What is this?
Jerry Gilfoyle The Hydrogen Atom 1 / 18
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
Jerry Gilfoyle The Hydrogen Atom 1 / 18
The Hydrogen Atom
Jerry Gilfoyle The Hydrogen Atom 1 / 18
The Hydrogen Atom
Jerry Gilfoyle The Hydrogen Atom 1 / 18
The Hydrogen Atom
1 λ = RH
n2
f
− 1 n2
i
Jerry Gilfoyle The Hydrogen Atom 1 / 18
En = −13.6 eV n2 Quantitative comparison for Balmer series hydrogen in units of σ. Line My Results (˚ A) NIST Results (˚ A) Normalized Percent Difference Difference α 6.64 ± 0.09 × 103 6.56280 × 103 0.95 1.2 β 4.85 ± 0.15 × 103 4.86133 × 103 0.11
γ 4.39 ± 0.06 × 103 4.34047 × 103 0.9 1.2 α : n = 3 → n = 2 β : n = 4 → n = 2 γ : n = 5 → n = 2
Jerry Gilfoyle The Hydrogen Atom 2 / 18
Jerry Gilfoyle The Hydrogen Atom 3 / 18
1 What is the mechanical energy? Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
2 What is the Schroedinger equation? Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
2 What is the Schroedinger equation?
− 2 2µ∇2ϕs( r) − e2 r ϕs( r) = Eϕs( r)
− 2 2µ 1 r2 ∂ ∂r r2 ∂ ∂r + 1 r2 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 r2 sin2 θ ∂2 ∂2φ
r)− e2 r ϕs( r) = Eϕs( r)
Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
2 What is the Schroedinger equation?
− 2 2µ∇2ϕs( r) − e2 r ϕs( r) = Eϕs( r)
− 2 2µ 1 r2 ∂ ∂r r2 ∂ ∂r + 1 r2 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 r2 sin2 θ ∂2 ∂2φ
r)− e2 r ϕs( r) = Eϕs( r)
3 What do we know about the solution? Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
2 What is the Schroedinger equation?
− 2 2µ∇2ϕs( r) − e2 r ϕs( r) = Eϕs( r)
− 2 2µ 1 r2 ∂ ∂r r2 ∂ ∂r + 1 r2 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 r2 sin2 θ ∂2 ∂2φ
r)− e2 r ϕs( r) = Eϕs( r)
3 What do we know about the solution?
ϕs( r) = R(r)Θ(θ)Φ(φ) = R(r)Y m
l (θ, φ)
Jerry Gilfoyle The Hydrogen Atom 4 / 18
1 What is the mechanical energy?
E = p2 2µ − e2 r = p2
r
2µ + L2 2µr2 − e2 r µ = mpme mp + me ≈ me
2 What is the Schroedinger equation?
− 2 2µ∇2ϕs( r) − e2 r ϕs( r) = Eϕs( r)
− 2 2µ 1 r2 ∂ ∂r r2 ∂ ∂r + 1 r2 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 r2 sin2 θ ∂2 ∂2φ
r)− e2 r ϕs( r) = Eϕs( r)
3 What do we know about the solution?
ϕs( r) = R(r)Θ(θ)Φ(φ) = R(r)Y m
l (θ, φ)
Jerry Gilfoyle The Hydrogen Atom 4 / 18
ϕnlm(r, θ, φ) = Rnl(r)Y m
l (θ, φ)
= (2κ)3/2 Anlρle−ρ/2Fnl(ρ)Y m
l (θ, φ)
Jerry Gilfoyle The Hydrogen Atom 5 / 18
ϕnlm(r, θ, φ) = Rnl(r)Y m
l (θ, φ)
= (2κ)3/2 Anlρle−ρ/2Fnl(ρ)Y m
l (θ, φ)
F(ρ) =
∞
= aiρi ⇒ ai+1 = (i + l + 1) − λ (i + 1)(i + 2l + 2)ai a0 = 1 En = −|E| ρ = 2κr κ =
2 λ = Ze2
2|E| Fnl(ρ) = L2l+1
n−l−1(ρ)
Anl =
2n[(n + l)!]3
Jerry Gilfoyle The Hydrogen Atom 5 / 18
En = −µ(e2)2 22n2 = −13.6 eV n2
Discrete States Continuum States 2 4 6 8
5 Energy (eV) Jerry Gilfoyle The Hydrogen Atom 6 / 18
ψEnlm(r, θ, φ) = Rnl(r)Y m
l (θ, φ)
= Anlρle−ρ kmax
bkρk
l (θ, φ) Jerry Gilfoyle The Hydrogen Atom 7 / 18
ψEnlm(r, θ, φ) = Rnl(r)Y m
l (θ, φ)
= Anlρle−ρ kmax
bkρk
l (θ, φ)
bk+1 = 2(k + l + 1) − λe2 (k + 1)(k + 2l + 2)bk b0 = 1 En = −W ρ = κr κ =
2 λ =
2W a0 = 2 me2 ψEnlm = 2 na0 3 (n − l − 1)! 2n[(n + l)!]3 e−r/na0 2r na0 l (n + l)! L2l+1
n−l−1
2r na0
l (θ, φ) Jerry Gilfoyle The Hydrogen Atom 7 / 18
Jerry Gilfoyle The Hydrogen Atom 8 / 18
Jerry Gilfoyle The Hydrogen Atom 9 / 18
Red - l=0 10 20 30 40 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 r P Hydrogen Probability Density (n=4)
Jerry Gilfoyle The Hydrogen Atom 10 / 18
Red - l=0 Blue - l=1 10 20 30 40 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 r P Hydrogen Probability Density (n=4)
Jerry Gilfoyle The Hydrogen Atom 11 / 18
Red - l=0 Blue - l=1 Green - l=2 10 20 30 40 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 r P Hydrogen Probability Density (n=4)
Jerry Gilfoyle The Hydrogen Atom 12 / 18
Red - l=0 Blue - l=1 Green - l=2 Gray - l=3 10 20 30 40 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 r P Hydrogen Probability Density (n=4)
Jerry Gilfoyle The Hydrogen Atom 13 / 18
0.0 0.2 0.4 0.6 0.8 1.0 0.000 0.002 0.004 0.006 0.008 0.010 0.012 r (angstroms) Probability Density Red : n=1, Blue: n=4
Jerry Gilfoyle The Hydrogen Atom 14 / 18
Jerry Gilfoyle The Hydrogen Atom 15 / 18
Jerry Gilfoyle The Hydrogen Atom 15 / 18
How are these plots related to what we know?
Jerry Gilfoyle The Hydrogen Atom 15 / 18
Jerry Gilfoyle The Hydrogen Atom 16 / 18
En = −13.6 eV n2 Quantitative comparison for Balmer series hydrogen in units of σ. Line My Results (˚ A) NIST Results (˚ A) Normalized Percent Difference Difference α 6.64 ± 0.09 × 103 6.56280 × 103 0.95 1.2 β 4.85 ± 0.15 × 103 4.86133 × 103 0.11
γ 4.39 ± 0.06 × 103 4.34047 × 103 0.9 1.2 α : n = 3 → n = 2 β : n = 4 → n = 2 γ : n = 5 → n = 2
Jerry Gilfoyle The Hydrogen Atom 17 / 18
Jerry Gilfoyle The Hydrogen Atom 18 / 18