Physical Chemistry II Exam 2 Solutions

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1 Chemistry 362 Spring 208 Dr Jean M Standard March 9, 208 Name KEY Physical Chemistry II Exam 2 Solutions ) (4 points) The harmonic vibrational frequency (in wavenumbers) of LiH is 4057 cm Based upon this information, predict the harmonic vibrational frequency of LiD in wavenumbers By definition, the harmonic frequency ν 0 is given by ν 0 $ k ' & ) 2π % µ ( / 2 Therefore, the ratio of frequencies can be written ν 0 (LiH) ν 0 (LiD) 2π 2π k LiH µ LiH k LiD µ LiD Assuming that k LiH k LiD, this equation simplifies to ν 0 (LiH) ν 0 (LiD) µ LiD µ LiH Using the isotopic masses m H 0078 amu, m D 2040 amu, and m Li 7060amu leads to the reduced mass values of µ LiH 08822amu and µ LiD 5648amu Substituting into the harmonic frequency ratio leads to ν 0 (LiH) ν 0 (LiD) ν 0 (LiH) ν 0 (LiD) 5648amu amu 3326 Solving for the unknown frequency of LiD yields, ν 0 (LiD) ν 0 (LiH) 3326

2 ) continued 2 The harmonic frequency we in wavenumbers is related to the harmonic frequency in Hz by the relation, ω e ν 0 c, or, solving for the harmonic frequency, ν 0 ω e c Substituting, the LiD frequency in wavenumbers is given by ω e (LiD)c ω e (LiH) c 3326, or ω e (LiD) ω e (LiH) cm 3326 ω e (LiD) 0549 cm

3 2) (5 points) 3 a) In an experiment, buckyballs (C 60, with mass kg) are accelerated to a velocity of 320 m/s with an uncertainty of 60 m/s Determine the uncertainty in the position Δx of the buckyballs in Å From the Heisenberg Uncertainty Principle, Δp x Δx! Solving for Δx, 2 Δx! 2Δp x Assuming there is no uncertainty in m (or that the uncertainty is very small compared to the uncertainty in p x ), we get the relation Δp x m Δv Substituting the mass along with uncertainty in the velocity into this relation, we get the uncertainty in the momentum, Δp x mδv Substituting into the uncertainty relation, ( ) 60m/s kg Δp x kgm/s Δx Δx! 2Δp x ( Js) ( ) kgm/s ( ) Δx m, or Å

4 2) continued 4 b) If instead of the situation in part (a), the position of one of the buckyballs is known exactly such that Δx 0, what restrictions are there (if any) on knowledge of the momentum of the buckyball? Explain From the Heisenberg Uncertainty Principle, Δp x Δx! 2 Solving for Δp x, Δp x! 2Δx Substituting Δx 0, Δp x! 0, or Δp x This result tells us that if we know the position exactly, then we know nothing about the momentum It is completely uncertain

5 3) (4 points) Consider the second excited state of the harmonic oscillator, with wavefunction given by 5 ψ 2 (x) N 2 ( 4αx 2 2) e α x2 /2, and N 2 α 64π a) Make a sketch of the second excited state wavefunction of the harmonic oscillator How many nodes does the function have? The second excited state wavefunction is shown below It possesses 2 nodes /4 b) Does the second excited state wavefunction possess even or odd symmetry? It has even symmetry, since f ( x) f ( x) c) Determine whether or not the second excited state of the harmonic oscillator is an eigenfunction of the linear momentum operator, ˆp x If it is an eigenfunction, give the eigenvalue In order to show that the function is an eigenfunction of the momentum operator, we start by operating ˆp x on it, ˆp x ψ 2 (x) ( ) e α x2 /2 i! d dx N 2 4αx 2 2

6 3 c) continued 6 Evaluating the first derivative (using the product and chain rules), d dx ( 4αx 2 2) e α x2 /2 { } 8αx Substituting into the eigenvalue equation, ( ) e α x2 /2 ( 0αx 4α 2 x 3 )e α x2 /2 + ( 4αx 2 2) ( αx)e α x2 /2 ˆp x ψ 2 (x) ( ) e α x2 /2 i! d dx N 2 4αx 2 2 { } i!n 2 ( 0αx 4α 2 x 3 )e α x2 /2 This result does not equal a constant times the original function; therefore, ψ 2 (x) is not an eigenfunction of the momentum operator

7 7 4) (5 points) Calculate the vibrational frequency (in cm ) that yields a population ratio N / N 0 equal to 0 (ie, 0%) at 300 K using the harmonic oscillator model The population of the v level relative to the v0 population ( N / N 0 ) is given by the Boltzmann factor, N N 0 exp { ΔE / k B T} For a harmonic oscillator, the energy difference between adjacent levels is given by ΔE E v+ E v hν 0 (v + 3 / 2) hν 0 (v +/ 2) ΔE hν 0 Thus the population ratio is given by N N 0 exp { hν 0 / k B T} Setting the population ratio equal to 0 yields the relation, 0 exp { hν 0 / k B T} To solve for the harmonic frequency, we need to take the natural log of both sides, ( ) ln exp { hν 0 / k B T} ln 0 Solving for the harmonic frequency gives, Substituting, ln( 0) hν 0 / k B T ν 0 k BT ln( 0) h ν 0 k BT ln( 0) h ( ) 300 K J/K ( )ln Js ν s ( )

8 4) continued 8 Converting the harmonic frequency to units of cm gives ω e ν 0 c s cm/s ω e 4453cm

9 5) (4 points) The ground state wavefunction of the harmonic oscillator is given by ψ 0 ( x ) α π /4 e α x2 /2, where α µk the harmonic oscillator in its ground state! 2 Determine the average value of the potential energy for 9 The expectation value of the potential energy is defined as V ψ * x V ψ x ( ) ˆ ( ) dx The potential energy operator ˆ V is defined as ˆ V 2 kx2 Substituting, and using the wavefunction for the ground state of the harmonic oscillator given above, the expectation value becomes V ψ * x ( ) ˆ 2 k α π V ψ x ( ) dx e α x2 /2 x 2 e α x2 /2 dx Combining the exponentials yields V 2 k α π x 2 α x2 e dx From the table of useful integrals, we have x 2 e bx2 dx 0 4b π b Note that this integral is given on the range [ 0, ] For the range [, ], the value of the integral must be doubled since the integrand corresponds to an even function In addition, the substitution b α must be made

10 5) continued 0 Substituting, V V 2 k α π 2 k α π k 4α π 2α α x 2 e α x2 dx This is sufficient as an answer However, the result may be simplified further by using the definition of the parameter α and the harmonic frequency ν 0 V k 4α k 4! 2 µk! 4 k µ V 4 h k 2π µ 4 hν 0

11 6) (4 points) Discuss the requirements for observing an absorption spectrum corresponding to transitions between vibrational energy levels in a molecule Your discussion should focus on items such as the type of electromagnetic radiation required for absorption, the quantum numbers involved in the vibrational transitions, selection rules, requirements for activity, intensities, etc In order to observe an absorption spectrum corresponding to vibrational transitions, the dipole moment of the mode must change during the vibration for a mode to exhibit activity The frequency of the electromagnetic radiation for vibrational transitions generally falls in the infrared region of the spectrum For a peak to appear in the vibrational spectrum, the frequency of infrared radiation must match the energy difference between the vibrational levels Typically, the most intense peaks in the spectrum correspond to fundamental transitions from the v0 level to the v level for a particular vibrational mode This is because the intensity is proportional to the population of molecules in a given level, and at room temperature most molecules are in the ground vibrational level The selection rules for vibrational transitions vary depending upon whether the system is treated using the harmonic oscillator model or whether the system is treated as an anharmonic oscillator Within the harmonic oscillator model, the selection rule for absorption is Δv + Thus, only the fundamental v 0 transition would be observed, given the lack of population in v and higher levels For the anharmonic oscillator model, the selection rules for absorption are Δv +, +2, +3, etc Thus, for an anharmonic oscillator (as for a "real" molecule), not only would the fundamental v 0 transitions be observed, but at lower intensity overtones such as v 0 2 also might be observed

12 7) (4 points) An unknown molecule is trapped on a smooth copper surface in a square area of length 00 Å on each side The transition wavelength observed when the system undergoes a transition from the ground state to the first excited state is 0404 cm Determine the mass of the unknown molecule in kg 2 Assuming the unknown molecule trapped on the surface behaves like a particle in a two-dimensional box, the energy levels are given by: E nx n y n 2 2 ( x + n y ) h 2 8mL 2, where n x and n y are the quantum numbers, m is the mass of the particle, and L is the length of the box For a transition from the ground state with n x, n y ( ), degenerate with ( n x, n y ) (, 2) or ( 2, ), the energy difference ΔE is ( ) to the first excited state, which is ΔE E 2 E ΔE ( ) h 2 8mL 2 h 2 ( ) 8mL h 2 8mL 2 A photon with an energy corresponding to ΔE has a frequency given by E photon ΔE hν Since for light we know that λν c, we can substitute ν λ c, and obtain an expression for the transition wavelength λ, λ hc ΔE hc 8mL2 3h 2 λ 8cmL2 3h Solving this expression for the mass m yields, m 3hλ 8cL 2

13 7) continued 3 Substituting, m 3hλ 8cL 2 ( )( m) ( )( m) Js m s - m kg Multiplying this by Avogadro's number (and converting to g) gives a molar mass of 202 g/mol Thus, the unknown molecule trapped on the copper surface is H 2

Physical Chemistry II Exam 2 Solutions

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