Electron beam results Jerry Gilfoyle The Configurations of CO - - PowerPoint PPT Presentation

▶
electron beam results jerry gilfoyle the configurations
SMART_READER_LITE
LIVE PREVIEW

Electron beam results Jerry Gilfoyle The Configurations of CO - - PowerPoint PPT Presentation

Jan 26, 2024 366 likes •782 views

More On Carbon Monoxide E = 0 . 25 0 . 05 eV Electron beam results Jerry Gilfoyle The Configurations of CO 1 / 28 More On Carbon Monoxide E = 0 . 25 0 . 05 eV Electron beam results Jerry Gilfoyle The

slide-1
SLIDE 1

More On Carbon Monoxide

− →

Electron beam results ∆E = 0.25 ± 0.05 eV

Jerry Gilfoyle The Configurations of CO 1 / 28

slide-2
SLIDE 2

More On Carbon Monoxide

− →

Electron beam results ∆E = 0.25 ± 0.05 eV

Jerry Gilfoyle The Configurations of CO 1 / 28

slide-3
SLIDE 3

More On Carbon Monoxide

− →

Electron beam results ∆E = 0.25 ± 0.05 eV

− →

Jerry Gilfoyle The Configurations of CO 1 / 28

slide-4
SLIDE 4

Even More On Carbon Monoxide

Electron beam results ∆E = 0.25 ± 0.05 eV

→

Incident light CO gas target Photon detector

CO Absorption Spectrum

Jerry Gilfoyle The Configurations of CO 2 / 28

slide-5
SLIDE 5

Even More On Carbon Monoxide

Electron beam results ∆E = 0.25 ± 0.05 eV

→

Incident light CO gas target Photon detector

CO Absorption Spectrum

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 2 / 28

slide-6
SLIDE 6

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How do you get the expression above for the rotational energy? Is CO a rigid rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 3 / 28

slide-7
SLIDE 7

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How do you get the expression above for the rotational energy? Is CO a rigid rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 3 / 28

slide-8
SLIDE 8

The Plan

1 What is the kinetic and potential energy between the carbon and

  • xygen atoms in CO in the CM frame in cartesian and spherical

coordinates?

2 How do you decompose the kinetic energy into radial and angular

parts?

3 What is the Schroedinger equation for the rigid rotator? 4 What is the solution of the rigid rotator Schroedinger equation? Jerry Gilfoyle The Configurations of CO 4 / 28

slide-9
SLIDE 9

Coordinates r2 r1

1

r=r − r

2

R x y

Jerry Gilfoyle The Configurations of CO 5 / 28

slide-10
SLIDE 10

Coordinates r2 r1

1

r=r − r

2

R x y r1

1

r=r − r

2

R x y m1 m r2

2

cm

Jerry Gilfoyle The Configurations of CO 5 / 28

slide-11
SLIDE 11

Angular Momentum

α µ

Jerry Gilfoyle The Configurations of CO 6 / 28

slide-12
SLIDE 12

Angular Momentum

α µ

x y p

Τ

p

r

p

α

µ r

Jerry Gilfoyle The Configurations of CO 6 / 28

slide-13
SLIDE 13

Going To 3D

The Laplacian ∇2ψ = 1 r2 ∂ ∂r

  • r2 ∂

∂r

  • +

1 r2 sin θ ∂ ∂θ

  • sin θ ∂

∂θ

  • +

1 r2 sin2 θ ∂2 ∂φ2

  • ψ

Jerry Gilfoyle The Configurations of CO 7 / 28

slide-14
SLIDE 14

Going To 3D

The Laplacian ∇2ψ = 1 r2 ∂ ∂r

  • r2 ∂

∂r

  • +

1 r2 sin θ ∂ ∂θ

  • sin θ ∂

∂θ

  • +

1 r2 sin2 θ ∂2 ∂φ2

  • ψ

The Schroedinger Equation in 3D − 2 2µ∇2ψ + V (r)ψ = Eψ − 2 2µ 1 r2 ∂ ∂r

  • r2 ∂

∂r

  • +

1 r2 sin θ ∂ ∂θ

  • sin θ ∂

∂θ

  • +

1 r2 sin2 θ ∂2 ∂φ2

  • ψ

+ V (r)ψ = Eψ

Jerry Gilfoyle The Configurations of CO 7 / 28

slide-15
SLIDE 15

Going To 3D

Steps along the way. − 1 sin θ d dθ

  • sin θdΘ

dθ

  • +

m2

ℓ

sin2 θΘ = AΘ

Jerry Gilfoyle The Configurations of CO 8 / 28

slide-16
SLIDE 16

Going To 3D

Steps along the way. − 1 sin θ d dθ

  • sin θdΘ

dθ

  • +

m2

ℓ

sin2 θΘ = AΘ Legendre’s Associated Equation (1 − z2)d2Θ dz2 − 2z dΘ dz +

  • A −

m2

ℓ

1 − z2

  • Θ = 0

where z = cos θ

Jerry Gilfoyle The Configurations of CO 8 / 28

slide-17
SLIDE 17

Going To 3D

Steps along the way. − 1 sin θ d dθ

  • sin θdΘ

dθ

  • +

m2

ℓ

sin2 θΘ = AΘ Legendre’s Associated Equation (1 − z2)d2Θ dz2 − 2z dΘ dz +

  • A −

m2

ℓ

1 − z2

  • Θ = 0

where z = cos θ And its recursion relationship when mℓ = 0 ak+2 = k(k + 1) − A (k + 2)(k + 1)ak

Jerry Gilfoyle The Configurations of CO 8 / 28

slide-18
SLIDE 18

Going To 3D

We have the recursion relationship when mℓ = 0 ak+2 = k(k + 1) − A (k + 2)(k + 1)ak Notice. Given a0 → a2 → a4 · · · and given a1 → a3 → a5 · · · so Θ(z) =

∞

  • k=0

akzk =

∞

  • even

akzk +

∞

  • dd

akzk and we choose a0 = a1 = 1.

Jerry Gilfoyle The Configurations of CO 9 / 28

slide-19
SLIDE 19

Recall the Harmonic Oscillator Solution

Red Dashed - ⅇ-ξ2H50

2 (ξ)

Green Dashed - offset × ⅇξ2 ξ log(f(ξ))

ξ = βx β2 = mω0/

Jerry Gilfoyle The Configurations of CO 10 / 28

slide-20
SLIDE 20

Another Convergence Problem

20 40 60 80 100 0.0 0.5 1.0 1.5 2.0 2.5 kmax Θ(z) Truncated Calculation of Θ(z=1), ml=0

Θ(z = 1) = kmax

k=0 ak(1)k

ak+2 =

k(k+1)−A (k+2)(k+1)ak

Jerry Gilfoyle The Configurations of CO 11 / 28

slide-21
SLIDE 21

Another Convergence Problem

200 400 600 800 1000 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 kmax Θ(z) Truncated Calculation of Θ(z=1), ml=0

Θ(z = 1) = kmax

k=0 ak(1)k

ak+2 =

k(k+1)−A (k+2)(k+1)ak

Jerry Gilfoyle The Configurations of CO 12 / 28

slide-22
SLIDE 22

Another Convergence Problem

2000 4000 6000 8000 10000 1 2 3 4 5 kmax Θ(z) Truncated Calculation of Θ(z=1), ml=0

Θ(z = 1) = kmax

k=0 ak(1)k

ak+2 =

k(k+1)−A (k+2)(k+1)ak

Jerry Gilfoyle The Configurations of CO 13 / 28

slide-23
SLIDE 23

Another Convergence Problem

Blue - Truncated Calculation of Θ(z=1), ml=0 Green - ln(kmax)

2000 4000 6000 8000 10000 1 2 3 4 5 kmax Θ(z) Truncated Calculation of Θ(z=1), ml=0

Θ(z = 1) = kmax

k=0 ak(1)k

ak+2 =

k(k+1)−A (k+2)(k+1)ak

Jerry Gilfoyle The Configurations of CO 14 / 28

slide-24
SLIDE 24

Legendre Polynomials (mℓ = 0)

ak+2 = k(k + 1) − A (k + 2)(k + 1)ak mℓ = 0 a0 = a1 = 1 Θ = Pℓ(z) =

ℓ

  • even/odd

akzk z = cos θ First few polynomials. P0(cos θ) = 1 P3(cos θ) = 1

2

  • 5 cos3 θ − 3 cos θ
  • P1(cos θ) = cos θ

P4(cos θ) = 1

8

  • 35 cos4 θ − 30 cos2 θ + 3
  • P2(cos θ) = 1

2

  • 3 cos2 θ − 1
  • P5(cos θ) = 1

8

  • 63 cos5 θ − 70 cos3 θ + 15 cos θ
  • Jerry Gilfoyle

The Configurations of CO 15 / 28

slide-25
SLIDE 25

Spherical Harmonics (mℓ = m)

Θ(θ)Φ(φ) = Y m

ℓ (θ, φ) =

  • 2ℓ + 1

4π (ℓ − m)! (ℓ + m)!Pm

ℓ (cos θ)eimφ

Y 0

0 (θ, φ) =

1 √ 4π Y 1

1 (θ, φ) = −

  • 3

8π sin θeiφ Y −1

1

(θ, φ) =

  • 3

8π sin θe−iφ Y 0

1 (θ, φ) =

  • 3

4π cos θ

Jerry Gilfoyle The Configurations of CO 16 / 28

slide-26
SLIDE 26

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How do you get the expression above for the rotational energy? Is CO a rigid rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 17 / 28

slide-27
SLIDE 27

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How do you get the expression above for the rotational energy? Is CO a rigid rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 17 / 28

slide-28
SLIDE 28

Summary So Far

p2

r

2µ = − 2 2µ 1 r2 ∂ ∂r

  • r2 ∂

∂r

  • L2

2µr2 = − 2 2µ

  • 1

r2 sin θ ∂ ∂θ

  • sin θ ∂

∂θ

  • +

1 r2 sin2 θ ∂2 ∂φ2

  • mℓ = 0, ±1

2, ±2 2, ±3 2, ±4 2, ±5 2, ...

  • − 1

sin θ ∂ ∂θ

  • sin θ ∂

∂θ

  • +

m2

ℓ

sin2 θ

  • Θ = AΘ

A = ℓ(ℓ + 1) L2|φs = 2ℓ(ℓ + 1)|φs

Jerry Gilfoyle The Configurations of CO 18 / 28

slide-29
SLIDE 29

The Eigenvalues of ˆ L2 and Lz

Jerry Gilfoyle The Configurations of CO 19 / 28

slide-30
SLIDE 30

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by passing light through a cell containing CO and measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy can is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How does one

  • btain the expression above for the rotational energy? Is CO a rigid

rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 20 / 28

slide-31
SLIDE 31

Rotational Kinetic Energy

Axis at the CM.

Jerry Gilfoyle The Configurations of CO 21 / 28

slide-32
SLIDE 32

Rotational Kinetic Energy

Axis at the CM.

Jerry Gilfoyle The Configurations of CO 21 / 28

slide-33
SLIDE 33

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by passing light through a cell containing CO and measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How does one

  • btain the expression above for the rotational energy? Is CO a rigid

rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 22 / 28

slide-34
SLIDE 34

Is Carbon Monoxide A Rigid Rotator?

Excited states of carbon monoxide (CO) can be observed by passing light through a cell containing CO and measuring the absorption spectrum shown below. The molecule can both vibrate and rotate at the same time. The rotational energy states of a rigid rotator are Eℓ = 2 2I ℓ(ℓ + 1) where I is the moment of inertia. The vibrational part of the energy is described by the harmonic oscillator so En = (n + 1

2)ω0 with

∆E = ω0 = 0.25 ± 0.05 eV from our previous results. How does one

  • btain the expression above for the rotational energy? Is CO a rigid

rotator?

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

How far apart are the atoms?

Jerry Gilfoyle The Configurations of CO 22 / 28

slide-35
SLIDE 35

Even More On Carbon Monoxide

Electron beam results ∆E = 0.25 ± 0.05 eV

→

Incident light CO gas target Photon detector

CO Absorption Spectrum

Carbon−Monoxide Spectrum Absorption Energy (eV)

0.2650 0.2600 0.2550 0.2700 0.2750

Jerry Gilfoyle The Configurations of CO 23 / 28

slide-36
SLIDE 36

CO Vibration-Rotation Level Scheme

n=0 n=1 n=2

ℓ-value

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0.00 0.02 0.04 0.06 0.08 E (eV) CO Rotator Level Scheme

Enl = (n + 1

2)ω0 + 2 2I ℓ(ℓ + 1)

∆En = ω0 = 0.025 ± 0.05 eV ∆Eℓ = 2

I = 0.44 ± 0.07 meV

Jerry Gilfoyle The Configurations of CO 24 / 28

slide-37
SLIDE 37

CO Vibration-Rotation Transitions

l=5 l=5 n=0 n=1

Jerry Gilfoyle The Configurations of CO 25 / 28

slide-38
SLIDE 38

CO Vibration-Rotation Transitions

l=5 l=5 n=0 n=1 l=5 n=0 n=1 l=6 l=6 l=7

Vibration−Rotation Adjacent Initial Angular Momenta l+1 l

Jerry Gilfoyle The Configurations of CO 26 / 28

slide-39
SLIDE 39

CO Vibration-Rotation Transitions

l=5 l=5 n=0 n=1 l=5 l=5 n=0 n=1 l=6 l=6 l=7 l=4

Vibration−Rotation Adjacent Initial Angular Momenta

Jerry Gilfoyle The Configurations of CO 27 / 28

slide-40
SLIDE 40

Carbon Monoxide Rotation Spectrum

ΔE Probability Rotational Spectra

Jerry Gilfoyle The Configurations of CO 28 / 28