Advanced kinetics Exercise 9 10/05/2019

Release date: 10/05/2019 onlineReturn date: 17/05/2019 blue box in front of HCI E241 or email. Please clearly mark yourfull name on your work!Questions: max.waters@phys.chem.ethz.ch

1 Angular distributions in reactive molecular collisions

Motivation

Last week, we were introduced to the concept of molecular reactions as scatteringevents. This week, we see that scattering is a powerful probe for collision dynamics.The angular distribution of scattering products, which reflects the differential scatteringcross section, can be measured in crossed molecular beam experiments. The distribu-tion of scattering angles θ and product velocities vAB in the centre-of-mass (CM) framecan be inferred from a Newton diagram (velocity diagram) as shown in Figure 1.

Figure 1: Newton diagram for the reaction A+BC −−→ AB+C (left), and reconstruction ofthe CM angular distribution from a crossed molecular beam experiment (right).

The CM product flux distributions are usually represented in a polar plot. The contourlines (Figure 2) indicate the product flux scattered into a certain angle θ with a givenvelocity v or kinetic energy E′t.

The reaction mechanism manifests itself directly in the angular distribution of the reactionproducts. Two important types of mechanisms can be distinguished:

I Direct mechanisms entail a direct scattering event.

II Indirect (or complex-forming) mechanisms entail the formation of an intermediaryreaction complex.

In this exercise sheet we will see how these can be distinguished. In order to conceptu-alise this, it is important to think of how an angular distribution might change for collisioncomplexes of varying lifetimes.

Let us consider the following reactions:

K + I2 −−→ KI + I

K + CH3I −−→ KI + CH3

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Advanced kinetics Exercise 9 10/05/2019

The KI product flux contour plots in the center-of-mass (CM) frame as a function of KI velocityfrom these two reactions are displayed in Figure 2.

Figure 2: KI product flux contour plots in the center-of-mass (CM) frame as a function of KI velocity (i.e.flux as a function of scattering angle and KI velocity) for the K + I2 (left) and K + CH3I (right)reactions.

1.1 Compare and contrast the plots for these two reactions and comment on the origin ofthese differences. The dotted outer rings in figure 2 show the maximum CM velocities ofKI in the two reactions. For discussing the K + I2 reaction, keep in mind what we learnedin the previous week, namely the large integral cross section and large impact parameter.

1.2 What does the fact that both reactions lead to a preferential scattering (forward or back-ward) suggest about the duration of the reaction (lifetime of the intermediate complex)?

1.3 Why is the velocity for the KI products different to the maximum velocity?

Figure 3: Theoretical (QM) and experimental differential cross section versus center-of-mass scatteringangle for S(1D)+H2 −−→ SH+H2. The collision energy is 9.37 kJmol−1. The QM contributionsof each vibrational state are also show as thin solid lines. (Honvault et al., Chem. Phys. Lett.370, 2003, 371.)

Contrast the scattering behavior observed for the above two reactions with that found for thereaction S(1D) + H2, the differential cross section for which is shown in figure 3.

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Advanced kinetics Exercise 9 10/05/2019

1.4 Assuming that the scattering amplitude has the simple analytical form f(θ, φ) = sin θ cosφ,find an expression for the differential cross section, dσ(θ,φ)

dΩ and evaluate also the totalcross section σ.

1.5 Evaluate the integral scattering cross section for a case in which the differential crosssection σ(θ, φ) is a constant C independent of the angles θ and φ.

1.6 Comment on what information about the S(1D)+H2 reaction can be learned from figure 3.

2 Density of states

Motivation

As was demonstrated in the prediscussion, the density of states can have a drasticimpact on the outcome of a collision between two particles. It’s also a useful quantityto estimate for a wide variety of sciences, such as in estimating the efficiency of a solarcell with Fermi’s Golden Rule: Γi→f = 2π

~ | 〈ψf | H |ψi〉 |2 ρ. The density of states ρ

is used to estimate the rate of internal conversion Γi→f (where 〈ψf | H |ψi〉 is the matrixelement that describes the degree of overlap between the initial and final states). In thisexercise, we will look at how the number of accessible states can vary as a function oftemperature.A strongly covalently bonded crystal (like diamond) may be considered to be a macro-molecule consisting ofN atoms. It has s = 3N−6 internal degrees of freedom which canbe approximated as harmonic oscillators (phonons) with frequencies νi, i ∈ [1, 3N − 6].The distribution of frequencies νi is given by Debye theory as

dρ(r) = g(ν)dν =3

ν3D

ν2 dν , (1)

where the Debye frequency νD = kTD/h is the maximum possible frequency for theoscillators, i.e. νi ∈ [0, νD]. The Debye temperature TD for diamond is 1900 K. Thedistribution of frequencies satisfies the relation∫ νD

0g(ν) dν =

∫ νD

0

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ν3D

ν2 dν = 1. (2)

2.1 Estimate the number of states W (E) of a diamond with a mass of 12mg (1mmol) forenergies of E1 = 5.4 Jmol−1, E2 = 530.0 Jmol−1, and E3 = 11.6 kJmol−1, which cor-respond to inner energies of T = 100K, 300K, and 1000K, and for the energy E4 =71.1 kJmol−1, which corresponds to the excitation of all vibrational degrees of freedomby 2 quanta.

Calculate the zero point energy EZ , where EZ = hν2 and use this to calculate the number

of states with and without EZ by setting a = 0 and a = 1 in the expression for the numberof states

W (E) =(E + aEZ)s

s!∏si=1 hνi

for 0 ≤ a ≤ 1. (3)

Hint : To evaluate W (E), multiply eq. (3) by 1 = (hνD)s/(hνD)s, take the logarithm ofW (E), replace the discrete values of νi by their continuous distribution of eq. (1) andintegrate. Additionally, recall Stirling’s formula: lnn! = n lnn − n + O(lnn), and that forthe continuous distribution g(ν) of ν: 1

s

∑si=1 →

∫ νD0 g(ν) dν.

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Advanced kinetics Exercise 9 10/05/2019

2.2 As a further simplification, estimate the number of states W (E) when all frequenciesνi are equal to the “average” frequency νE =

√3/5νD (root-mean-square of given fre-

quency distribution). Again, calculate EZ and compare the number of states with andwithout the zero point energy.

2.3 Suggest an alternative way to obtain the number of states W (E) of diamond, e.g. basedon the entropy.

3 Density of states, part 2

In the lecture, the density of states for simple model systems was discussed. To understandthe concepts better, numerical simulations may be used. How this can be done is illustrated inthe article Mulhall & Moelter, Am. J. Phys. 82, 665 (2014).1

3.1 Read the article. Make sure that you can explain the meaning of figures 1 and 2. Howwould figure 1 look like for a particle in an isotropic two-dimensional harmonic oscillator,i.e., for a particle feeling a potential that is a product of two harmonic oscillators with thesame force constant?

3.2 Explain the difference between the numerical and the theoretical results forN(ε) in figures3, 4, 5, and 6.

3.3 Summarize what you think are the most important messages of the study.

3.4 If you have access to Matlab (you may also use Octave as open source alternative), runthe Matlab script given in the supplementary information of the article to recreate figure5. Reproduce also the other figures of the article.

Note that the article also provides additional problems. Working on them will help you to under-stand the concept of the density of states and its behavior for different model systems better.

1 The American Journal of Physics is a journal of the American Association of Physics Teachers which has manyeducational articles on basic and advanced physical concepts.

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