2m dx 2. The particle in a one dimensional box (of size L) energy levels are

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1 Name: Chem 3322 test #1 solutions, out of 40 marks I want complete, detailed answers to the questions. Show all your work to get full credit. indefinite integral : sin 2 (ax)dx = x 2 sin(2ax) 4a (1) with The Schrödinger time dependent equation is, for ψ(x, t) ψ(x, t) i t The Schrödinger time independent equation is, for ψ(x) = 2 2 ψ(x, t) + V (x)ψ(x, t) (2) 2m x 2 2 d 2 ψ(x) + V (x)ψ(x) = Eψ(x) (3) 2m dx 2 The particle in a one dimensional box (of size L) energy levels are E n = n2 π 2 2 2mL 2 n = 1, 2, 3,... (4) The particle in a two-dimensional box (of side lengths a and b) energy levels are E = π2 2 2m ( ) n 2 x a + n2 y 2 b 2 ground state wavefunction of the harmonic oscillator: ψ(x) = n x, n y = 1, 2, 3,... (5) ( ) 1/4 2α e αx2 (6) π α = mω 2 and ω = (k/m) 1/2 (7) where k is the force constant and m is the mass. energy levels of the harmonic oscillator: E n = (n + 1/2) ω, n = 0, 1, 2,... energy levels of the particle on a ring: Speed of light = c = m/s E n = n2 2 2mR 2 n = 0, 1, 2,... (8) electron mass = kg. proton mass = kg. 1

2 Avogadro s constant = N A = mol 1 Planck constants: h = Js and = Js π = ev = J (electron volt to joule conversion) mass conversion from amu to kg: kg/amu Relationship between the wavelength of a photon and its energy λ = hc/e 2

3 Problem 1 7 marks a) 4 marks Define: classically forbidden region (of space). Illustrate your answer with a sketch. FIG. 1: Classically forbidden regions in the particle in a finite box model (left side); classically forbidden regions in the harmonic oscillator model (right side). A region of space where the potential energy is greater than the particle energy. b) 3 marks Under what circumstances (if any) does a quantum mechanical particle exist in a classically forbidden region of space? If the potential energy is not infinity, in the classically forbidden region the probability density decays exponentially or faster than exponentially. This is the behavior predicted by the Schroedinger Equation. This phenomenon leads to the concept of tunneling in which quantum mechanical particles can go through barriers. This is seen in certain proton or electron transfer reactions in chemistry and biochemistry, and it also explains how the scanning tunneling microscope functions. 3

4 Problem 2 14 marks a) 3 marks Sketch the ground state wavefunction for the particle in a box problem across all of space ( < x < ). FIG. 2: Ground state wavefunction for the particle in a box. b) 2 marks Identify the cusp behavior at x = 0 by evaluating the slope of the wavefunction at x = 0 approaching from the left, and separately approaching from the right. Approaching from negative x values, the wavefunction is identically zero and so the slope is zero at x = 0. Approaching from positive x values, the wavefunction is 2/L sin πx/l making the derivative 2/L(π/L) cos πx/l; thus at x = 0 the slope is 2/L(π/L). Since these two slopes are different, we have identified a cusp at x = 0. c) 2 marks In equation (3) for the particle in a box problem, state what the function V (x) is across all of space ( < x < ). V (x) is 0 for x (0, L) and V (x) is for x < 0 and x > L. d) 4 marks Given the behavior identified in part b, and putting aside the answer to part c, comment on the validity of the ground state wavefunction as being a solution to equation (3). The ground state wavefunction cannot solve equation (3) because equation (3) is a differential equation, however the wavefunction is not differentiable. 4

5 e) 3 marks In light of the answer to part c, comment further on the validity of the ground state wavefunction as being a solution to equation (3). There are at least two ways to answer this. First, we can argue that V (x), since it switches from 0 to, is pathological and so if we wish to find a solution to equation (3) we must relax our criteria. Second, we can argue that the wavefunction must be zero in the V = regions of space, and since this gives us two boundary conditions (at x = 0 and x = L) this uniquely yields a solution. 5

6 Problem 3 6 marks a) 4 marks Define the term zero point energy in quantum mechanics and give some examples from class. The zero point energy is the lowest energy a particle can have among its stable (or stationary) states. In quantum mechanics, it is not necessarily zero. The particle may have some residual energy which, however, cannot be used to do work. In this case, even at absolute zero, the particle is in motion. From class, the particle in a box, the harmonic oscillator, and the Morse oscillator (and the hydrogen atom) have some residual energy, whereas the particle on a ring does not. b) 2 marks Does this concept exist for a particle obeying the laws of classical mechanics? Discuss. Classically, the zero point energy is zero (the particle is at rest at its potential energy minimum). Therefore, we could equally well argue that this concept doesn t exist for a classical particle. 6

Problem 4 8 marks β-carotene is a linear conjugated molecule in which 10 C-C single bonds and 11 C=C double bonds alternate along a chain of 22 carbon atoms.

7 Problem 4 8 marks β-carotene is a linear conjugated molecule in which 10 C-C single bonds and 11 C=C double bonds alternate along a chain of 22 carbon atoms. The π electrons from these bonds are assumed to behave like particles in a box. FIG. 3: β-carotene molecule a 1 mark) If we take each C=C double bond length to be 130 pm and each C-C single bond length to be 151 pm, show that the length of β-carotene available to the delocalized π electrons can be estimated as L = 2.94 nm. 11(130) + 10(151) = 2940 pm = 2.94 nm. b 3 marks) Using the fact that each carbon atom in the conjugated system contributes one π-electron, draw the energy level diagram for this system, showing which levels are occupied by electrons and which are vacant. Label the HOMO (highest occupied molecular orbital) and the LUMO (lowest unoccupied molecular orbital) energy levels. The particle in a box model has n = 1 has the lowest level and no degeneracies. Therefore the HOMO is n = 11 and the LUMO is n = 12. c 4 marks) Determine the wavelength of light (in nm) corresponding to the lowest energy absorption (first electronic transition). E = E 12 E 11 = ( )π 2 ( Js) 2 (2)( kg)( m) 2 (9) Then use λ = hc/ E to get λ = nm. 7

8 Problem 5 5 marks Compare and contrast the harmonic oscillator and the Morse oscillator models. Give at least two differences between them. The Morse oscillator is an accurate (yet simple) model for the quantized vibrational energy in a chemical bond. It allows the bond to dissociate at high energy. The harmonic oscillator is obtained by a Taylor (power series) expansion of the Morse oscillator up to second order, with the expansion being centered at the equilibrium bond length. The HO does not allow the bond to dissociate. Also, the HO allows the bond to compress too much because a parabola does not have vertical asymptotes. In terms of the energy levels, the ground state for both models are the same (each having a zero point energy), but the HO has equally spaced levels whereas in the Morse model the energy levels get closer together as the energy increases. In terms of the wavefunctions, the HO wavefunctions are symmetrical about the equilibrium bond length, whereas at higher energy the Morse wavefunctions are skewed towards the bond being stretched. (In addition, a sketch never hurts.) 8

Physics 43 Exam 2 Spring 2018

Physics 43 Exam 2 Spring 2018 Print Name: Conceptual Circle the best answer. (2 points each) 1. Quantum physics agrees with the classical physics limit when a. the total angular momentum is a small multiple