Fun With Carbon Monoxide. p. 1/2

Transcription

1 Fun With Carbon Monoxide p. 1/2

2 p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results

3 p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results C V (J/K-mole) Specific Heats of Gases SO 2 CO 2CH4 H 2 O Cl 2 H 2 N 2 O 2 CO He Ar Ne Kr 5 0 Molecule

4 p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results C V (J/K-mole) Specific Heats of Gases SO 2 CO 2CH4 H 2 O Cl 2 H 2 N 2 O 2 CO He Ar Ne Kr 5 0 Molecule

5 p. 2/2 Carbon Monoxide Electron beam results E = 0.25 ± 0.05 ev CO Absorption Spectrum Incident light Photon detector CO gas target

6 p. 2/2 Carbon Monoxide Electron beam results E = 0.25 ± 0.05 ev CO Absorption Spectrum Incident light Photon detector CO gas target Absorption Carbon Monoxide Spectrum Energy (ev)

7 p. 3/2 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 E l = 2 l(l + 1) 2I where I is the moment of inertia. The vibrational part of the energy can is described by the harmonic oscillator so E n = (n ) ω 0 with E = ω 0 = 0.25 ± 0.05 ev. How does one arrive at the expression above for the rotational energy? Is CO a rigid rotator? Absorption Carbon Monoxide Spectrum Energy (ev)

8 p. 4/2 The Plan 1. What is the kinetic and potential energy between the carbon and oxygen 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?

9 p. 5/2 Coordinates y r=r r 1 2 r 1 R x r 2

10 p. 5/2 Coordinates y m 1 r=r r 2 1 r 1 R cm m r 2 2 x

11 Angular Momentum p. 6/2

12 p. 6/2 Angular Momentum y p rel p θ φ m p r r x

13 p. 7/2 A Convergence Problem Truncated Calculation of z 1 z k max

14 p. 7/2 A Convergence Problem z Truncated Calculation of z k max

15 p. 8/2 Legendre Polynomials (m l = 0) Θ(θ) = P l (cosθ) P 0 (cosθ) = 1 P 1 (cosθ) = cos θ P 2 (cosθ) = 1 2 ( 3 cos 2 θ 1 ) P 3 (cosθ) = 1 2 ( 5 cos 3 θ 3 cos θ )

16 p. 9/2 Spherical Harmonics (m l = m) Θ(θ)Φ(φ) = Y m l (θ,φ) = 2l + 1 4π (l m)! (l + m)! P m l (cosθ)e imφ Y 0 0 (θ,φ) = 1 4π 3 Y1 1 (θ,φ) = 8π sin θeiφ Y 1 1 (θ,φ) = 3 8π sin θe iφ Y 0 1 (θ,φ) = 3 4π cosθ

17 p. 10/2 Summary So Far L 2 2µr 2 = 2 2µ p 2 r 2µ = 2 2µ [ 1 r 2 sin θ θ 1 r 2 r ( r 2 ) r ( sin θ ) θ + 1 r 2 sin 2 θ 2 ] φ 2 m l = 0, ± 1 2, ±2 2, ±3 2, ±4 2, ±5 2,... [ 1 sin θ θ ( sin θ ) θ ] + m2 l sin 2 Θ = AΘ A = l(l + 1) θ L 2 φ s = 2 l(l + 1) φ s

18 p. 11/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk

19 p. 11/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk

20 p. 11/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk = [( y d i dz z d ) ( î + z d dy dx x d ) ( ĵ + x d dz dy y d ) dx ] ˆk

21 p. 11/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk = [( y d i dz z d ) ( î + z d dy dx x d ) ( ĵ + x d dz dy y d ) dx ] ˆk Transformation from Cartesian to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cosθ

22 p. 12/2 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 E l = 2 l(l + 1) 2I where I is the moment of inertia. The vibrational part of the energy can is described by the harmonic oscillator so E n = (n ) ω 0 with E = ω 0 = 0.25 ± 0.05 ev from our previous results. How does one arrive at the expression above for the rotational energy? Is CO a rigid rotator? Absorption Carbon Monoxide Spectrum Energy (ev)

23 p. 13/2 Summary So Far L 2 2µr 2 = 2 2µ p 2 r 2µ = 2 2µ [ 1 r 2 sin θ θ 1 r 2 r ( r 2 ) r ( sin θ ) θ + 1 r 2 sin 2 θ 2 ] φ 2 m l = 0, ± 1 2, ±2 2, ±3 2,... Φ(φ) = Φ 0e ±im lφ [ 1 sinθ θ ( sin θ ) θ ] + m2 l sin 2 Θ = AΘ A = l(l + 1) θ L 2 φ = 2 l(l + 1) φ L z φ = ±m l 2 φ Θ(θ)Φ(φ) = Y lm (θ,φ)

24 p. 14/2 Rotational Kinetic Energy Axis at the CM.

25 p. 14/2 Rotational Kinetic Energy Axis at the CM.

26 Angular Momentum p. 15/2

27 p. 15/2 Angular Momentum y p rel p θ φ m p r r x

28 p. 16/2 Is CO a Rgid Rotator? Electron beam results E = 0.25 ± 0.05 ev CO Absorption Spectrum Incident light Photon detector CO gas target Absorption Carbon Monoxide Spectrum Energy (ev)

29 p. 17/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk

30 p. 17/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk

31 p. 17/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk = [( y d i dz z d ) ( î + z d dy dx x d ) ( ĵ + x d dz dy y d ) dx ] ˆk

32 p. 17/2 The Eigenvalues of ˆL 2 and ˆL z L = r p = L x î + L y ĵ + L zˆk = (yp z zp y )î + (zp x xp z )ĵ + (xp y yp x ) ˆk = [( y d i dz z d ) ( î + z d dy dx x d ) ( ĵ + x d dz dy y d ) dx ] ˆk Transformation from Cartesian to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cosθ

33 The Eigenvalues of ˆL 2 and L z p. 18/2

34 p. 19/2 CO Atomic Transitions n=1 l=5 l=5 n=0

35 p. 20/2 Carbon Monoxide Rotation Spectrum Rotational Spectra Probability E