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Chem 4502 & Spectroscopy (Jason Goodpaster) Main Topics

Exam 2 Review in a box 1. Derivation of energy levels and wave functions of the Schrödinger equation for the 1-D PIB Homework 4: 2. Comparing the predictions of the S. eqn for the PIB to Particle-in-a-box (in Chapter 4, pp. 80 - 95) classical predictions; Correspondence principle Probability and Statistics (MathChapter B, pp. 63 - 70) 3. Use of the "free " model to predict the energies of the π* ← π absorption (or π* → π emission) Homework 5: electronic transitions in linear polyenes, whose Harmonic oscillator (Chapter 5, Sections 1-7, pp. 157-173) occupy delocalized orbitals; predicting the Postulates and principles of quantum mechanics (Chapter 4, pp. 115 - 133) wavelength (or frequency or cm-1) of light absorbed or emitted in these transitions. The exam (like exam 1), will be made up of multiple choice, short and long answer.

Last names beginning A - M Akerman 225; N – Z Tate 110.

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Main Topics Main Topics

Harmonic Oscillator: Postulates of Quantum Mechanics: 1. Quantum mechanical HO, a model for vibrations of atoms 1. contains all the information available about in molecules a system 2. Setting up the Schrödinger equation 2. Single ideal measurements of a property corresponding 3. Showing that a Gaussian function solves this equation and to a given operator can observe only eigenvalues of that calculating the corresponding energy operator 4. Discussion of this wave function (zero point 3. Orthonormal wave functions vibrational energy, most probable displacement)

Particle-in-a-Box (1 Dimension) Particle-in-a-Box (1 Dimension)

∞ ∞ U(x) is the particles . Solve the S. equation to obtain Ψ(x) inside the box. 0 !2 ∂2 We assume U = ∞ for x < 0 and x > a − ⎡Ψ(x)⎤+U (x)Ψ(x) = EΨ(x) U 2m x2 ⎣ ⎦ so particle cant get out of box ∂ 0 0 a Define U = 0 for 0 ≤ x ≤ a ∂2 ⎛ 2mE ⎞ 2 x ⎡ (x)⎤ (x) 0 ∂ Ψ(x) 2 2 ⎣Ψ ⎦+⎜ 2 ⎟Ψ = − k Ψ(x) = 0 ∂x ⎝ ! ⎠ ∂x2

What is Ψ(x), the wave function for the particle, Solutions: for x < 0 and x > a ? (x) Acos(kx) Bsin(kx) Ψ = 0, since the particle cant be found outside the box. Ψ = +

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Particle-in-a-Box (1 Dimension) ∞ ∞ Particle-in-a-Box (1 Dimension) ∞ ∞

We had: Ψ(x) = Acos(kx) + Bsin(kx) U We had: Ψ(x) = Acos(kx) + Bsin(kx) U Ψ(x) must be single-valued and continuous. A = 0 k a = n π 0 0 0 a 0 a Apply boundary conditions: x x Ψ(x) = B sin (k x) = B sin (n π x / a) Ψ(x)=0 at x=0 and x=a Wave functions Normalization constant Ψ(0)=0 = A cos 0 + B sin 0 = A(1) + B(0) = A so A = 0 E = n2 h2 / (8 m a2) n = 1, 2, 3.... quantum # Ψ(a)=0 = 0 + B sin ka so sin ka = 0 Why not start with n=0 ? Ψ(x) = B sin (0) = 0 k a = n π recall k2 = 2mE / ℏ2 so k = √2mE / ℏ This would mean that there is no particle in the box. √2mE a / = n π E = n2 2 π2 / (2 m a2) Why not have negative quantum numbers?

Ψn = -Ψ same wavefunction, same energy E = n2 h2 / (8 m a2) n = 1, 2, 3.... quantum # -n arises naturally upon solving S. eqn. Energy quantization with boundary conditions 7 8 2 2 2 2 2 2 En = n h / (8 ma ) Particle-in-a-Box Ψn = B sin (n π x / a) En = n h / (8 ma ) Particle-in-a-Box Ψn = B sin (n π x / a)

Normalization: Evaluate B Normalization: Evaluate B 2 Ψn (x)dx =1 2 ∫ Recall: Ψ dx gives the probability of finding the particle a All Space 2 2 ⎛ nπ x ⎞ between x and x+dx (for real Ψ). B ∫ sin ⎜ ⎟dx =1 0 ⎝ a ⎠ Lets say a probability of 0 means no chance Problem 3-10 (not assigned) 2 ⎛ a ⎞ 1 its certain to be there B ⎜ ⎟ =1 ⎝ 2⎠

In each state (with its quantum number, n), 2 ⎛ 2⎞ 2 B is the normalization B = ⎜ ⎟ B = the particle is certain to be somewhere : ⎝ a ⎠ a constant So, the normalized wave functions for the 1-D PIB are: Ψ2 (x)dx =1 Normalization ∫ n condition 2 ⎛ nπ x ⎞ All Space Ψ = sin⎜ ⎟ n a a 9 ⎝ ⎠ 10

Particle in a Cube (Fig. 3.6 p. 94) Multiple Choice E = h2 (n 2 + n 2 + n 2) nxnynz x y x 1. When we solved the Schrödinger equation for the particle 8 m a2 in a one-dimensional box, the quantization of energy levels resulted when we required that the wave functions:

A. be normalized B. be finite for all possible values of x C. have n-1 nodes, where n is an integer D. be zero at the walls E. be mutually orthogonal

11 Multiple Choice Multiple Choice

1. When we solved the Schrödinger equation for the particle What is the potential energy operator in the Schrödinger in a one-dimensional box, the quantization of energy levels equation for the harmonic oscillator? resulted when we required that the wave functions: A. zero A. be normalized B. ½ kx2 B. be finite for all possible values of x C. (1/2π) (k/µ)½ C. have n-1 nodes, where n is an integer D. -ħ2/(2µ) (d2/dx2) D. be zero at the walls E. (v + ½) h ν E. be mutually orthogonal F. (µ k / ħ )½

Multiple Choice Multiple Choice

What is the potential energy operator in the Schrödinger 8. In view of the selection rules for vibrational equation for the harmonic oscillator? spectroscopy, which of the following molecules are not expected to absorb infrared light? Choose the best answer. A. zero 2 B. ½ kx N2, H2O, CO2 C. (1/2π) (k/µ)½ 2 2 2 D. -ħ /(2µ) (d /dx ) A. N2 only E. (v + ½) h ν B. H2O only ½ F. (µ k / ħ ) C. CO2 only

D. N2 and H2O E. N2 and CO2 F. H2O and CO2 G. All 3 molecules will absorb infrared light H. None of these molecules will absorb infrared light

Multiple Choice Short Answer

8. In view of the selection rules for vibrational For butadiene (H2C=CH–CH=CH2), which has four π electrons, some of the electronic transitions can be approximated using the particle-in-a-box model. If spectroscopy, which of the following molecules are not the length of the molecule is 6.0 Å (that is, 6.0 x 10-10 m), what photon energy is expected to absorb infrared light? Choose the best answer. required to promote an electron from the π HOMO (highest occupied molecular orbital) to the π* LUMO (lowest unoccupied molecular orbital)?

N , H O, CO 2 2 2 2 2 2 En = n h / (8ma ) −34 −31 h = 6.626 x 10 J·s me = 9.109 x 10 kg

A. N2 only

B. H2O only C. CO2 only D. N2 and H2O E. N2 and CO2 F. H2O and CO2 G. All 3 molecules will absorb infrared light H. None of these molecules will absorb infrared light

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Short Answer Short Answer

35 For butadiene (H2C=CH–CH=CH2), which has four π electrons, some of the The constant of D Cl is about 480 N/m. Calculate its vibrational electronic transitions can be approximated using the particle-in-a-box model. If frequency, in units of cm-1 (give your answer to 2 or more significant figures). the length of the molecule is 6.0 Å (that is, 6.0 x 10-10 m), what photon energy is Note: D, deuterium, is the 2.0 amu isotope of hydrogen (having one neutron). required to promote an electron from the π HOMO (highest occupied molecular orbital) to the π* LUMO (lowest unoccupied molecular orbital)? h = 6.626 x 10−34 J·s amu = 1.661 x 10-27 kg ½ Ev = (v+½) hνo where νo = (1 / (2π)) (k/µ) and µ = m1m2 / (m1+m2) 2 2 2 En = n h / (8ma ) (diatomic) −34 −31 h = 6.626 x 10 J·s me = 9.109 x 10 kg

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35 The force constant of D Cl is about 480 N/m. Calculate its vibrational The line below represents the "x" axis for the harmonic oscillator, with the frequency, in units of cm-1 (give your answer to 2 or more significant figures). classical turning points labeled as +xTP and -xTP. On this line, draw the square Note: D, deuterium, is the 2.0 amu isotope of hydrogen (having one neutron). of the wave function for the harmonic oscillator in its v = 1 state.

h = 6.626 x 10−34 J·s amu = 1.661 x 10-27 kg ½ Ev = (v+½) hνo where νo = (1 / (2π)) (k/µ) and µ = m1m2 / (m1+m2) (diatomic)

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Short Answer Short Answer

The line below represents the "x" axis for the harmonic oscillator, with the The line below represents the "x" axis for the harmonic oscillator, with the classical turning points labeled as +xTP and -xTP. On this line, draw the square classical turning points labeled as +xTP and -xTP. On this line, draw the square of the wave function for the harmonic oscillator in its v = 1 state. of the wave function for the harmonic oscillator in its v = 1 state. Longer Answer Longer Answer

For a quantum harmonic oscillator of m, show that: For a quantum harmonic oscillator of mass m, show that:

−αx2 /2 mk −αx2 /2 mk f (x) = xe α = f (x) = xe α = !2 !2 Is an eigenfunction of the Hamiltonian for this system. Give the Is an eigenfunction of the Hamiltonian for this system. Give the eigenvalue. eigenvalue.

Longer Answer

Assuming the harmonic oscillator approximation, use Postulate #4 (which tells us how to calculate an average or expectation value) to calculation the root mean square displacement:

1 x2 2

For a molecular in its zero point vibrational level. You can leave the answer is terms of a.

1 4 ⎛ a ⎞ ax2 /2 Ψ = ⎜ ⎟ e− ⎝π ⎠ ∞ 1 π 2 ax2 = ∫ x e− 2a a −∞ 27 Longer Answer

2 1 ∞ ⎛ 4 ⎞ 2 2 ⎛ a ⎞ ax2 /2 x = x ⎜⎜ ⎟ e− ⎟ dx ∫ ⎜ π ⎟ −∞ ⎝⎝ ⎠ ⎠ 1 2 ∞ 2 ⎛ a ⎞ 2 ax2 x = ⎜ ⎟ ∫ x (e− )dx ⎝π ⎠ −∞

2 a 1 π 1 x = = π 2a a 2a 1/2 1 x2 = 2a