Chapter 11, Problem 11C.18P

Interpretation Introduction

Interpretation:

The values of $\widetilde{B_0}$ and $\widetilde{B_1}$ for ${}^{\text{123}}{\text{C}}^{\text{16}}\text{O}$ has to be estimated.

Concept introduction:

Spectroscopy method is used to identify the structure of the substance.  It is based on the interactions between matter and electromagnetic radiations.  Each line of the high-resolution vibrational spectrum of a gas phase heteronuclear diatomic molecule consists of closely space rotational levels.

Answer to Problem 11C.18P

The value of $\widetilde{B_0}$ is $\underset{\_}{1.923c{m}^{-1}}$.  The value of $\widetilde{B_1}$ is $\underset{\_}{1.905c{m}^{-1}}$.

Explanation of Solution

The expression for the calculation of ${\tilde{\nu }}_{\text{R}}\left(J\right)$ is given below.

    ${\tilde{\nu }}_{\text{R}}\left(J\right)=\tilde{\nu }+\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(J+1\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(J+1\right)}^2$        (1)

Where,

• $\tilde{\nu }$ is the wavenumber of the fundamental vibration.
• ${\tilde{\nu }}_{\text{R}}\left(J\right)$ is the wavenumber of the R branch.
• $J$ is the rotational quantum number.
• $\widetilde{B_0}$ is the rotational quantum number of the rotational states associated with $\tilde{\nu }=0$.
• $\widetilde{B_1}$ is the rotational quantum number of the rotational states associated with $\tilde{\nu }=1$.

The value of ${\tilde{\nu }}_{\text{R}}\left(0\right)$ is $2147.084{\text{cm}}^{-1}$.

The value of ${\tilde{\nu }}_{\text{R}}\left(1\right)$ is $2150.858{\text{cm}}^{-1}$.

Substitute the value of ${\tilde{\nu }}_{\text{R}}\left(0\right)$ in equation (1).

    $\begin{array}{c}2147.084{\text{cm}}^{-1}=\tilde{\nu }+\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(0+1\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(0+1\right)}^2\\ \tilde{\nu }+\widetilde{B_1}+\widetilde{B_0}+\widetilde{B_1}-\widetilde{B_0}=2147.084{\text{cm}}^{-1}\\ \tilde{\nu }+2\widetilde{B_1}=2147.084{\text{cm}}^{-1}\end{array}$        (2)

Substitute the value of ${\tilde{\nu }}_{\text{R}}\left(1\right)$ in equation (1).

    $\begin{array}{c}2150.858{\text{cm}}^{-1}=\tilde{\nu }+\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(1+1\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(1+1\right)}^2\\ \tilde{\nu }+2\widetilde{B_1}+2\widetilde{B_0}+4\widetilde{B_1}-4\widetilde{B_0}=2150.858{\text{cm}}^{-1}\\ \tilde{\nu }+6\widetilde{B_1}-2\widetilde{B_0}=2150.858{\text{cm}}^{-1}\end{array}$        (3)

The expression for the calculation of ${\tilde{\nu }}_{\text{P}}\left(J\right)$ is given below.

    ${\tilde{\nu }}_{\text{P}}\left(J\right)=\tilde{\nu }-\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(J\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(J\right)}^2$        (4)

Where,

• $\tilde{\nu }$ is the wavenumber of the fundamental vibration.
• ${\tilde{\nu }}_{\text{P}}\left(J\right)$ is the wavenumber of the P branch.
• $J$ is the rotational quantum number.
• $\widetilde{B_0}$ is the rotational quantum number of the rotational states associated with $\tilde{\nu }=0$.
• $\widetilde{B_1}$ is the rotational quantum number of the rotational states associated with $\tilde{\nu }=1$.

The value of ${\tilde{\nu }}_{\text{P}}\left(1\right)$ is $2139.427{\text{cm}}^{-1}$.

The value of ${\tilde{\nu }}_{\text{P}}\left(2\right)$ is $2135.548{\text{cm}}^{-1}$.

Substitute the value of ${\tilde{\nu }}_{\text{P}}\left(1\right)$ in equation (4).

    $\begin{array}{c}2139.427{\text{cm}}^{-1}=\tilde{\nu }-\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(1\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(1\right)}^2\\ \tilde{\nu }-\widetilde{B_1}-\widetilde{B_0}+\widetilde{B_1}-\widetilde{B_0}=2139.427{\text{cm}}^{-1}\\ \tilde{\nu }-2\widetilde{B_0}=2139.427{\text{cm}}^{-1}\end{array}$        (5)

Substitute the value of ${\tilde{\nu }}_{\text{P}}\left(2\right)$ in equation (4).

    $\begin{array}{c}2135.548{\text{cm}}^{-1}=\tilde{\nu }-\left(\widetilde{B_1}+\widetilde{B_0}\right)\left(2\right)+\left(\widetilde{B_1}-\widetilde{B_0}\right){\left(2\right)}^2\\ \tilde{\nu }-2\widetilde{B_1}-2\widetilde{B_0}+4\widetilde{B_1}-4\widetilde{B_0}=2135.548{\text{cm}}^{-1}\\ \tilde{\nu }+2\widetilde{B_1}-6\widetilde{B_0}=2135.548{\text{cm}}^{-1}\end{array}$        (6)

Subtract equation (5) from (3).

    $\begin{array}{c}\tilde{\nu }+6\widetilde{B_1}-2\widetilde{B_0}-\left(\tilde{\nu }-2\widetilde{B_0}\right)=2150.858{\text{cm}}^{-1}-2139.427{\text{cm}}^{-1}\\ \tilde{\nu }+6\widetilde{B_1}-2\widetilde{B_0}-\tilde{\nu }+2\widetilde{B_0}=11.431{\text{cm}}^{-1}\\ 6\widetilde{B_1}=11.431{\text{cm}}^{-1}\\ \widetilde{B_1}=\underset{\_}{1.905c{m}^{-1}}\end{array}$

Therefore, the value of $\widetilde{B_1}$ is $\underset{\_}{1.905c{m}^{-1}}$.

Subtract equation (2) from (6).

    $\begin{array}{c}\tilde{\nu }+2\widetilde{B_1}-6\widetilde{B_0}-\left(\tilde{\nu }+2\widetilde{B_1}\right)=2135.548{\text{cm}}^{-1}-2147.084{\text{cm}}^{-1}\\ \tilde{\nu }+2\widetilde{B_1}-6\widetilde{B_0}-\tilde{\nu }-2\widetilde{B_1}=-11.536{\text{cm}}^{-1}\\ -6\widetilde{B_0}=-11.536{\text{cm}}^{-1}\\ \widetilde{B_0}=\underset{\_}{1.923c{m}^{-1}}\end{array}$

Therefore, the value of $\widetilde{B_0}$ is $\underset{\_}{1.923c{m}^{-1}}$.