Lecture25.pptx - 3/11/11...

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3/11/11 1 1 PHYS 360 Quantum Mechanics Fri Mar 11, 2011 Lecture 25: What do the hydrogen electron orbitals look like? HW 7 due on Monday Mar 22: Ch 4, #2, 11, 13, 14, 15. 2 The Hydrogen Atom: Insert the Coulomb PotenOal V ( r ) = e 2 4 πε 0 1 r The goal: solve for the eigenfuncOons and eigenvalues, i.e. the wave funcOons and the energies. The radial equaOon: d dr r 2 dR dr 2 mr 2 2 V ( r ) E R = l ( l + 1) R The exact soluOon: ψ nlm = 2 na 3 ( n l 1)! 2 n [( n + l )!] 3 e r / na 2 r na l L n l 1 2 l + 1 (2 r / na ) Y l m ( θ , φ ) 3 100 ( r , , ) = R 10 ( r ) Y 0 0 ( , ) Note: n=1 requires that l=m=0 . The ground state wave funcOon: R 10 ( r ) = c 0 a e r / a 100 ( r , , ) = 1 π a 3 e r / a 100 ( r , , ) = 1 a 3 e r / a QuesOon: What is the value of the square of the wave funcOon at the origin? 100 ( r , , ) = 1 a 3 e r / a QuesOon: What is the probability of ±nding the electron at the origin?
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This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue University-West Lafayette.

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Lecture25.pptx - 3/11/11...

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