Chapter 12, Problem 12.56E

Interpretation Introduction

Interpretation:

The value of ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ as well as $E_1$ and $E_2$ for $R=1.00$, $1.15$, $1.45$, and $1.60\stackrel{\circ }{\text{A}}$ is to be determined. These values are to be combined with the determinations from Example 12.14 and a simple potential energy diagram for this system is to be constructed.

Concept introduction:

The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ are given as follows:

$\begin{array}{rll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2S_{12}}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2S_{12}}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$

Answer to Problem 12.56E

The values of ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ as well as $E_1$ and $E_2$ for $R=1.00$, $1.15$, $1.45$, and $1.60\stackrel{\circ }{\text{A}}$ is given below.

$R\left(\stackrel{\text{o}}{\text{A}}\right)$	${\varphi }_{{\text{H}}_2^{+},1}$	${\varphi }_{{\text{H}}_2^{+},2}$	$E_1$	$E_2$
$1.00$	$0.558\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$1.141\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.23$	$3.83$
$1.15$	$0.57\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$1.03\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.61$	$4.21$
$1.45$	$0.59\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$0.91\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.75$	$3.21$
$1.60$	$0.61\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$0.87\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.66$	$2.60$

The simple potential energy diagram for this system is shown in Figure 1 and Figure 2.

Explanation of Solution

The value of $S_{12}$ is calculated as follows:

$S_{12}=e^{-R/a_0}\left(1+\frac{R}{a_0}+\frac{R^2}{3a_0^2}\right)$ …$\left(1\right)$

Where,

• $a_0$ is a Bohr’s radius $0.529\stackrel{\circ }{\text{A}}$.

• $R$ is the distance between two nuclei.

The value of $H_{12}$ is calculated as follows:

$H_{12}=K\left[\frac{a_0{S}_{12}}{R}-e^{-R/a_0}\left(1+\frac{R}{a_0}-\frac{S_{12}}{2}\right)\right]$ …$\left(2\right)$

Where,

• $K$ is the conversion factor from atomic to SI units.

The value of $E_1$ and $E_2$ are given by the formula,

$\begin{array}{rll}{E}_1& =& \frac{H_{11}+H_{12}}{1+S_{12}}\\ E_2& =& \frac{H_{22}-H_{12}}{1-S_{12}}\end{array}$ …$\left(3\right)$

The wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ is given as follows:

$\begin{array}{rll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2S_{12}}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2S_{12}}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$ …$\left(4\right)$

The value of $R=1.00\stackrel{\circ }{\text{A}}$, $K=27.09\text{eV}$ and $H_{11}=-0.258\text{eV}$ is given.

Substitute the value of $R$ and $K$ in the equation $\left(1\right)$.

$\begin{array}{rll}{S}_{12}& =& e^{-1.00\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.00\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}+\frac{{\left(1.00\stackrel{\circ }{\text{A}}\right)}^2}{3{\left(0.529\stackrel{\circ }{\text{A}}\right)}^2}\right)\\ & =& 0.151\left(1+1.89+1.194\right)\\ & =& 0.151\times 4.084\\ & =& 0.616\end{array}$

Thus, the value of $S_{12}$ is $0.616$.

Substitute the value of $R$, $S_{12}$ and $K$ in the equation $\left(2\right)$.

$\begin{array}{rll}{H}_{12}& =& 27.09\text{eV}\left[\frac{0.529\stackrel{\circ }{\text{A}}\times 0.616}{1.00\stackrel{\circ }{\text{A}}}-e^{-1.00\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.00\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}-\frac{0.616}{2}\right)\right]\\ & =& 27.09\text{eV}\left[0.325-0.151\left(1+1.89-0.308\right)\right]\\ & =& 27.09\text{eV}\left[0.325-0.151\times 2.582\right]\\ & =& -1.73\text{eV}\end{array}$

Thus, the value of $H_{12}$ is $-1.73\text{eV}$.

Substitute the value of $H_{12}$, $S_{12}$ and $H_{11}$ in the equation $\left(3\right)$.

$\begin{array}{rllll}{E}_1& =& \frac{-0.258+\left(-1.73\right)}{1+0.616}& =& -1.230\text{eV}\\ E_2& =& \frac{-0.258-\left(-1.73\right)}{1-0.616}& =& 3.833\text{eV}\end{array}$

Thus, the value of $E_1$ and $E_2$ are $-1.230\text{eV}$ and $3.833\text{eV}$.

Substitute the value of $S_{12}$ in the equation $\left(4\right)$.

$\begin{array}{rllll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2\times 0.616}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.558\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2\times 0.616}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)& =& 1.141\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$

Thus, the wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ is $0.558\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$ and $1.141\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$.

The value of $R=1.15\stackrel{\circ }{\text{A}}$, $K=27.09\text{eV}$ and $H_{11}=-0.258\text{eV}$ is given.

Substitute the value of $R$ and $K$ in the equation $\left(1\right)$.

$\begin{array}{rll}{S}_{12}& =& e^{-1.15\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.15\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}+\frac{{\left(1.15\stackrel{\circ }{\text{A}}\right)}^2}{3{\left(0.529\stackrel{\circ }{\text{A}}\right)}^2}\right)\\ & =& 0.113\left(1+2.17+1.57\right)\\ & =& 0.535\end{array}$

Thus, the value of $S_{12}$ is $0.535$.

Substitute the value of $R$, $S_{12}$ and $K$ in the equation $\left(2\right)$.

$\begin{array}{rll}{H}_{12}& =& 27.09\text{eV}\left[\frac{0.529\stackrel{\circ }{\text{A}}\times 0.535}{1.15\stackrel{\circ }{\text{A}}}-e^{-1.15\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.15\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}-\frac{0.535}{2}\right)\right]\\ & =& 27.09\text{eV}\left[0.246-0.113\left(1+2.173-0.267\right)\right]\\ & =& -2.22\text{eV}\end{array}$

Thus, the value of $H_{12}$ is $-2.22\text{eV}$.

Substitute the value of $H_{12}$, $S_{12}$ and $H_{11}$ in the equation $\left(3\right)$.

$\begin{array}{rllll}{E}_1& =& \frac{-0.258+\left(-2.22\text{eV}\right)}{1+0.535}& =& -1.61\text{eV}\\ E_2& =& \frac{-0.258-\left(-2.22\text{eV}\right)}{1-0.535}& =& 4.21\text{eV}\end{array}$

Thus, the value of $E_1$ and $E_2$ are $-1.61\text{eV}$ and $4.21\text{eV}$.

Substitute the value of $S_{12}$ in the equation $\left(4\right)$.

$\begin{array}{rllll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2\times 0.535}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.57\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2\times 0.535}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)& =& 1.03\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$

Thus, the wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ is $0.57\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$ and $1.03\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$.

The value of $R=1.45\stackrel{\circ }{\text{A}}$, $K=27.09\text{eV}$ and $H_{11}=-0.258\text{eV}$ is given.

Substitute the value of $R$ and $K$ in the equation $\left(1\right)$.

$\begin{array}{rll}{S}_{12}& =& e^{-1.45\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.45\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}+\frac{{\left(1.45\stackrel{\circ }{\text{A}}\right)}^2}{3{\left(0.529\stackrel{\circ }{\text{A}}\right)}^2}\right)\\ & =& 0.064\left(1+2.74+2.50\right)\\ & =& 0.399\end{array}$

Thus, the value of $S_{12}$ is $0.399$.

Substitute the value of $R$, $S_{12}$ and $K$ in the equation $\left(2\right)$.

$\begin{array}{rll}{H}_{12}& =& 27.09\text{eV}\left[\frac{0.529\stackrel{\circ }{\text{A}}\times 0.399}{1.45\stackrel{\circ }{\text{A}}}-e^{-1.45\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.45\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}-\frac{0.399}{2}\right)\right]\\ & =& 27.09\text{eV}\left[0.145-0.064\left(1+2.741-0.1995\right)\right]\\ & =& -2.19\text{eV}\end{array}$

Thus, the value of $H_{12}$ is $-2.19\text{eV}$.

Substitute the value of $H_{12}$, $S_{12}$ and $H_{11}$ in the equation $\left(3\right)$.

$\begin{array}{rllll}{E}_1& =& \frac{-0.258+\left(-2.19\right)}{1+0.399}& =& -1.75\text{eV}\\ E_2& =& \frac{-0.258-\left(-2.19\right)}{1-0.399}& =& 3.21\text{eV}\end{array}$

Thus, the value of $E_1$ and $E_2$ are $-1.75\text{eV}$ and $3.21\text{eV}$.

Substitute the value of $S_{12}$ in the equation $\left(4\right)$.

$\begin{array}{rllll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2\times 0.399}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.59\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2\times 0.399}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.91\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$

Thus, the wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ is $0.59\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$ and $0.91\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$.

The value of $R=1.60\stackrel{\circ }{\text{A}}$, $K=27.09\text{eV}$ and $H_{11}=-0.258\text{eV}$ is given.

Substitute the value of $R$ and $K$ in the equation $\left(1\right)$.

$\begin{array}{rll}{S}_{12}& =& e^{-1.60\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.60\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}+\frac{{\left(1.60\stackrel{\circ }{\text{A}}\right)}^2}{3{\left(0.529\stackrel{\circ }{\text{A}}\right)}^2}\right)\\ & =& 0.048\left(1+3.02+3.05\right)\\ & =& 0.339\end{array}$

Thus, the value of $S_{12}$ is $0.339$.

Substitute the value of $R$, $S_{12}$ and $K$ in the equation $\left(2\right)$.

$\begin{array}{rll}{H}_{12}& =& 27.09\text{eV}\left[\frac{0.529\stackrel{\circ }{\text{A}}\times 0.339}{1.60\stackrel{\circ }{\text{A}}}-e^{-1.60\stackrel{\circ }{\text{A}}/0.529\stackrel{\circ }{\text{A}}}\left(1+\frac{1.60\stackrel{\circ }{\text{A}}}{0.529\stackrel{\circ }{\text{A}}}-\frac{0.339}{2}\right)\right]\\ & =& 27.09\text{eV}\left[0.112-0.048\left(1+3.02-0.16\right)\right]\\ & =& -1.977\text{eV}\end{array}$

Thus, the value of $H_{12}$ is $-1.977\text{eV}$.

Substitute the value of $H_{12}$, $S_{12}$ and $H_{11}$ in the equation $\left(3\right)$.

$\begin{array}{rllll}{E}_1& =& \frac{-0.258+\left(-1.977\text{eV}\right)}{1+0.339}& =& -1.66\text{eV}\\ E_2& =& \frac{-0.258-\left(-1.977\text{eV}\right)}{1-0.339}& =& 2.60\text{eV}\end{array}$

Thus, the value of $E_1$ and $E_2$ are $-1.66\text{eV}$ and $2.60\text{eV}$.

Substitute the value of $S_{12}$ in the equation $\left(4\right)$.

$\begin{array}{rllll}{\varphi }_{{\text{H}}_2^{+},1}& =& \frac{1}{\sqrt{2+2\times 0.339}}\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.61\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)\\ {\varphi }_{{\text{H}}_2^{+},2}& =& \frac{1}{\sqrt{2-2\times 0.339}}\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)& =& 0.87\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)\end{array}$

Thus, the wavefunction ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ is $0.61\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$ and $0.87\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$.

The values of potential energy and $R$ is given below.

$R\left(\stackrel{\text{o}}{\text{A}}\right)$	Energy $\left(\text{eV}\right)$
$1.00$	$3.83$
$1.15$	$4.21$
$1.32$	$16.23$
$1.45$	$3.21$
$1.60$	$2.60$

The plot between the potential energy with $R$ is shown below.

Figure 1

The values of potential energy and $R$ is given below.

$R\left(\stackrel{\text{o}}{\text{A}}\right)$	Energy $\left(\text{eV}\right)$
$1.00$	$-1.23$
$1.15$	$-1.61$
$1.32$	$-6.37$
$1.45$	$-1.75$
$1.60$	$-1.66$

The plot between the potential energy with $R$ is shown below.

Figure 2

Conclusion

The values of ${\varphi }_{{\text{H}}_2^{+},1}$ and ${\varphi }_{{\text{H}}_2^{+},2}$ as well as $E_1$ and $E_2$ for $R=1.00$, $1.15$, $1.45$, and $1.60\stackrel{\circ }{\text{A}}$ is given below.

$R\left(\stackrel{\text{o}}{\text{A}}\right)$	${\varphi }_{{\text{H}}_2^{+},1}$	${\varphi }_{{\text{H}}_2^{+},2}$	$E_1$	$E_2$
$1.00$	$0.558\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$1.141\left({\Psi }_{\text{H}\left(1\right)}-{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.23$	$3.83$
$1.15$	$0.57\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$1.03\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.61$	$4.21$
$1.45$	$0.59\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$0.91\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.75$	$3.21$
$1.60$	$0.61\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$0.87\left({\Psi }_{\text{H}\left(1\right)}+{\Psi }_{\text{H}\left(2\right)}\right)$	$-1.66$	$2.60$

The simple potential energy diagram for this system is shown in Figure 1 and Figure 2.