Chapter 11, Problem 11F.5BE

Interpretation Introduction

Interpretation:

The Franck-Condon factor for vibrational state transitions has to be stated.

Concept introduction:

The Franck-Condon principle describes the intensity of the vibronic transitions. The vibronic transition is the simultaneous change in the electronic and vibrational energy levels because of emission or absorption of a photon.

Answer to Problem 11F.5BE

The Franck-Condon factor for the vibrational transition is shown below.

  ${\left|S\left(\upsilon ,0\right)\right|}^2=a^2{x}_0^2{e}^{-\frac{a}{2}{x}_0^2}$

Explanation of Solution

The Franck-Condon factor is given by the expression as shown below.

  ${\left|S\left({\upsilon }_{\text{f}},{\upsilon }_{\text{i}}\right)\right|}^2={\left({\displaystyle \int {\psi }_{\upsilon ,\text{f}}^{*}{\psi }_{\upsilon ,i}\text{d}{\tau }_{\text{n}}}\right)}^2$        (1)

Where,

• ${\upsilon }_i$ is the initial vibrational states.
• ${\upsilon }_{\text{f}}$ is the final vibrational states.
• $\psi$ is the wavefunction.

The wave function expression for the vibrational ground state of a molecule in one-dimensional box is shown below.

  ${\psi }_0=N_0{e}^{-a{x}^2/2}$

The normalization constant $\left(N_0\right)$ for this wavefunction is calculated as shown below.

  $\begin{array}{c}{N}_0^2=\frac{1}{{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-a{x}^2/2}{e}^{-a{x}^2/2}}\text{d}x}\\ N_0^2=\frac{1}{{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-a{x}^2}}\text{d}x}\\ N_0^2=\frac{1}{{\left(\frac{\pi }{a}\right)}^{\frac{1}{2}}}\\ N_0^2={\left(\frac{a}{\pi }\right)}^{\frac{1}{2}}\end{array}$

The wave function expression for the other vibrational state of a molecule in one-dimensional box is shown below.

  ${{\psi }^{\prime }}_0=N_{1}x{e}^{-a{\left(x-x_0\right)}^2/2}$

The normalization constant $\left(N_1\right)$ for this wavefunction is calculated as shown below.

  $\begin{array}{c}{N}_1^2=\frac{1}{{\displaystyle {\int }_{-\infty }^{\infty }x{e}^{-a{\left(x-x_0\right)}^2/2}x{e}^{-a{\left(x-x_0\right)}^2/2}}\text{d}x}\\ N_1^2=\frac{1}{{\displaystyle {\int }_{-\infty }^{\infty }{x}^2{e}^{-a{\left(x-x_0\right)}^2}}\text{d}x}\\ N_1^2=\frac{1}{{\displaystyle {\int }_{-\infty }^{\infty }{x}^2{e}^{-2b{\left(x-x_0\right)}^2}}\text{d}x}\left(b=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\end{array}$

The above equation is further solved as shown below.

  $\begin{array}{c}{N}_1^2=\frac{1}{\left(\frac{{\pi }^{1/2}/2}{{\left(2b\right)}^{3/2}}\right)}\\ N_1^2=\left(\frac{{\left(2b\right)}^{3/2}}{{\pi }^{1/2}/2}\right)\\ N_1^2=\frac{2{\left(2b\right)}^{3/2}}{{\pi }^{1/2}}\end{array}$

The vibration overlap integral between the vibrational wavefunction in the upper and lower electronic states is shown below.

   $\begin{array}{c}S\left(\upsilon ,0\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{N}_0{e}^{-a{x}^2/2}{N}_{1}x{e}^{-a{\left(x-x_0\right)}^2/2}}\text{d}x\\ =N_0{N}_1{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2/2}x{e}^{-a{\left(x-x_0\right)}^2/2}}\text{d}x\\ =N_0{N}_1{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{e}^{-\left(\frac{a{x}^2+a{\left(x-x_0\right)}^2}{2}\right)}}\text{d}x\\ =N_0{N}_1{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{e}^{-\left(b{x}^2+b{\left(x-x_0\right)}^2\right)}}\text{d}x\end{array}$        (2)

The above equation is further solved as shown below.

Let $z={\left(2b\right)}^{1/2}x-\frac{b}{{\left(2b\right)}^{1/2}}{x}_0$

Then

  $\begin{array}{l}x=\left(\frac{z+\frac{b}{{\left(2b\right)}^{1/2}}{x}_0}{{\left(2b\right)}^{1/2}}\right)\\ \text{d}z={\left(2b\right)}^{1/2}\text{dx}\\ \text{dx}=\frac{\text{d}z}{{\left(2b\right)}^{1/2}}\end{array}$

The equation $b{x}^2+b{\left(x-x_0\right)}^2$ is written as shown below.

  $b{x}^2+b{\left(x-x_0\right)}^2=z^2+\frac{b}{2}{x}_0^2$

Substitute the values of $x,\text{dx}$ and $b{x}^2+b{\left(x-x_0\right)}^2$ in equation (2) as shown below.

  $\begin{array}{c}S\left(\upsilon ,0\right)=N_0{N}_1{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(\frac{z+\frac{b}{{\left(2b\right)}^{1/2}}{x}_0}{{\left(2b\right)}^{1/2}}\right)e^{-\left(z^2+\frac{b}{2}{x}_0^2\right)}}\frac{\text{d}z}{{\left(2b\right)}^{1/2}}\\ =N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(\frac{z+\frac{b}{{\left(2b\right)}^{1/2}}{x}_0}{{\left(2b\right)}^{1/2}}\right)e^{-z^2}}\frac{\text{d}z}{{\left(2b\right)}^{1/2}}\\ =\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}}{{\left(2b\right)}^{1/2}}\left\{\frac{b}{{\left(2b\right)}^{1/2}}{x}_0{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-z^2}\text{d}z+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}z{e}^{-z^2}\text{d}z}}\right\}\\ =\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}}{{\left(2b\right)}^{1/2}}\left\{b{x}_0{\left(\frac{\pi }{2b}\right)}^{1/2}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}z{e}^{-z^2}\text{d}z}\right\}\end{array}$        (3)

The integral of $z{e}^{-z^2}\text{d}z$ which is an ungerade function is zero because the integrals of ungerade function is always zero.

The equation (3) is further solved as shown below.

  $\begin{array}{c}S\left(\upsilon ,0\right)=\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}}{{\left(2b\right)}^{1/2}}\left\{b{x}_0{\left(\frac{\pi }{2b}\right)}^{1/2}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}z{e}^{-z^2}\text{d}z}\right\}\\ =\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}}{{\left(2b\right)}^{1/2}}\left\{b{x}_0{\left(\frac{\pi }{2b}\right)}^{1/2}+0\right\}\\ =\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}{x}_0{\pi }^{1/2}}{2}\end{array}$        (4)

The Franck-Condon factor is calculated as shown below.

  $\begin{array}{c}{\left|S\left(\upsilon ,0\right)\right|}^2={\left(\frac{N_0{N}_1{e}^{-\frac{b}{2}{x}_0^2}{x}_0{\pi }^{1/2}}{2}\right)}^2\\ =\frac{N_0^2{N}_1^2{e}^{-b{x}_0^2}{x}_0^2\pi }{4}\end{array}$        (5)

Substitute the value of $N_0$, $b$ and $N_1$ in equation (5) as shown below.

  $\begin{array}{c}{\left|S\left(\upsilon ,0\right)\right|}^2=\frac{N_0^2{N}_1^2{e}^{-b{x}_0^2}{x}_0^2\pi }{4}\\ =\frac{{\left(\frac{a}{\pi }\right)}^{\frac{1}{2}}\left(\frac{2{\left(2\times \frac{a}{2}\right)}^{3/2}}{{\pi }^{1/2}}\right)e^{-\frac{a}{2}{x}_0^2}{x}_0^2\pi }{4}\\ =\frac{\left(\frac{2a^2}{\pi }\right)e^{-\frac{a}{2}{x}_0^2}{x}_0^2\pi }{4}\\ =a^2{x}_0^2{e}^{-\frac{a}{2}{x}_0^2}\end{array}$

Therefore, the Franck-Condon factor for the vibrational transition described by the two wave functions is shown below

  ${\left|S\left(\upsilon ,0\right)\right|}^2=a^2{x}_0^2{e}^{-\frac{a}{2}{x}_0^2}$