2010
DOI: 10.1021/jp103504a
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Rovibrational Energy Transfer in Ne−Li2(A1Σu+,v=0): Comparison of Experimental Data and Results from Classical and Quantum Calculations

Abstract: Absolute rate constants for rotational and rovibrational energy transfer in the system Ne-Li2(A1Sigmau+) were measured by a dispersed fluorescence technique following excitation of the (v = 0,j = 18) initial level of Li2(A1Sigmau+). The rate coefficients for Deltav = 0 processes decline monotonically with increasing |Deltaj|. The Deltav = 1 rate coefficients are also peaked near Deltaj = 0 but show a broad shoulder extending to approximately Deltaj = 30. Classical trajectory calculations and accurate quantum m… Show more

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Cited by 12 publications
(18 citation statements)
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“…where µ is the B-C reduced mass, L is the angular momentum of the molecule, r is the distance between the two atoms, and k is the spring constant of the molecule. Then the e↵ective potential of the molecule is given by 3) and the Lagrangian is given by…”
Section: Given Arbitrary Initial Positions Smentioning
confidence: 99%
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“…where µ is the B-C reduced mass, L is the angular momentum of the molecule, r is the distance between the two atoms, and k is the spring constant of the molecule. Then the e↵ective potential of the molecule is given by 3) and the Lagrangian is given by…”
Section: Given Arbitrary Initial Positions Smentioning
confidence: 99%
“…This will allow us to study T-shaped collisions, in which A collides with the molecule on its side, and incites vibrational energy by pushing B and C apart from one another other. It has been shown in our lab [3] that this mechanism dominates the vibrational energy transfer process for Li 2 -X collisions where X is a non-reactive noble gas atom, and thus it is important to understand it in the context of hard spheres.…”
Section: Introductionmentioning
confidence: 99%
“…If the molecule undergoes an inelastic collision, there is satellite fluorescence. [3] If the molecule undergoes a rotationally inelastic collision, ∆j, the change in the rotational quantum state is always even. [4] As we see in figure 2, fluorescence is not emitted as a series of delta functions, but rather is emitted as a spectral line shape with finite width.…”
Section: Background and Motivationmentioning
confidence: 99%
“…Here, n f is the population of a given collisionally populated final state, n i is the population of the initial excited state and n X is the population of the collision partner; Γ f is the rate of decay from the final state f and k if is the thermally averaged rate constant of the collision leading to final state f . [3] If we assume a steady state solution, we can solve for the ratio of final state population to initial state population in terms of the rate constant and two quantities as below:…”
Section: Background and Motivationmentioning
confidence: 99%
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