Physical Chemistry II Exam 2 Solutions

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1 Chemistry 362 Spring 2017 Dr Jean M Standard March 10, 2017 Name KEY Physical Chemistry II Exam 2 Solutions 1) (14 points) Use the potential energy and momentum operators for the harmonic oscillator to evaluate the commutator V, ˆ ˆpx Do the V ˆ and ˆp x operators commute? To evaluate the commutator, we determine the effects of operating it on a general function, V, ˆ ˆp x f (x) ( V ˆ ˆpx ˆp V ˆ ) f (x) x 1 2 kx2 i! d dx i! d dx 1 2 kx2 f (x) i!k 2 i!k 2 i!k 2 i!k 2 x 2 x 2 x 2 V, ˆ ˆp x f (x) i!k x f (x) d dx d dx x2 d dx f (x) f ʹ(x) Therefore, we see that the commutator V, ˆ ˆpx is given by f (x) d ( dx x2 f (x)) d ( dx x2 f (x)) ( x 2 f ʹ(x) 2x f (x) x 2 f ʹ(x) ) V, ˆ ˆp x i!kx Since the commutator is not equal to zero, the operators ˆ V and ˆp x do not commute

2 2) (14 points) Compare and contrast the harmonic and anharmonic oscillator models of vibrational motion for diatomic molecules You might want to focus on topics such as the shapes of the potential energy curves, the energy levels, and energy level spacings 2 The anharmonic oscillator is most closely representative of the real behavior of a bond as it is vibrating Most importantly, the anharmonic oscillator accounts for the dissociation of the bond as it is stretched The harmonic oscillator is a simplified model that works at low energies, but because its functional form is parabolic, V ( x ) 1 2 kx2, it cannot account for bond dissociation and so deviates from the true behavior of a chemical bond at higher energies and large bond displacements A comparison of an anharmonic oscillator vs a harmonic oscillator potential is shown below on the left Harmonic Osc V (cm -1 ) Energy (cm -1 ) x (Å) The energy levels of the harmonic oscillator are evenly spaced, with a separation equal to the harmonic vibrational frequency In contrast, the energy levels of an anharmonic oscillator are lower than those of the harmonic oscillator, and the difference increases as the energy increases The spacing between the anharmonic oscillator energy levels decreases as the energy increases These effects are illustrated in the figure above on the right

3 3) (15 points) The ground state wavefunction of the harmonic oscillator is ψ 0 (x) N e α x2 /2, where N α kinetic energy 1/4 and α µk! 2 For the ground state, determine the average value of the 3 The expectation value of the kinetic energy is defined as T ψ * x T ψ x ( ) ˆ ( ) dx The kinetic energy operator ˆ T is defined as d 2 T ˆ!2 dx 2 Substituting, and using the wavefunction for the ground state of the harmonic oscillator given above, the expectation value becomes T ψ * x!2 Evaluating the first derivative gives Evaluating the second derivative gives ( ) ˆ T ψ x ( ) dx e α x 2 /2 d 2 2 /2 dx dx 2 e α x d d α x 2 / 2 - αxe α x 2 / 2 d 2 dx 2 e α x 2 / 2 (-α + α 2 x 2 ) e α x 2 / 2 Substituting, and splitting the expression into two integrals yields T ψ * x!2 T α!2 ( ) ˆ T ψ x ( ) dx e α x 2 /2 ( α + α 2 x 2 ) e α x 2 /2 dx e α x 2 dx α 2! 2 x 2 e α x 2 dx

4 3) continued 4 From the table of integrals, e bx 2 dx 1 π 2 b 0 and x 2 e bx 2 dx 0 1 π 4b b Note that these integrals are given on the range [ 0,] For the range [,], the values of the integrals must be doubled since the integrands correspond to even functions Substituting, T α!2 α!2 e α x 2 dx π α α 2! 2 α 2! 2 1 π 2α α x 2 e α x 2 dx α!2 α!2 4µ T α!2 4µ This is sufficient as an answer However, the result may be simplified further by using the definition of the parameter α, T α!2 4µ µk! 2! k 4 µ!2 4µ 1 4 h k 2π µ T 1 4 hν 0

5 4) (15 points) Consider the molecule C 4 H 2 (diacetylene or 1,3-butadiyne), which has the linear structure shown below 5 H C C C C H a) How many vibrational modes does C 4 H 2 possess? For a linear molecule, we expect 3N 5 vibrational modes, where N is the number of atoms For C 4 H 2, N6, so # vibrational modes b) The following frequencies correspond to the only infrared active vibrational modes of C 4 H 2 : 630, 2020, and 3329 cm 1 The mode at 630 cm 1 is doubly degenerate, while the modes at 2020 and 3329 cm 1 are non-degenerate Assign these IR active modes in the table below with the corresponding types of vibrational motion (ie, things like symmetric or antisymmetric stretch or bend, etc) Frequency (cm 1 ) Degeneracy Assignment antisymmetric C-C-H bend antisymmetric C C stretch antisymmetric C-H stretch c) How many of the modes of C 4 H 2 are Raman active? Describe 2 distinct vibrational modes of C 4 H 2 that are Raman active and give their degeneracies Since there are only 4 IR active modes and the molecule has a center of symmetry, all the other modes must be Raman active Thus, there are Raman active modes The Raman active modes include: central C-C stretch (deg 1) symmetric C C stretch (deg 1) symmetric C-H stretch (deg 1) symmetric C-C-H bend (deg 2) symmetric C-C-C bend (deg 2) antisymmetric C-C-C bend (deg 2) (Listing any two of these is sufficient for the answer)

6 5) (14 points) The infrared spectrum of NaF exhibits two peaks, the fundamental v0 1 transition at 5284 cm 1 and the overtone v0 2 transition at cm 1 Treat the NaF molecule as an anharmonic oscillator, and use the infrared spectral data to determine the harmonic vibrational frequency in cm 1 ( ) and the anharmonicity constant ( ) 6 For an anharmonic oscillator, the vibrational energy in wavenumbers is given by ω v E v hc ( v + 1 2) ( v + 1 2) 2 The fundamental transition corresponds to v 0 v 1 This transition is ω 0 1 ω 1 ω 0 ω 3 e 2 ω ( ) x 3 e ( 2 ) 2 1 ( 2 ) x 1 e 2 ( ) 2 The overtone transition corresponds to v 0 v 2 This transition is ω 0 2 ω 2 ω 0 ω 5 e 2 ω ( ) x 5 e ( 2 ) 2 1 ( 2 ) x 1 e 2 ( ) 2 The numerical values of the v 0 v 1 and v 0 v 2 transitions are given above The unknown quantities are and : 5284 cm cm We can solve two equations for two unknowns in a variety of ways In this case, it is easy to solve the first equation for, Substituting this relation into the second equation, we have cm 1 2 6, or ( ) cm 1

7 5 continued 7 Now that is known, it can be substituted back into the first equation to obtain we, 5284 cm 1 2 2( 38 cm 1 ) 5360 cm 1 And finally, we can get as the ratio, 38 cm cm

8 6) (14 points) Consider a "quantum cube" of gold atoms 120 Å on each side Electrons inside the gold quantum cube are free to move throughout the cubic region inside, but are forbidden from moving outside the cubic region 8 a) Determine the energies (in Joules) of the ground and first excited states of an electron trapped in the gold quantum cube The energy levels of the particle in a 3D box are E nx,n y,n z n x 2 h 2 8mL 2 + n 2 y h 2 8mL + 2 n 2 z h 2 8mL 2 The lowest quantum number in each direction is 1, so substituting n x 1, n y 1, and n z 1 into the expression yields the ground state energy, E 111, E h 2 8mL h 2 8mL h 2 8mL 2, or E 111 3h 2 8mL 2 Substituting the box length, the ground state energy is E 111 3h 2 8mL 2 3( Js) 2 ( )( m) kg E J The first excited state corresponds to the (1,1,2) or (1,2,1) or (2,1,1) states; these are degenerate in energy Substituting n x 1, n y 1, and n z 2 into the energy expression, the energy of the first excited state is E h 2 8mL + 12 h 2 2 8mL + 22 h 2 2 8mL 2 6h 2 8mL 2 6( Js) 2 ( )( m) kg E J

9 9 6) continued b) Determine the wavelength for a transition from the ground to first excited state of the gold quantum cube For a transition from the ground state to the first excited state of the quantum cube, the energy difference ΔE is ΔE E 112 E 111 A photon with an energy corresponding to ΔE would have a frequency given by E photon ΔE hν Since, for light, λν c, we can substitute ν λ, and solve c for the wavelength to give λ hc ΔE hc E 112 E 111 Inserting numerical values leads to λ ( Js) ms J J ( ) ( ) λ m or λ 1580nm

10 7) (14 points) A diatomic molecule XH has a harmonic vibrational frequency in wavenumbers of cm 1 and a force constant given by k kg/s 2 What atom is X? 10 The relation for the harmonic vibrational frequency ν 0 is, ν 0 1 k 2π µ In this case, the harmonic frequency ν 0 and force constant k are known Thus, we can solve for the unknown reduced mass µ, µ k 4π 2 ν 0 2, or in terms of wavenumbers, µ k 4π 2 c 2 2 Substituting, µ ( kgs 2 ) 4π 2 ( cms 1 ) cm 1 ( ) 2 µ kg In units of amu, µ amu kg kg µ 08812amu The definition of the reduced mass for the XH molecule is µ m H m X m H + m X Solving for m X, m X µ m H m H µ

11 7) continued 11 Substituting the value calculated for µ and m H 10078amu, m X ( amu) ( 10078amu) ( 10078amu amu), m X 7015amu ; therefore, X Li