Peer instructions questions for basic quantum mechanics
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Transcript of Peer instructions questions for basic quantum mechanics
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
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)
−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)
−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)
−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)
−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
End of video slides
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Room number 9076
−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
−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
−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
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?
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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
−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
−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
-‐4.5 eV
+4.2 eV
-‐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)
Ψ(x) = 2L
⎛⎝⎜
⎞⎠⎟1/2
sin 2π xL
⎛⎝⎜
⎞⎠⎟
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)
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
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
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
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
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
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
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)
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