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Second exam review

# Second exam review - The Hydrogen Atom The H atom is one of...

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The Hydrogen Atom The H atom is one of the problems we discussed two lectures ago. The Schrodinger equation is The angular terms are solved identically to last time. The separation constant β is so the radial equation becomes. ) , , ( ) , , ( ) ( ) , , ( ) ( sin 1 ) sin( ) sin( 1 1 2 2 2 2 2 2 2 2 2 φ θ ψ φ θ ψ φ θ ψ φ θ θ θ θ θ μ r E r r V r r r r r r r = + + + h ) 1 ( + l l ( ) r e r V r R E r R r V r dr r dR r dr r R d r E r V r dr r dR r dr d r R 0 2 2 2 2 2 2 2 2 2 2 4 ) ( where ) ( ) ( ) ( ) 1 ( 2 ) ( 2 ) ( 2 or ) 1 ( ) ( 2 ) ( ) ( 1 ε π μ μ μ = = + + + + + = + l l h h l l h Since V(r) is a function only of r, the potential energy commutes with both L 2 and L z , and therefore so does the Hamiltonian operator itself. All have simultaneous eigenvalues and eigenfunctions. The constant ε 0 = 8.854·10 12 just makes things work in the infuriating MKS electrical system. Note also that this equation will solve for any one-electron ion if we replace e 2 by Ze 2 where Z is the atomic number of the nucleus, though answers become poor much above Kr because the electron motion starts becoming relativistic.

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This is an important equation, so lets choose it to discuss how these things are solved or rather were solved 160 years ago. As r becomes large, the terms in 1/r and 1/r 2 become progressively smaller, and at asymptotically large r we have whose general solution is If c is imaginary we find that E is positive. These are not the solutions we want, as they represent incoming and outgoing spherical wavefronts. If c is real and positive the term with A blows up as r gets large, which is wrong. So c is real and positive and As r goes to zero the term with 1/r 2 in it dominates (even the term in E becomes unimportant) and whose solution is of the form We then define a function v(r): and try to solve the equation for v(r) . large) r very (for ) ( ) ( 2 2 2 2 r u E dr r u d = μ h imaginary or real be may c where ) ( cr cr Be Ae r u + r Be r u cr as ) ( 0 2 2 2 2 2 2 4 where ) ( ) ( ) 1 ( 2 ) ( 2 becomes equation above the and define First we ε π μ μ e V r u E r u r V r dr r u d rR(r) u(r) = = + + + = l l h h ) ( ) 1 ( ) ( 2 2 2 r u r dr r u d + = l l r small for ) ( so 0 r as up blows but ) ( 1 1 + + + = l l l l Cr r u r D r D Cr r u ) ( ) ( 1 r v e r r u cr + = l