0, 1, 2, 3 Adding to those states found in (a), For = 2, there are 2 + 1 = 5 m states and 10 total, including spin. For = 3, there are 2 + 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) = 1 | L | = (b) = 2 | L | =
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Chapter 7 Atomic Physics 157 (Problem 7-12 continued) (c) = 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 = 1, (b) For = 1, m = 1, 0, = 1
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Chapter 7 Atomic Physics 158 (Problem 7-14 continued) (c) Z 1 1 (d) For = 3, and m = 3. 2, 1, 0, 1, 2, 3. Z 3 2 1 0 1 2 3 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 = 3, n = 4, 5, 6, ... and m = 3, 2, 1, 0, 1, 2, 3 (b) For = 4, n = 5, 6, 7,... and m = 4, 3, 2, 1, 0, 1, 2, 3, 4 (c) For = 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, = 3 (b) (c) (d) L Z = 3 , 2 , 1 , 0, 1 , 2 , 3 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 Compare with Figure 7-10(a). 7-19. (a) For the ground state n = 1, = 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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