Chapter 11, Problem 11C.6P

Interpretation Introduction

Interpretation:

The Morse potential energy curve from $\text{50}\text{pm}$ to $\text{800}\text{pm}$ around $R_e=236.7\text{pm}$ has to be plotted. The explanation for the way by which the rotation of a molecule may weaken its bond through the permission for the kinetic energy of rotation of a molecule has to be stated. The graph for $J=40,80$ and $100$ is to be drawn. The affect on the dissociation energy through the rotation is to be stated.

Concept introduction:

The curve that represents the interatomic interaction model corresponding to the diatomic molecule’s potential energy is known as Morse potential curve. The curve gives the approximation for the vibrational structure of a diatomic molecule.

Answer to Problem 11C.6P

The Morse potential energy curve from $\text{50}\text{pm}$ to $\text{800}\text{pm}$ around $R_e=236.7\text{pm}$ is shown below.

The rotation of a molecule may weaken its bond through the permission for the kinetic energy of rotation of a molecule as the value of $V^{\ast }$ is attained after considering the kinetic energy. So, at high value of $V^{\ast }$, the increase in the value of $R$ is observed.

The graph for $J=40,80$ and $100$ is shown below.

The dissociation energy through the rotation is not affected by rotation.

Explanation of Solution

It is given that in the study of ${}^{\text{85}}{\text{Rb}}^{\text{1}}\text{H}$, the value of wave number is $\tilde{v}=936.8{\text{cm}}^{-1}$ and $x_e\tilde{v}=14.15{\text{cm}}^{-1}$.

The value of $R_e$ is $236.7\text{pm}$.

The value of rotational constant, $\tilde{B}=3.020{\text{cm}}^{-1}$.

The given mass of ${}^{85}\text{Rb}$ is $84.911m_u$.

The value of $1m_u$ is $1.66054\times {10}^{-27}\text{kg}$.

The value of Planck’s constant is $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

The value of speed of light is $2.998\times {10}^{10}\text{cm/s}$.

The expression that to calculate the Morse potential energy is mentioned as follows.

    $V\left(R\right)=hc\widetilde{D_e}{\left\{1-e^{-a\left(R-R_e\right)}\right\}}^2$        (1)

Where,

• $\widetilde{D_e}$ is the molecular partition function.
• $h$ is the Planck’s constant.
• $c$ is the speed of light.
• $R_e$ is the equilibrium bond length.
• $R$ is the bond length.

The expression for calculating the wave number is given below.

    $\begin{array}{c}\tilde{v}=\frac{\omega }{2\pi c}\\ =936.8{\text{cm}}^{-1}\end{array}$        (2)

Where,

• $\omega$ is the Wave number.

The expression for calculating the anharmonicity constant is given below.

    $\begin{array}{c}{x}_e=\frac{\hslash a^2}{2m_{eff}\omega }\\ \frac{\hslash a^2}{2m_{eff}\omega }=\frac{\tilde{v}}{4\widetilde{D_e}}\\ =\frac{\tilde{v}}{4x_e}\end{array}$        (3)

Where,

• $m_{eff}$ is the effective mass.

The expression for calculating the effective mass of $\text{RbH}$ is given below.

    $m_{eff}\left(\text{RbH}\right)=\frac{m_{\text{H}}{m}_{\text{Rb}}}{m_{\text{H}}+m_{\text{Rb}}}$        (4)

Where,

• $m_{\text{H}}$ is the molar mass of hydrogen.
• $m_{\text{Rb}}$ is the molar mass of rubidium.

Substitute $m_{\text{H}}$ as $\text{1}\text{.008}\text{u}$, and $m_{\text{Rb}}$ as $85.47\text{u}$ in the above equation.

    $\begin{array}{c}{m}_{eff}\left(\text{RbH}\right)=\frac{\text{1}\text{.008}\text{u}\times 85.47\text{u}}{\text{1}\text{.008}\text{u}+85.47\text{u}}\\ =\frac{86.153}{86.478}\text{u}\times 1.66\times {10}^{-27}\text{kg}\\ \text{=1}\text{.654}\times {10}^{-27}\text{kg}\end{array}$

Thus, the effective mass is $\text{1}\text{.654}\times {10}^{-27}\text{kg}$.

The expression for calculating the dissociation constant is given below.

    $\widetilde{D_e}=\frac{\overline{v}}{4x_e\tilde{v}}$        (5)

Substitute the values of $\widetilde{v^2}$ as ${\left(936.8{\text{cm}}^{-1}\right)}^2$, and $x_e\tilde{v}=14.15{\text{cm}}^{-1}$ in the above equation.

    $\begin{array}{c}\widetilde{D_e}=\frac{{\left(936.8{\text{cm}}^{-1}\right)}^2}{4\times 14.15{\text{cm}}^{-1}}\\ =\frac{877594.2}{56.6}{\text{cm}}^{-1}\\ =15505{\text{cm}}^{-1}\end{array}$

As $v=c\overline{v}$, so, the value of $a$ is calculated by the expression given below.

    $a=2\pi c\overline{v}\times {\left(\frac{\left(m_{eff}\right)}{2hc\widetilde{D_e}}\right)}^{1/2}$        (6)

Substitute the value of $h$, $c$, $m_{eff}$ and $\widetilde{D_e}$ in the above equation.

    $\begin{array}{c}a=2\times 3.14\times 2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\times 936.8{\text{cm}}^{-1}\\ \times {\left(\frac{\left(\text{1}\text{.654}\times {10}^{-27}\text{kg}\right)}{2\times 15505{\text{cm}}^{-1}\times 6.626\times {10}^{-34}\text{Kg}{\text{m}}^2{\text{s}}^{-2}\times 2.998\times {10}^{10}\text{cm/s}}\right)}^{1/2}\\ =17637.54\times {10}^{10}{\text{s}}^{-1}\times {\left(\frac{\left(\text{1}\text{.654}\times {10}^{-27}\text{kg}\right)}{616005.83\text{Kg}{\text{m}}^2{\text{s}}^{-2}}\right)}^{1/2}\\ =9.140\times {10}^9{\text{m}}^{-1}\end{array}$

Thus, the value of $a$ is $9.140\times {10}^9{\text{m}}^{-1}$. The reciprocal value of $\begin{array}{c}\frac{1}{a}=\frac{1}{9.140\times {10}^9{\text{m}}^{-1}}\\ =0.1094\times {10}^{-9}{\text{m}}^1\end{array}$, In $\text{nm}$, it becomes $a=\frac{1}{0.1094\text{nm}}$.

Substitute the value of $a$ in equation (1).

    $\begin{array}{c}V\left(R\right)=hc\widetilde{D_e}{\left\{1-e^{-\left(R-236.7\text{pm}\right)/0.1094\text{nm}}\right\}}^2\\ \frac{V\left(R\right)}{hc\widetilde{D_e}}={\left\{1-e^{-\left(R-236.7\text{pm}\right)/0.1094\text{nm}}\right\}}^2\end{array}$

The value of $\frac{V\left(R\right)}{hc\widetilde{D_e}}$ at different values of $R$ is given below in the table.

$R/\text{pm}$	$50$	$100$	$200$	$300$	$400$	$500$	$600$	$700$	$800$
$\frac{V\left(R\right)}{hc\widetilde{D_e}}$	$20.4$	$6.20$	$0.159$	$0.193$	$0.601$	$0.828$	$0.929$	$0.971$	$0.988$

The graph between $\frac{V\left(R\right)}{hc\widetilde{D_e}}$ and $R$ is shown below.

Figure 1

The expression for  rotational constant is given below.

    $\tilde{B}\propto \frac{1}{R^2}$

So, the given expression, $V^{\ast }=V+hcBJ\left(J+1\right)$ can be written as follows.

    $V^{\ast }=V+hc{B}_{e}J\left(J+1\right)\left(\frac{R_e^2}{R^2}\right)$

At the value of $B_e$ equal to $3.020{\text{cm}}^{-1}$, $R_e=236.4$ and $R=50\text{pm}$ to $\text{800}\text{pm}$. The different values are given below.

$R_e/\text{R}$	$\frac{V}{hc\widetilde{D_e}}$ (for $J=40,80$ and $100$)	$\left(R_e^2/R^2\right)$	$hc{B}_{e}40{\left(40+1\right)}^{}$	$hc{B}_{e}80{\left(80+1\right)}^{}$	$hc{B}_{e}100{\left(100+1\right)}^{}$
$4.734$	$20.34093148$	$22.410756$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$2.367$	$6.19386711$	$5.602689$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$1.1835$	$0.158875764$	$1.40067225$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$0.789$	$0.193004849$	$0.622521$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$0.59175$	$0.60098496$	$0.3501681$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$0.3945$	$0.92905995$	$0.15563025$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$0.295875$	$0.988423223$	$0.087542016$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$
$0.2367$	$0.99813495$	$0.05602689$	$9.84518\times {10}^{-20}$	$3.89005\times {10}^{-19}$	$6.06319\times {10}^{-19}$

Thus, the graph for $V^{\ast }=V+hcBJ\left(J+1\right)$ with $\tilde{B}=h/4\pi c\mu R^2$ at $J=40,80$ and $100$ is shown below.

Figure 2

The value of $V^{\ast }$ is attained after considering the kinetic energy. Thus, at high value of $V^{\ast }$, the increase in the value of $R$ is observed. In case of weak bond, the bond length increases. So, the increased bond length corresponds to the weaker bond. The weaker bond is the result of the involved kinetic energy.

In the above shown graph, the three lines coincides each other. The equation for rotational constant $B$ is given below.

    $\begin{array}{c}B=\frac{h}{8{\pi }^{2}cI}\\ =B{R}_e^2\end{array}$        (7)

Where,

• $I$ is the moment of inertia equals to $\mu R^2$
• $R$ is the bond length.
• $\mu$ is the reduced mass.

So, the above equation can be written as follows.

    $\frac{h}{8{\pi }^{2}c\mu }=B{R}_e^2$

Substitute the value of $B$ as $3.020{\text{cm}}^{-1}$ and $R_e$ as $236.7\times {10}^{-10}\text{cm}$ in the above equation.

    $\begin{array}{c}\frac{h}{8{\pi }^{2}c\mu }=3.020{\text{cm}}^{-1}\times {\left(236.7\times {10}^{-10}\text{cm}\right)}^2\\ =1.692\times {10}^{-10}\text{cm}\end{array}$

Therefore, the equation (1) can be written as follows.

    $V\left(R\right)=hc{D}_e{\left\{1-e^{-a\left(R-R_e\right)}\right\}}^2+\frac{hc}{R^2}1.692\times {10}^{-10}\text{cm}\left(J+1\right)$

The above equation suggests that a non linear area will be obtained in the curve. Therefore, there will be no effect on the dissociation energy through rotation.