Sample_Problem_Math_Chpt_D

# Sample_Problem_Math_Chpt_D - integrals. This is not true in...

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Chem 370 February 20, 2008 Sample Problem from Math Chapter D The H atom 2 2p orbital is a wavefunction with the following form in Cartesian coordinates ( 29 ( 29 ( 29 ( 29 1 2 2 2 2 1 2 2 2 2 2p 2p 1 2 2 2 2 z z x y z z N x y z exp 2 x y z + + ψ = + + - + + r (1) and the following much simpler form in spherical polar coordinates ( 29 r 2 2p 2p z z N re cos . - ψ = θ r (2) Our problem is to choose 2p z N so that ( 29 2p z ψ r is normalized. This may be done by choosing 2p z N as ( 29 ( 29 z 1 2 2p 2p 2p 2p 2p z z z z N N N - = ψ ψ 6 dr r r . (3) Using the Cartesian form for h dr dx dy dz h - - - I = dr (4) and Eqs. (1) and (3) yields an intractable integral form for 2p z N . However using the spherical polar form for h dr 2 2 0 0 0 r dr sin d d h π π = θ θ φ dr (5) and Eqs. (2) and (3) yields a much simpler integral for 2p z N z 1 2 2p N I - = . (6) where } 3 2 1 2 4 r 2 0 0 0 I r e dr sin cos d d h π π - = θ θ θ φ 647 48 6 4 47 4 48 . (7) Note from Eq. (7) the spherical polar form for I reduces to a product of three-one-dimensional

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Unformatted text preview: integrals. This is not true in Cartesian coordinates. Let us perform the three integrals in Eq. (7). Integral 1. 2 1 I d 2 = = . (8a) Integral 2. 2 2 I sin cos d . = Letting y cos = and thus 2 dy sin d , I = - becomes 1 2 2 1 2 I y dy 3-= = . (8b) Integral 3 We need the result at the bottom of text page 244, which is n ax n 1 n! x e dx a h-+ = . Using this relation we have that 4 r 3 I r e dr 4! 24 h-= = = . (8c) Thus comparing Eq. (7) and (8) gives that 1 2 3 I I I I 32 = = . (9) Thus from Eq. (6) T ....
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## This note was uploaded on 06/27/2008 for the course CHEM 370 taught by Professor Oldsleepyman during the Fall '08 term at Purdue University.

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Sample_Problem_Math_Chpt_D - integrals. This is not true in...

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