G r1 r2 1 r1 1 r2 4 exp z r1 r2

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Unformatted text preview: nergy of He •  For the ground state, both electrons are in the (n=1,l=0,m=0) (ortho vanishes) Z3 ! G (r1 , r2 ) = ! 1 (r1 )! 1 (r2 ) = 4 exp[ " Z (r1 + r2 ) / a0 ] a0 Recall from Lecture 2, the ground state energy of the Bohr H atom was: E0 = !13.6 Z 2 eV Z 2 E0 Z 2 E0 And so the addi;on of the two single par;cle energies: E = ! !2 2 n1 n2 So the ground state of the He atom ignoring e ­e repulsion: E0 = !13.6 " ( 4 ) ! 13.6 " 4 = !108.8 eV The a...
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This document was uploaded on 02/04/2014.

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