# Apartthatdependsonandapart thatdependon

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Unformatted text preview: to hydrogen atom  The equa0on below suggests that we can write the wavefunc0on as a  product. We see independent part. A part that depends only on the disance  and a par that depends on the orienta0on  1 ∂⎛ ∂⎞ 1 ∂2 ⎤ ⎛ ∂⎛ 2 ∂⎞ ⎞ 2⎡ − ⎜ ⎜ r ψ ( r,θ , φ )⎟ − ⎢ ⎜ sin (θ ) ⎟ + 2 ⎥ ψ ( r,θ , φ ) ⎠ ⎝ ∂r ⎝ ∂r ⎟ ⎠ sin (θ ) ∂θ ⎝ ∂θ ⎠ sin (θ ) ∂φ 2 ⎦ ⎣ 2 ⎡ e2 ⎤ −2 me r ⎢ + E ⎥ ψ ( r,θ , φ ) = 0 ⎣ 4πε 0 r ⎦ 2 And the sugges0on  ψ ( r,θ , φ ) = R ( r )Y (θ , φ ) = R ( r ) Θ (θ ) Φ (φ ) Dividing by the wavefunc0on we have  Angular momentum: Hydrogen atom  2 ⎛ ∂ ⎛ 2 ∂ ⎞ 2 ⎡ 1 ∂⎛ ∂⎞ 1 ∂2 ⎤ ⎞ − r sin (θ ) ⎟ + 2 ⎟ R ( r )⎟ − ⎥ Y (θ , φ ) ⎝ ⎝⎜ ⎠ Y (θ , φ ) ⎢ sin (θ ) ∂θ ⎜ R ( r ) ⎜ ∂r ⎝ ∂r ⎠ ∂θ ⎠ sin (θ ) ∂φ 2 ⎦ ⎣ ⎡ e2 ⎤ −2 me r ⎢ + E⎥ = 0 ⎣ 4πε 0 r ⎦ 2 We have a sum of terms. Two terms depending on the distance and one  depends only on the orienta0on. For the equality to hold each of he term  must be a constant. We already know the solu0on for the angular part  ⎡1 ∂⎛ ∂⎞ 1 ∂2 ⎤ ⎜ sin (θ ) ⎟ + 2 ⎢ ⎥ Y (θ , φ ) = l (l + 1)Y (θ , φ ) l = 0,1, 2, ... sin (θ ) ∂θ ⎝ ∂θ ⎠ sin (θ ) ∂φ 2 ⎦ ⎣ We can actually do more than that since we can also separate the angular  momentum...
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## This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas at Austin.

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