Chapter 14, Problem 14.2IA

(a)

Interpretation Introduction

Interpretation:

The Lennard-Jones parameters $r_0$ and $\epsilon$ has to be calculated for ${\text{He}}_{\text{2}}$.  The plot of Lennard-Jones potential for $\text{He}-\text{He}$ has to be drawn.

Concept introduction:

Two neutral atoms or molecule first experience an attractive force when they come in contact with each other.  After a certain distance these particles experience repulsion force.  This type of interaction can be represented by Lennard-Jones potential equation.  The Lennard-Jones potential energy $\left(V\right)$ is given by the expression as shown below.

    $V=4\epsilon \left\{{\left(\frac{r_0}{r}\right)}^{12}-{\left(\frac{r_0}{r}\right)}^6\right\}$

Answer to Problem 14.2IA

The Lennard-Jones parameters $r_0$ and $\epsilon$ for ${\text{He}}_{\text{2}}$ are $\underset{\_}{264.5969pm}$ and $\underset{\_}{1.51\times 10^{-23}J}$ respectively.

The plot of Lennard-Jones potential for $\text{He}-\text{He}$ is shown below.

Explanation of Solution

The Lennard-Jones potential energy $\left(V\right)$ is given by the expression as shown below.

    $V=4\epsilon \left\{{\left(\frac{r_0}{r}\right)}^{12}-{\left(\frac{r_0}{r}\right)}^6\right\}$        (1)

Where,

• $\epsilon$ is the dept of the well.
• $r$ is the distance between the distance between the atoms.
• $r_0$ is the distance between the atoms at $V=0$.

The value of $hc{\tilde{D}}_{\text{e}}$ is $1.51\times {10}^{-23}\text{ J}$.

The value of $hc{\tilde{D}}_{\text{0}}$ is $2\times {10}^{-26}\text{ J}$.

The value of distance between the atoms at well minima $\left(R_{\text{e}}\right)$ is $297\text{ pm}$.

The dept of the well is give by the expression as shown below.

    $\epsilon =hc{\tilde{D}}_{\text{e}}$

Where,

• $h$ is the Plank’s constant.
• $c$ is the speed of light.
• ${\tilde{D}}_{\text{e}}$ is the wavelength corresponding to the dissection of molecule.

Substitute the value of $hc{\tilde{D}}_{\text{e}}$ in the above equation.

    $\epsilon =\underset{\_}{1.51\times 10^{-23}J}$

The distance between the atoms at which potential energy is zero is given by the expression as shown below.

    $r_0=2^{-1/6}{R}_{\text{e}}$

Substitute the value of $R_{\text{e}}$ in the above equation.

    $\begin{array}{c}{r}_0=\left(297\text{ pm}\right)2^{-1/6}\\ =\underset{\_}{264.5969pm}\end{array}$

Therefore, the Lennard-Jones parameters $r_0$ and $\epsilon$ for ${\text{He}}_{\text{2}}$ are $\underset{\_}{264.5969pm}$ and $\underset{\_}{1.51\times 10^{-23}J}$ respectively.

The Lennard-Jones potential equation for ${\text{He}}_{\text{2}}$ is written as shown below.

    $V=4\left(1.51\times {10}^{-23}\text{ J}\right)\left\{{\left(\frac{264.5969\text{ pm}}{r}\right)}^{12}-{\left(\frac{264.5969\text{ pm}}{r}\right)}^6\right\}$        (2)

Substitute the value of $r=0\text{ pm}$ in the above equation.

    $\begin{array}{c}V=4\left(1.51\times {10}^{-23}\text{ J}\right)\left\{{\left(\frac{264.5969\text{ pm}}{0\text{ pm}}\right)}^{12}-{\left(\frac{264.5969\text{ pm}}{\text{0 pm}}\right)}^6\right\}\\ =\infty \end{array}$

Similarly, the Lennard-Jones potential energy is from $r=0\text{ pm}$ to $r=600\text{ pm}$ by the use of equation (2). The calculated values of Lennard-Jones potential energy is shown below in the table.

$r\left(\text{pm}\right)$	Lennard-Jones potential energy $\left(\text{J}\right)$
$0\text{ pm}$	$\infty \text{ J}$
$\text{100 pm}$	$7.09\times {10}^{-18}\text{ J}$
$200\text{ pm}$	$1.41\times {10}^{-21}\text{ J}$
$300\text{ pm}$	$-1.5\times {10}^{-23}\text{ J}$
$400\text{ pm}$	$-4.6\times {10}^{-24}\text{ J}$
$500\text{ pm}$	$-1.3\times {10}^{-24}\text{ J}$
$600\text{ pm}$	$-4.4\times {10}^{-25}\text{ J}$

The plot of Lennard-Jones potential for $\text{He}-\text{He}$ is shown below.

Figure 1

(b)

Interpretation Introduction

Interpretation:

The plot of Morse potential for for $\text{He}-\text{He}$ is to be drawn.

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 14.2IA

The plot of Morse potential for $\text{He}-\text{He}$ is shown below.

Explanation of Solution

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

    $V\left(R\right)=hc\widetilde{D_{\text{e}}}{\left\{1-e^{-a\left(R-R_{\text{e}}\right)}\right\}}^2$        (3)

Where,

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

The value of $hc{\tilde{D}}_{\text{e}}$ is $1.51\times {10}^{-23}\text{ J}$.

The value of $hc{\tilde{D}}_{\text{0}}$ is $2\times {10}^{-26}\text{ J}$.

The value of distance between the atoms at well minima $\left(R_{\text{e}}\right)$ is $297\text{ pm}$.

The value of $a$ is $5.79\times {10}^{10}{\text{ m}}^{-1}$.

The conversion of $a$ in ${\text{cm}}^{-1}$ is shown below.

    $\begin{array}{c}a=\left(5.79\times {10}^{10}{\text{ m}}^{-1}\right)\left(\frac{1\text{ m}}{{10}^{12}\text{ pm}}\right)\\ =5.79\times {10}^{-2}{\text{ pm}}^{-1}\end{array}$

Substitute the values of $hc{\tilde{D}}_{\text{e}}$, $R_{\text{e}}$, and $a$ in the above equation.

    $V\left(R\right)=\left(1.51\times {10}^{-23}\text{ J}\right){\left\{1-e^{-\left(5.79\times {10}^{-2}{\text{ pm}}^{-1}\right)\left(R-297\text{ pm}\right)}\right\}}^2$        (4)

Substitute the value of $r=0\text{ pm}$ in the above equation.

    $\begin{array}{c}V\left(R\right)=\left(1.51\times {10}^{-23}\text{ J}\right){\left\{1-e^{-\left(5.79\times {10}^{-2}{\text{ pm}}^{-1}\right)\left(0\text{ pm}-297\text{ pm}\right)}\right\}}^2\\ =\left(1.51\times {10}^{-23}\text{ J}\right){\left\{1-29393966.79\right\}}^2\\ =\left(1.51\times {10}^{-23}\text{ J}\right){\left\{-29393965.79\right\}}^2\\ =1.3046\times {10}^{-8}\text{ J}\end{array}$

Similarly, the Morse potential energy is from $r=100\text{ pm}$ to $r=600\text{ pm}$ by the use of equation (4). The calculated values of Morse potential energy is shown below in the table.

$r\left(\text{pm}\right)$	Morse potential energy $\left(\text{J}\right)$
$\text{100 pm}$	$1.21998\times {10}^{-13}\text{ J}$
$200\text{ pm}$	$1.13258\times {10}^{-18}\text{ J}$
$300\text{ pm}$	$3.83912\times {10}^{-25}\text{ J}$
$400\text{ pm}$	$1.50225\times {10}^{-23}\text{ J}$
$500\text{ pm}$	$1.50998\times {10}^{-23}\text{ J}$
$600\text{ pm}$	$1.51\times {10}^{-23}\text{ J}$

The plot of Morse potential for $\text{He}-\text{He}$ is shown below.

Figure 2