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Lecture25

# Lecture25 - TheHydrogenAtom: FriMar11,2011 Lecture25 V(r =...

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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 ψ 100 ( r , θ , φ ) = R 10 ( r ) Y 0 0 ( θ , φ ) Note: n=1 requires that l=m=0 . The ground state wave funcOon: R 10 ( r ) = c 0 a e r / a ψ 100 ( r , θ , φ ) = 1 π a 3 e r / a ψ 100 ( r , θ , φ ) = 1 π a 3 e r / a QuesOon: What is the value of the square of the wave funcOon at the origin? ψ 100 ( r , θ , φ ) = 1 π a 3 e r / a QuesOon: What is the probability of finding the electron at the origin?

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