Chapter 11, Problem 11.40E

What are the energies and angular momenta of the first five energy levels of benzene in the $\text{2-D}$ rotational motion approximation? Use the mass of the electron and a radius of $1.51\stackrel{\circ }{\text{A}}$ to determine $I$.

Interpretation Introduction

Interpretation:

The energies and angular momenta of the first five energy levels of benzene in the $\text{2-D}$ rotational motion are to be stated.

Concept introduction:

The wavefunction for $\text{2-D}$ rotational motion is given below.

${\Psi }_m=\frac{1}{\sqrt{2\pi }}{e}^{im\varphi }$

Where, ${\Psi }_m$ represents the wavefunction dependent on quantum number $m$. The values of $m$ start from $0$. Also, it can take up both negative and positive values integral values.

Answer to Problem 11.40E

The energies and angular momenta of the first five energy levels of benzene in the $\text{2-D}$ rotational motion are given below.

$m$	Energy	Angular momentum
$0$	$0$	$0$
$\pm 1$	$2.67\times {10}^{-19}\text{J}$	$\pm 1.05\times {10}^{-34}\text{J}\text{s}$
$\pm 2$	$1.07\times {10}^{-18}\text{J}$	$\pm 2.11\times {10}^{-34}\text{J}\text{s}$
$\pm 3$	$2.41\times {10}^{-18}\text{J}$	$\pm 3.16\times {10}^{-34}\text{J}\text{s}$
$\pm 4$	$4.28\times {10}^{-18}\text{J}$	$\pm 4.22\times {10}^{-34}\text{J}\text{s}$

Explanation of Solution

The formula to calculate energy is given below.

$E=\frac{m^2{h}^2}{8{\pi }^{2}I}$ …(1)

Where,

• $m$ is the quantum number.

• $h$ is Planck’s constant.

• $I$ is moment of inertia.

The formula to calculate moment of inertia is given as follows.

$I=m{r}^2$

Where,

• $m$ is the mass of particle.

• $r$ is the radius.

Substitute the values in the above equation as follows.

$\begin{array}{rll}I& =& m{r}^2\\ & =& \left(9.109\times {10}^{-31}\text{kg}\right){\left(1.51\times {10}^{-10}\text{m}\right)}^2\\ & =& 2.08\times {10}^{-50}\text{kg}{\text{m}}^2\end{array}$

Substitute the values in equation (1) for $m=0$ as follows.

$\begin{array}{rll}{E}_0& =& \frac{m^2{h}^2}{8{\pi }^{2}I}\\ & =& \frac{{\left(0\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times {\left(3.14\right)}^2\times 2.08\times {10}^{-50}\text{kg}{\text{m}}^2}\\ & =& 0\end{array}$

Substitute the values in equation (1) for $m=\pm 1$ as follows.

$\begin{array}{rll}{E}_1& =& \frac{m^2{h}^2}{8{\pi }^{2}I}\\ & =& \frac{{\left(1\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times {\left(3.14\right)}^2\times 2.08\times {10}^{-50}\text{kg}{\text{m}}^2}\\ & =& 2.67\times {10}^{-19}\text{J}\end{array}$

Substitute the values in equation (1) for $m=\pm 2$ as follows.

$\begin{array}{rll}{E}_2& =& \frac{m^2{h}^2}{8{\pi }^{2}I}\\ & =& \frac{{\left(2\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times {\left(3.14\right)}^2\times 2.08\times {10}^{-50}\text{kg}{\text{m}}^2}\\ & =& 1.07\times {10}^{-18}\text{J}\end{array}$

Substitute the values in equation (1) for $m=\pm 3$ as follows.

$\begin{array}{rll}{E}_3& =& \frac{m^2{h}^2}{8{\pi }^{2}I}\\ & =& \frac{{\left(3\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times {\left(3.14\right)}^2\times 2.08\times {10}^{-50}\text{kg}{\text{m}}^2}\\ & =& 2.41\times {10}^{-18}\text{J}\end{array}$

Substitute the values in equation (1) for $m=\pm 4$ as follows.

$\begin{array}{rll}{E}_4& =& \frac{m^2{h}^2}{8{\pi }^{2}I}\\ & =& \frac{{\left(4\right)}^2{\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}^2}{8\times {\left(3.14\right)}^2\times 2.08\times {10}^{-50}\text{kg}{\text{m}}^2}\\ & =& 4.28\times {10}^{-18}\text{J}\end{array}$

The formula to calculate angular momentum is given below.

$L=\frac{mh}{2\pi }$ …(2)

Where,

• $m$ is the quantum number.

• $h$ is Planck’s constant.

Substitute the values in equation (2) for $m=0$ as follows.

$\begin{array}{rll}{L}_0& =& \frac{mh}{2\pi }\\ & =& \frac{\left(0\right)\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}{2\times \left(3.14\right)}\\ & =& 0\end{array}$

Substitute the values in equation (2) for $m=\pm 1$ as follows.

$\begin{array}{rll}{L}_{\pm 1}& =& \frac{mh}{2\pi }\\ & =& \frac{\left(\pm 1\right)\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}{2\times \left(3.14\right)}\\ & =& \pm 1.05\times {10}^{-34}\text{J}\text{s}\end{array}$

Substitute the values in equation (2) for $m=\pm 2$ as follows.

$\begin{array}{rll}{L}_{\pm 2}& =& \frac{mh}{2\pi }\\ & =& \frac{\left(\pm 2\right)\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}{2\times \left(3.14\right)}\\ & =& \pm 2.11\times {10}^{-34}\text{J}\text{s}\end{array}$

Substitute the values in equation (2) for $m=\pm 3$ as follows.

$\begin{array}{rll}{L}_{\pm 3}& =& \frac{mh}{2\pi }\\ & =& \frac{\left(\pm 3\right)\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}{2\times \left(3.14\right)}\\ & =& \pm 3.16\times {10}^{-34}\text{J}\text{s}\end{array}$

Substitute the values in equation (2) for $m=\pm 4$ as follows.

$\begin{array}{rll}{L}_{\pm 4}& =& \frac{mh}{2\pi }\\ & =& \frac{\left(\pm 4\right)\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}\right)}{2\times \left(3.14\right)}\\ & =& \pm 4.22\times {10}^{-34}\text{J}\text{s}\end{array}$

Conclusion

The energies and angular momenta of the first five energy levels of benzene in the $\text{2-D}$ rotational motion are given below.

$m$	Energy	Angular momentum
$0$	$0$	$0$
$\pm 1$	$2.67\times {10}^{-19}\text{J}$	$\pm 1.05\times {10}^{-34}\text{J}\text{s}$
$\pm 2$	$1.07\times {10}^{-18}\text{J}$	$\pm 2.11\times {10}^{-34}\text{J}\text{s}$
$\pm 3$	$2.41\times {10}^{-18}\text{J}$	$\pm 3.16\times {10}^{-34}\text{J}\text{s}$
$\pm 4$	$4.28\times {10}^{-18}\text{J}$	$\pm 4.22\times {10}^{-34}\text{J}\text{s}$