What is this? Jerry Gilfoyle The Hydrogen Atom 1 / 18 What is - PowerPoint PPT Presentation

What is this? Jerry Gilfoyle The Hydrogen Atom 1 / 18

What is this? The Hydrogen Atom Jerry Gilfoyle The Hydrogen Atom 1 / 18

What is this? The Hydrogen Atom Jerry Gilfoyle The Hydrogen Atom 1 / 18

What is this? The Hydrogen Atom � � 1 1 1 = R H R H - Rydberg constant − n 2 n 2 λ f i Jerry Gilfoyle The Hydrogen Atom 1 / 18

Hydrogen Eigenvalues E n = − 13 . 6 eV n 2 Quantitative comparison for Balmer series hydrogen in units of σ . My Results (˚ NIST Results (˚ Line A) A) Normalized Percent Difference Difference 6 . 64 ± 0 . 09 × 10 3 6 . 56280 × 10 3 0.95 1.2 α 4 . 85 ± 0 . 15 × 10 3 4 . 86133 × 10 3 β 0.11 -0.2 4 . 39 ± 0 . 06 × 10 3 4 . 34047 × 10 3 0.9 1.2 γ α : n = 3 → n = 2 β : n = 4 → n = 2 γ : n = 5 → n = 2 Jerry Gilfoyle The Hydrogen Atom 2 / 18

n = 8 , l = 3 , m = 1 Jerry Gilfoyle The Hydrogen Atom 3 / 18

How do we build the quantum model? 1 What is the mechanical energy? Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r 2 What is the Schroedinger equation? Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r 2 What is the Schroedinger equation? − � 2 r ) − e 2 2 µ ∇ 2 ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � r ) � 1 − � 2 ∂ 2 r ) − e 2 ∂ r r 2 ∂ ∂ 1 ∂θ sin θ ∂ ∂ 1 � ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � ∂ r + ∂θ + r ) r 2 sin θ r 2 sin 2 θ 2 µ r 2 ∂ 2 φ Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r 2 What is the Schroedinger equation? − � 2 r ) − e 2 2 µ ∇ 2 ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � r ) � 1 − � 2 ∂ 2 r ) − e 2 ∂ r r 2 ∂ ∂ 1 ∂θ sin θ ∂ ∂ 1 � ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � ∂ r + ∂θ + r ) r 2 sin θ r 2 sin 2 θ 2 µ r 2 ∂ 2 φ 3 What do we know about the solution? Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r 2 What is the Schroedinger equation? − � 2 r ) − e 2 2 µ ∇ 2 ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � r ) � 1 − � 2 ∂ 2 r ) − e 2 ∂ r r 2 ∂ ∂ 1 ∂θ sin θ ∂ ∂ 1 � ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � ∂ r + ∂θ + r ) r 2 sin θ r 2 sin 2 θ 2 µ r 2 ∂ 2 φ 3 What do we know about the solution? r ) = R ( r )Θ( θ )Φ( φ ) = R ( r ) Y m ϕ s ( � l ( θ, φ ) Jerry Gilfoyle The Hydrogen Atom 4 / 18

How do we build the quantum model? 1 What is the mechanical energy? E = p 2 2 µ − e 2 r = p 2 2 µ r 2 − e 2 L 2 m p m e r 2 µ + µ = ≈ m e m p + m e r 2 What is the Schroedinger equation? − � 2 r ) − e 2 2 µ ∇ 2 ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � r ) � 1 − � 2 ∂ 2 r ) − e 2 ∂ r r 2 ∂ ∂ 1 ∂θ sin θ ∂ ∂ 1 � ϕ s ( � r ϕ s ( � r ) = E ϕ s ( � ∂ r + ∂θ + r ) r 2 sin θ r 2 sin 2 θ 2 µ r 2 ∂ 2 φ GO SOLVE IT! 3 What do we know about the solution? r ) = R ( r )Θ( θ )Φ( φ ) = R ( r ) Y m ϕ s ( � l ( θ, φ ) Jerry Gilfoyle The Hydrogen Atom 4 / 18

Hydrogen Bound State Eigenfunctions ϕ nlm ( r , θ, φ ) = R nl ( r ) Y m l ( θ, φ ) = (2 κ ) 3 / 2 A nl ρ l e − ρ/ 2 F nl ( ρ ) Y m l ( θ, φ ) Jerry Gilfoyle The Hydrogen Atom 5 / 18

Hydrogen Bound State Eigenfunctions ϕ nlm ( r , θ, φ ) = R nl ( r ) Y m l ( θ, φ ) = (2 κ ) 3 / 2 A nl ρ l e − ρ/ 2 F nl ( ρ ) Y m l ( θ, φ ) ∞ ( i + l + 1) − λ = a i ρ i ⇒ a i +1 = � F ( ρ ) = ( i + 1)( i + 2 l + 2) a i a 0 = 1 i =0 � µ � λ = Ze 2 2 µ | E | E n = −| E | ρ = 2 κ r κ = � 2 � 2 | E | � ( n − l − 1)! F nl ( ρ ) = L 2 l +1 n − l − 1 ( ρ ) A nl = 2 n [( n + l )!] 3 Jerry Gilfoyle The Hydrogen Atom 5 / 18

Hydrogen Eigenvalues (Energy Levels) 5 Continuum States 0 E n = − µ ( e 2 ) 2 2 � 2 n 2 = − 13 . 6 eV Discrete n 2 Energy ( eV ) States - 5 - 10 - 15 0 2 4 6 8 Jerry Gilfoyle The Hydrogen Atom 6 / 18

Hydrogen Bound State Eigenfunctions ψ Enlm ( r , θ, φ ) = R nl ( r ) Y m l ( θ, φ ) � k max � = A nl ρ l e − ρ � b k ρ k Y m l ( θ, φ ) k =0 Jerry Gilfoyle The Hydrogen Atom 7 / 18

Hydrogen Bound State Eigenfunctions ψ Enlm ( r , θ, φ ) = R nl ( r ) Y m l ( θ, φ ) � k max � = A nl ρ l e − ρ � b k ρ k Y m l ( θ, φ ) k =0 b k +1 = 2( k + l + 1) − λ e 2 ( k + 1)( k + 2 l + 2) b k b 0 = 1 � � � 2 2 µ W 2 µ E n = − W ρ = κ r κ = λ = a 0 = � 2 � 2 W me 2 �� 2 � 2 r � 2 r � 3 ( n − l − 1)! � l � 2 n [( n + l )!] 3 e − r / na 0 ( n + l )! L 2 l +1 Y m ψ Enlm = l ( θ, φ ) n − l − 1 na 0 na 0 na 0 Jerry Gilfoyle The Hydrogen Atom 7 / 18

Recall the Solid Angle Jerry Gilfoyle The Hydrogen Atom 8 / 18

Spherical Differential Volume Element Jerry Gilfoyle The Hydrogen Atom 9 / 18

Hydrogen Eigenfunctions Hydrogen Probability Density ( n = 4 ) 0.14 Red - l = 0 0.12 0.10 0.08 P 0.06 0.04 0.02 0.00 0 10 20 30 40 r Jerry Gilfoyle The Hydrogen Atom 10 / 18

Hydrogen Eigenfunctions Hydrogen Probability Density ( n = 4 ) 0.14 Red - l = 0 0.12 Blue - l = 1 0.10 0.08 P 0.06 0.04 0.02 0.00 0 10 20 30 40 r Jerry Gilfoyle The Hydrogen Atom 11 / 18

Hydrogen Eigenfunctions Hydrogen Probability Density ( n = 4 ) 0.14 Red - l = 0 0.12 Blue - l = 1 0.10 Green - l = 2 0.08 P 0.06 0.04 0.02 0.00 0 10 20 30 40 r Jerry Gilfoyle The Hydrogen Atom 12 / 18

Hydrogen Eigenfunctions Hydrogen Probability Density ( n = 4 ) 0.14 Red - l = 0 0.12 Blue - l = 1 0.10 Green - l = 2 Gray - l = 3 0.08 P 0.06 0.04 0.02 0.00 0 10 20 30 40 r Jerry Gilfoyle The Hydrogen Atom 13 / 18

Do the peaks line up? Red : n = 1, Blue: n = 4 0.012 0.010 Probability Density 0.008 0.006 0.004 0.002 0.000 0.0 0.2 0.4 0.6 0.8 1.0 r ( angstroms ) Jerry Gilfoyle The Hydrogen Atom 14 / 18

Old Orbitals Jerry Gilfoyle The Hydrogen Atom 15 / 18

Old Orbitals - New Orbitals Jerry Gilfoyle The Hydrogen Atom 15 / 18

Old Orbitals - New Orbitals How are these plots related to what we know? Jerry Gilfoyle The Hydrogen Atom 15 / 18

More Hydrogen Eigenfunctions Jerry Gilfoyle The Hydrogen Atom 16 / 18

Hydrogen Eigenvalues E n = − 13 . 6 eV n 2 Quantitative comparison for Balmer series hydrogen in units of σ . My Results (˚ NIST Results (˚ Line A) A) Normalized Percent Difference Difference 6 . 64 ± 0 . 09 × 10 3 6 . 56280 × 10 3 0.95 1.2 α 4 . 85 ± 0 . 15 × 10 3 4 . 86133 × 10 3 β 0.11 -0.2 4 . 39 ± 0 . 06 × 10 3 4 . 34047 × 10 3 0.9 1.2 γ α : n = 3 → n = 2 β : n = 4 → n = 2 γ : n = 5 → n = 2 Jerry Gilfoyle The Hydrogen Atom 17 / 18

Some Plots Jerry Gilfoyle The Hydrogen Atom 18 / 18