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p = h
λ= k
De Broglie:
∂2Ψ(x,t)∂t 2
= v2 ∂2Ψ(x,t)∂x2
Classical wave equa3on:
E = hν = ω
Planck/Einstein:
∂2
∂x2Ψ = −k2Ψ ⇒ −
2
2m∂2
∂x2Ψ =
p2
2m⎛⎝⎜
⎞⎠⎟Ψ
∂∂t
Ψ = −iωΨ ⇒ i ∂∂t
Ψ = EΨ
Schrödinger
Ψ(x,t) = Aei(kx−ω t )
∂∂xeax = aeax
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i ∂∂t
Ψ = EΨ =p2
2m+V
⎛⎝⎜
⎞⎠⎟Ψ
Time dependent Schrödinger equa3on
Time independent Schrödinger equa3on (standing wave solu0on)
Ψ(x,t) = Ψ(x)e− iEt /
i ∂∂t
Ψ = −2
2m∂2
∂x2+V
⎛⎝⎜
⎞⎠⎟Ψ
i ∂∂t
Ψ = HΨ
i ∂∂t
Ψ = EΨ
HΨ(x) = EΨ(x)
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−2
2m∇2 +V
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
V = −1r
H atom
V =0 0 < x < LC x < 0 or x > L
Par3cle in a box
V = 12 kx
2
Harmonic oscillator (vibra0onal spectroscopy)
∇2 =1r2
1sinθ
∂∂θsinθ ∂
∂θ+
1sin2θ
∂2
∂φ 2⎛⎝⎜
⎞⎠⎟
V = 0
Rigid Rotor (rota0onal spectroscopy)
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
Ψ2,0 =18π
1− r2
⎛⎝⎜
⎞⎠⎟e−r /2 2s
Ψ2,1 =1
4 2πxe−r /2 2p
P(x) = Ψ(x) 2 dxProbability
Ψ(x) 2
Probability density (amplitude)
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−2
2m∇2 +V
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
V = −1r
H atom
V =0 0 < x < LC x < 0 or x > L
Par3cle in a box
V = 12 kx
2
Harmonic oscillator (vibra0onal spectroscopy)
Ψ(x) 2
Probability density (amplitude)
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
Ψ2,0 =18π
1− r2
⎛⎝⎜
⎞⎠⎟e−r /2 2s
Ψ2,1 =1
4 2πxe−r /2 2p
P(x) = Ψ(x) 2 dxProbability
Ψ(x) 2Probability density
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End of video slides
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Please start your Socra0ve app
or go to m.socra3ve.com in your browser
Room number 9076
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
What is the most probable posi0on of an electron in the 1s orbital of H atom?
A inside the nucleus
B outside the nucleus
C don’t know
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
What is the most probable posi0on of an electron in the 1s orbital of H atom?
A inside the nucleus
B outside the nucleus
C don’t know
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
What is the most probable posi0on of an electron in the 1s orbital of H atom?
P(x) = Ψ(x) 2 dxProbability
Very small for nucleus
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Par0cle in a box Harmonic oscillator
Why is the probability density higher at the edges than in the middle for high energy solu0ons to the Schrödinger equa0on for the harmonic oscillator?
Enter your answer on Socra0ve
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
Ψ2,0 =18π
1− r2
⎛⎝⎜
⎞⎠⎟e−r /2 2s
What is the lowest excita3on energy of the H atom?
A 13.6 eV
B 10.2 eV
C 6.8 eV
D don’t know
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
Ψ2,0 =18π
1− r2
⎛⎝⎜
⎞⎠⎟e−r /2 2s
What is the lowest excita3on energy of the H atom?
A 13.6 eV
B 10.2 eV
C 6.8 eV
D don’t know
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−2
2m∇2 −
1r
⎛⎝⎜
⎞⎠⎟Ψn = EnΨn
En =−me4
22 4πε0( )2 n2= −
13.6 eVn2 n = 1,2,3,...
Ψ1 =1πe−r 1s
Ψ2,0 =18π
1− r2
⎛⎝⎜
⎞⎠⎟e−r /2 2s
What is the lowest excita3on energy of the H atom?
A 13.6 eV
B 10.2 eV
C 6.8 eV
D don’t know
ΔE = E2 − E1
=−13.6
4−−13.6
1= 10.2 eV
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-‐4.5 eV
+4.2 eV
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-‐4.5 eV
+4.2 eV
L = 2.94 nm
−2m
∇2 +V⎛⎝⎜
⎞⎠⎟Ψn = EnΨn V =
0 0 < x < LC x < 0 or x > L
Par3cle in a box
En =h2n2
8mL2n = 1,2,3...
E12 − E11 = (122 −112 ) h2
8mL2 = 1.60 ×10−19 J (1.0 eV)
Experiment = 2.5 eV (497 nm / blue green)
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Ψ(x) = 2L
⎛⎝⎜
⎞⎠⎟1/2
sin 2π xL
⎛⎝⎜
⎞⎠⎟
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Par3cle in a box: some useful predic3ons for nano sized systems
ΔE = En+1 − En = (2n +1)h2
8mL2
Excita0on energy (band gap) Increases with n
Decreases faster with L
ΔE decreases with molecular size
Absorp0on wave length (λ) increases with molecular size
λ =hcΔE
λ =8mch
L2
(2n +1)
= 3.30 ×1012 m−1( ) L2
(2n +1)
= 3.30 ×1012 m−1 ×1 m
109 nm⎛⎝⎜
⎞⎠⎟
L2
(2n +1)
= 3300 nm−1( ) L2
(2n +1)
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Where does 1,3,5-‐hexatriene absorb light? (you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 3300 nm−1( ) L2
(2n +1)
Based on
A ca 50 nm
B ca 100 nm
C ca 300 nm
D don’t know
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Where does 1,3,5-‐hexatriene absorb light? (you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 3300 nm−1( ) L2
(2n +1)
Based on
A ca 50 nm
B ca 100 nm
C ca 300 nm
D don’t know
λ = 3300 nm−1( ) L2
(2n +1)
= 3300 nm−1( ) 0.82
7= 3300 nm−1( ) 0.09( )= 302 nm
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Where does 1,3,5-‐hexatriene absorb light? (you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 3300 nm−1( ) L2
(2n +1)
Based on
λ = 3300 nm−1( ) L2
(2n +1)
= 3300 nm−1( ) 0.82
7= 3300 nm−1( ) 0.09( )= 302 nm
Experiment: 258 nm
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Where does 1,3,5-‐hexatriene absorb light? (you can use h[p://dgu.ki.ku.dk/molcalc/editor to compute orbital energies)
Based on
Orbital theory
A ca 50 nm
B ca 100 nm
C ca 300 nm
D don’t know
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Where does 1,3,5-‐hexatriene absorb light? (you can use h[p://dgu.ki.ku.dk/molcalc/editor to compute orbital energies)
Based on
Orbital theory
A ca 50 nm
B ca 100 nm
C ca 300 nm
D don’t know
λ =hcΔE
=1240 eV nm
(5.72 − (−6.28)) eV= 103 nm
Experiment: 258 nm
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m
As the length is increased for what value of m does trans-‐polyacetylene stop absorbing visible light?
(you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 3300 nm−1( ) L2
(2n +1)
Based on
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m
As the length is increased for what value of m does trans-‐polyacetylene stop absorbing visible light?
(you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 401 nm−1( ) L2
(2n +1)
Based on
1.044
= 0.260 nm/n
0.83
= 0.267 nm/n
L = 0.264n
λ = 3300 nm−1( ) 0.0697n2
(2n +1)
= 230 nm−1( ) n2
(2n +1)
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m
Visible light stops at about 800 nm
800 = 230 nm−1( ) n2
(2n +1)⇒ n ≈ 7 or 8
m ≈ 5 or 6
λ = 230 nm−1( ) n2
(2n +1)
λ
n
As the length is increased for what value of m does trans-‐polyacetylene stop absorbing visible light?
(you can use h[p://dgu.ki.ku.dk/molcalc/editor to es0mate length)
λ = 401 nm−1( ) L2
(2n +1)
Based on