PS6-08-Solution - Chem 442 Homework Set #6 DUE: Wednesday...

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Chem 442 Homework Set #6 DUE: Wednesday November 12, 2008 1) McQ&S 6-46 () B 24 1 B B z 24 1 B The magnitude of the spiltting is . 9.2740154 10 (Front cover of book - its denoted as ) B 15 in this problem. So, the magnitude of the splitting is 9.2740154 10 15 1 z z Bm JT T J T T m m β βμ −− = = 22 18 18 21 22 .4 10 11 1 2.1798736 10 1 1.6349 10 4 We see that the magnitude of the splitting is very small, about 4 orders of magnigude smaller than the 2p 1s energy level spac ps H J EE R J J × ⎛⎞ −= = × = × ⎜⎟ ⎝⎠ 18 22 18 18 18 18 22 18 ing. The three transitions from 2p 1s will take place at the following energies: 1 1.6349 10 1.4 10 1.6350 10 0 1.6349 10 0 1.6349 10 1 1.6349 10 1.4 10 1.6348 10 mE J J =→ Δ = × + × = × Δ = × + = × =− →Δ = × × = × We see that these transitions are very close together. J 2) McQ&S 6-47 m=2 m=1 m=0 m=-1 m=-2 E l =2 m=2 m=1 m=0 m=-1 m=-2 m=3 m=-3 l =3 If there is an external field, the states with different values of l will be split. Based on this picture, it is easy to see that there are of 5 possible initial states and 7 possible final states. Therefore, there are 5 × 7=35 total possible transitions. Using the selection rule allows 5 total transitions (for each initial state, there is exactly one final state possible). In the case when the selection rule is , each of the 5 initial states has 2 possible states. So, there are a total of 5 × 2=10 total transitions. 0 m Δ= 1 m ± Problem Set #6 - Solutions - 1 -
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3) a) () 22 1/4 2 /2 2 4/ 2 2 2/ 24 / 2 2 We know already that this function is normalized: | 1 ˆ 2 ˆ 2 ˆ x x xx Trial Function e d HD x e mdx d Hx e D x e H β ββ π −− ⎛⎞ =Φ= ⎜⎟ ⎝⎠ ΦΦ = Φ=− + Φ= + = = 2 2 2 / 2 4 / 2 2 4 2 2 4 1/2 2 42 2 4 ˆ ˆ ˆ 2 2 x x x x xe e D x e mm He x D x EH e x D x d x e x m −∞ ⎞⎛ + ⎟⎜ ⎠⎝ + + + + == = 2 0 2 2 2 34 2 44 2 13 ˆ 2 2 4 2 33 ˆ 2 84 8 4 4 Dx dx m D DD m βπ πβ ++ ⎜− + + Φ= − + + = + = = 55 2 1/6 2 12 6 03 2 6 opt dE D Dm D dm m Dm =− = →= = = = = 6 b) Just plug opt into the expression for E: 1/3 2 2 2/3 4 2 4/3 1/3 4/3 4 2 2 6 4 6 4 63 6 26 6 3 6 9 88 8 1 6 opt opt opt opt opt Dm E Dm Dm D E mD m DDD D E m m =+ = + = = = = = = = = Problem Set #6 - Solutions - 2 -
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4) a, b, c McQ&S 6-25 0 1 22 *2 2 * 2 2 00 0 0 3/2 2
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This note was uploaded on 02/12/2011 for the course CHEMISTRY 442 taught by Professor Lizzy during the Spring '10 term at Hanoi University of Technology.

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PS6-08-Solution - Chem 442 Homework Set #6 DUE: Wednesday...

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