Chapter 11, Problem 11C.15P

Interpretation Introduction

Interpretation:

The value of ${\tilde{B}}_0$ and ${\tilde{B}}_1$ have to be calculated.  The wave numbers of the transitions originating from $J=3$ of the P branches and R branches have to be predicted.  The dissociation energy of the $\text{IF}$ molecule has to be predicted.

Concept introduction:

The internal mode of vibration of molecule leads to a periodic change in the internal internuclear separation of a molecule. Due to this change, the molecules are considered as harmonic oscillators. The strength of a bond in a bond is measured by force constant. The force constant of a molecule is related to the effective mass and vibrational frequency.  The vibrational energy of a molecule is given by the equation as shown below.

  $E_{\nu }=\left(\nu +\frac{1}{2}\right)\hslash \omega$.

Answer to Problem 11C.15P

The value of ${\tilde{B}}_0$ is $\underset{\_}{0.278775c{m}^{-1}}$.

The value of ${\tilde{B}}_1$ is $\underset{\_}{0.276905c{m}^{-1}}$.

The wave numbers of the transitions originating from $J=3$ of the P branches is $\underset{\_}{601.724c{m}^{-1}}$.

The wave numbers of the transitions originating from $J=3$ of the R branches is $\underset{\_}{605.626c{m}^{-1}}$.

The dissociation energy of the $\text{IF}$ molecule is $\underset{\_}{29641.42c{m}^{-1}}$.

Explanation of Solution

The rotational constant for a diatomic molecule in vibrational state with quantum number $v$ is given by the expression shown below.

    ${\tilde{B}}_v={\tilde{B}}_e-a\left(v+\frac{1}{2}\right)$        (1)

Where,

• ${\tilde{B}}_e$ is the rotational constant corresponding to the equilibrium bond length.
• $v$ is the quantum number.
• $a$ is morse potential energy.

The values of ${\tilde{B}}_e$ and $a$ for $\text{IF}$ molecule are $0.2797\text{1}{\text{cm}}^{-1}$ and $0.18\text{7}{\text{m}}^{-1}$ respectively.

Substitute the value of of ${\tilde{B}}_e$ and $a$ in equation (1) to calculate the value of ${\tilde{B}}_0$ at $v=0$, as shown below.

    $\begin{array}{c}{\tilde{B}}_v={\tilde{B}}_e-a\left(v+\frac{1}{2}\right)\\ {\tilde{B}}_0=0.2797\text{1}{\text{cm}}^{-1}-0.18\text{7}{\text{m}}^{-1}\times \left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\left(0+\frac{1}{2}\right)\\ =0.2797\text{1}{\text{cm}}^{-1}-0.000935{\text{cm}}^{-1}\\ =\underset{\_}{0.278775c{m}^{-1}}\end{array}$

Substitute the value of of ${\tilde{B}}_e$ and $a$ in equation (1) to calculate the value of ${\tilde{B}}_1$ at $v=1$, as shown below.

    $\begin{array}{c}{\tilde{B}}_v={\tilde{B}}_e-a\left(v+\frac{1}{2}\right)\\ {\tilde{B}}_1=0.2797\text{1}{\text{cm}}^{-1}-0.18\text{7}{\text{m}}^{-1}\times \left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\left(1+\frac{1}{2}\right)\\ =0.2797\text{1}{\text{cm}}^{-1}-0.002805{\text{cm}}^{-1}\\ =\underset{\_}{0.276905c{m}^{-1}}\end{array}$

The wave numbers of the transitions originating from $J=3$ of the P branches is calculated by the formula given below.

    ${\tilde{v}}_{\text{p}}\left(J\right)=\tilde{v}-\left({\tilde{B}}_1+{\tilde{B}}_0\right)J+\left({\tilde{B}}_1-{\tilde{B}}_0\right)J^2$        (2)

Where,

• $\tilde{v}$ is wave number.
• ${\tilde{B}}_0$ is rotational constant at $v=0$.
• ${\tilde{B}}_1$ is rotational constant at $v=1$.
• $J$ is vibrational state.

When anharmonicities are present $\tilde{v}$ is replaced by the term $\Delta \tilde{G}\left(v\right)$.

    $\Delta \tilde{G}\left(v\right)=\tilde{v}-2\left(v+1\right)x_{\text{e}}\tilde{v}$

Where,

• $x_{\text{e}}$ is anharmonicity constant.

For $v=0$,

    $\begin{array}{c}\Delta \tilde{G}\left(v\right)=\tilde{v}-2\left(v+1\right)x_{\text{e}}\tilde{v}\\ =\tilde{v}-2\left(0+1\right)x_{\text{e}}\tilde{v}\\ =\tilde{v}-2x_{\text{e}}\tilde{v}\end{array}$

In equation (2) $\tilde{v}$ is replaced by the term $\Delta \tilde{G}\left(v\right)$, which is equal to $\tilde{v}-2x_{\text{e}}\tilde{v}$.  Therefore, the equation (2) is written as shown below.

    ${\tilde{v}}_{\text{p}}\left(J\right)=\tilde{v}-2x_{\text{e}}\tilde{v}-\left({\tilde{B}}_1+{\tilde{B}}_0\right)J+\left({\tilde{B}}_1-{\tilde{B}}_0\right)J^2$        (3)

For P branches, the value of $\tilde{v}$, $x_{\text{e}}\tilde{v}$, $J$, ${\tilde{B}}_1$ and ${\tilde{B}}_0$ are $610.25\text{8}{\text{cm}}^{-1}$, $3.14\text{1}{\text{cm}}^{-1}$, $4$, $0.27690\text{5}{\text{cm}}^{-1}$, and $0.278775{\text{cm}}^{-1}$.

Substitute the value of $\tilde{v}$, $x_{\text{e}}\tilde{v}$, $J$, ${\tilde{B}}_1$ and ${\tilde{B}}_0$ in equation (3).

    $\begin{array}{c}{\tilde{v}}_{\text{p}}\left(J\right)=\left(\begin{array}{l}610.25\text{8}{\text{cm}}^{-1}-2\times 3.14\text{1}{\text{cm}}^{-1}-\left(0.27690\text{5}{\text{cm}}^{-1}+0.278775{\text{cm}}^{-1}\right)4\\ +\left(0.27690\text{5}{\text{cm}}^{-1}-0.278775{\text{cm}}^{-1}\right)4^2\end{array}\right)\\ =610.25\text{8}{\text{cm}}^{-1}-6.282{\text{cm}}^{-1}-2.222{\text{cm}}^{-1}-0.02999{\text{cm}}^{-1}\\ =610.25\text{8}{\text{cm}}^{-1}-7.4004{\text{cm}}^{-1}\\ =\underset{\_}{601.724c{m}^{-1}}\end{array}$

The wave numbers of the transitions originating from $J=3$ of the R branches is calculated by the formula given below.

    ${\tilde{v}}_{\text{R}}\left(J\right)=\tilde{v}-2x_{\text{e}}\tilde{v}+\left({\tilde{B}}_1+{\tilde{B}}_0\right)\left(J+1\right)+\left({\tilde{B}}_1-{\tilde{B}}_0\right){\left(J+1\right)}^2$        (4)

For R branches, the value of $\tilde{v}$, $x_{\text{e}}\tilde{v}$, $J$, ${\tilde{B}}_1$ and ${\tilde{B}}_0$ are $610.25\text{8}{\text{cm}}^{-1}$, $3.14\text{1}{\text{cm}}^{-1}$, $2$, $0.27690\text{5}{\text{cm}}^{-1}$, and $0.278775{\text{cm}}^{-1}$.

Substitute the value of $\tilde{v}$, $x_{\text{e}}\tilde{v}$, $J$, ${\tilde{B}}_1$ and ${\tilde{B}}_0$ in equation (4).

  $\begin{array}{c}{\tilde{v}}_{\text{R}}\left(J\right)=\left(\begin{array}{l}610.25\text{8}{\text{cm}}^{-1}-2\times 3.14\text{1}{\text{cm}}^{-1}+\left(0.27690\text{5}{\text{cm}}^{-1}+0.278775{\text{cm}}^{-1}\right)\left(2+1\right)\\ +\left(0.27690\text{5}{\text{cm}}^{-1}-0.278775{\text{cm}}^{-1}\right){\left(2+1\right)}^2\end{array}\right)\\ =610.25\text{8}{\text{cm}}^{-1}-6.282{\text{cm}}^{-1}+1.667{\text{cm}}^{-1}-0.0168{\text{cm}}^{-1}\\ =610.25\text{8}{\text{cm}}^{-1}-7.9658{\text{cm}}^{-1}\\ =\underset{\_}{605.626c{m}^{-1}}\end{array}$

The dissociation energy of the $\text{IF}$ molecule is calculated by the formula given below.

    $D_{\text{e}}=\frac{{\tilde{v}}^2}{4x_{\text{e}}\tilde{v}}$        (5)

Where,

• $x_{\text{e}}$ is anharmonicity constant.
• $\tilde{v}$ is wave number.

Substitute the value of $\tilde{v}$ and $x_{\text{e}}\tilde{v}$, in equation (5).

    $\begin{array}{c}{D}_{\text{e}}=\frac{{\tilde{v}}^2}{4x_{\text{e}}\tilde{v}}\\ =\frac{{\left(610.25\text{8}{\text{cm}}^{-1}\right)}^2}{4\times 3.14\text{1}{\text{cm}}^{-1}}\\ =\frac{372414.826}{12.564}{\text{cm}}^{-1}\\ =\underset{\_}{29641.42c{m}^{-1}}\end{array}$