Homework 12
- Page ID
- 204093
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Q1
Franck-Condon factors are usually difficult to calculate, but often are approximated by simple harmonic oscillator models. Imagine a diatomic with two electronic states that are identically harmonic except for the location of their minima. Imagine one has a minimum displacement at a distance \(\delta\) from each other. We will consider only the \(v=0\) levels of each state so that vibrational wavefunction are just the zero-point harmonic oscillator functions:
\[ | v =0 \rangle^{g} = \dfrac{1}{\pi^{1/4}} e ^{-q_1^2/2} = \dfrac{1}{\pi^{1/4}} \exp\left[-\dfrac{k}{2\hbar \omega} (R-R_{eq})^2\right]\]
and
\[ | v =0 \rangle^{e} = \dfrac{1}{\pi^{1/4}} e ^{-q_2^2/2} = \dfrac{1}{\pi^{1/4}} \exp\left[-\dfrac{k}{2\hbar \omega} (R-R_{eq}-\delta)^2\right]\]
Where \(R_{eq}\) is the bond length of ground electronic state and \(R_{eq} + δ\) is the bond length of the excited electronic state.
- Show that the integral for the Franck-Condon overlap \[S_{00}= \sqrt{\dfrac{\alpha}{\pi}} e^{-\alpha(R_{eq} - R_{eq}-\delta )^2/4} \int_{-\infty}^{\infty} e ^{-\alpha\{R - 1/2(R_e + R_{eq} -\delta )\}^2} dR\] where \[\alpha = \dfrac{\sqrt{mk}}{\hbar}\]
- This is a Gaussian Integral and is known in standard integral tables. Solve for the analytical value of \(S_{00}\) for the 0-0 vibrational transition.
- What value of \(\delta\) gives \(S=1\) for the 0-0 transition?
- How does \(S\) depends on the absolute positions (i.e., electronic energies\) of the two electronic states?
Q2
Calculate the Frank-Condon factors for the 0-0 transition using the model in Q1 with the following displacements:
- \(δ=0.0 \; Å\)
- \(δ=0.05 \; Å\)
- \(δ=0.1 \; Å\)
- \(δ=0.5\; Å\)
for the vibration \(\dfrac{k}{\hbar \omega} =200 \; Å^{-2}\). These numbers should convince you of the importance of the Franck-Condon factor in controlling one’s access to vibration levels of excited electronic states.