6-Hydrogen Atom Wavefunctions

6-Hydrogen Atom Wavefunctions - 5.111 Lecture Summary #6...

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5.111 Lecture Summary #6 September 21, 2009 7.1 Reading for today: Section 1.9 Reading for Lecture 7: Sections 1.10-1.11 HYDROGEN ATOM WAVEFUNCTIONS When solving ψ = ψ ˆ HE for H atom, we get energy values E n and wavefunctions ψ (r, θ,φ ) Solutions to differential equation ψ = ψ ˆ with () ⎡⎤ ⎛⎞ θφ = + θ + + ⎜⎟ ⎢⎥ θθ θ θϕ ⎝⎠ ⎣⎦ = 2 2 2 22 2 2 2 e 1d d 1 d d 1 d ˆ Hr ,, r s in U (r ) 2m r dr dr r sin d d r sin d =− πε 2 0 e Ur 4r Energy values == ε 4 H n 2 2 0 R 1me E n8h n ψ (r, θ,φ ) stationary state wavefunction: time-independent In solutions for ψ (r, θ,φ ), two new quantum numbers appear! A total of 3 quantum numbers needed to describe a wavefunction in 3D. 1. n principal quantum number n = 1, 2, 3 … … determines binding energy 2. A angular momentum quantum number A = 0, 1, 2, 3 … … n – 1 A is related to n largest value of A = n – 1 determines angular momentum of e 3. m magnetic quantum number m 0, ±1, ±2, ±3, … … ± A m is related to A largest value is + A , smallest is – A determines behavior of atom in magnetic field
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5.111 Lecture Summary #6 September 21, 2009 7.2 State label wavefunction orbital E n E n [J] n = 1 A = 0 100 ψ 100 1s –R H /1 2 –2.18 × 10 –18 J m = 0 n = 2 A = 0 200 ψ 200 2s –R H /2 2 -5.45 × 10 –19 J m = 0 n = 2 A = 1 211 ψ 211 2p +1 –R H /2 2 -5.45 × 10 –19 J m = +1 n = 2 A = 1 210 ψ 210 2p 0 –R H /2 2 -5.45 × 10 –19 J m = 0 n = 2 A = 1 21-1 ψ 21-1 2p -1 –R H /2 2 -5.45 × 10 –19 J
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6-Hydrogen Atom Wavefunctions - 5.111 Lecture Summary #6...

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