PS4-08-Solution - Chem 442B Homework Set #4 - Solutions...

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Chem 442B Homework Set #4 - Solutions DUE: Saturday, Nov. 8, 2008 1)a) () 22 2 2 2 2 1 ˆ cos sin 1 cos 2 1 ˆ sin cos 1 sin 2 n nn n dd Hn Id n nE I n n I n n n θ θθ ππ π + + ⎛⎞ Ψ= = ⎜⎟ ⎝⎠ == Ψ = Ψ = = b) 2 2 00 2 2 0 2 2 0 2 2 1 cos We can use symmetry: We know that: sin cos 2 1 So, cos 2 2 1 cos 1 1 In a similar proof: sin 1 We could also have used n n dn d d d nd + + += = = ∫∫ d = ( ) ( ) the formulas 11 cos cos 2 sin cos 2 =+ =− c) Problem Set #4 Solution - 1 -
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() /2 /2 2 n+ n+ 0 0 Using the results from Lecture 9, slide 5 we have: 11 Prob(1st 1/4) sin 2 24 1 sin sin 0 44 4 1 independent of n 4 n n n nn π θ = + ⎡⎤ =+ ⎢⎥ ⎣⎦ = d) 2 0 2 0 22 2 2 2 0 2 0 2 2 cos cos cos sin 0 cos cos cos z z n z n Lz z n n id Ln n d in d d d n nd n LL n 2 d n d θθ πθ σ =− == = = = = = = = 2)a) 1/4 2 0 1/2 2 0 2 2 2 2 2 2 even function 1 2 (from back cover of book) 2 1 = 2 1 2 x xx x xe x x ed x x x β ββ ππ βπ ∞∞ −− −∞ ⎛⎞ Ψ= ⎜⎟ ⎝⎠ = = ∫∫ Problem Set #4 Solution - 2 -
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b) () 22 1/2 / 2 / 2 2 2 / 2 2 / 2 2 2
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PS4-08-Solution - Chem 442B Homework Set #4 - Solutions...

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