Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

Answer to Problem 14P

The moment of inertia is $1.88\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

  $I=6m_c{r}_c^2+6m_h{r}_h^2$

Here, $I$ is the moment of inertia, $m_c$ is the mass of carbon atom, $r_c$ is the distance from carbon atom to point O, $m_h$ is the mass of hydrogen atom, and $r_h$ is the distance from hydrogen atom to point O.

Conclusion:

Substitute $1.99\times {10}^{-26}\text{kg}$ for $m_c$, $0.110\text{nm}$ for $r_c$, $1.67\times {10}^{-27}\text{kg}$ for $m_h$, and $0.210\text{nm}$ for $r_h$ in the above equation to find $I$.

  $\begin{array}{c}I=6\left(1.99\times {10}^{-26}\text{kg}\right){\left(0.110\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2+6\left(1.67\times {10}^{-27}\text{kg}\right){\left(0.210\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2\\ =1.44\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}+0.442\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}\\ =1.88\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Therefore, the moment of inertia is $1.88\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}$.

(b)

To determine

The possible rotational energies about the axis.

Answer to Problem 14P

The possible rotational energies about the axis are $0\text{μeV,}36.9\text{μeV, 111}\text{μeV, 221}\text{μeV, and 369}\text{μeV}$.

Explanation of Solution

Write the equation to calculate the rotational energy.

  $E_{\text{rot}}=\frac{{\hslash }^2}{2I}J\left(J+1\right)$                                                                                               (I)

Here, $E_{\text{rot}}$ is the rotational energy, $\hslash$ is the reduced Plank’s constant, $I$ is the rotational moment of inertia, and $J$ is the rotational quantum umber.

Conclusion:

Substitute $1.055\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ and $1.88\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}$ for $I$ in the equation for $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\frac{{\left(1.055\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{2\left(1.88\times {10}^{-45}\text{ kg}\cdot {\text{m}}^{\text{2}}\right)}J\left(J+1\right)\\ =\left(2.95\times {10}^{-24}\text{ J}\left(\frac{1.6\times {10}^{-19}\text{eV}}{1\text{J}}\right)\right)J\left(J+1\right)\\ =\left(18.4\times {10}^{-6}\text{ eV}\right)J\left(J+1\right)\end{array}$

The possible values of $J$ are $0,1,2,3,\mathrm{4.....}$.

Substitute $0$ for $J$ in the above equation to find $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\left(18.4\times {10}^{-6}\text{ eV}\right)\left(0\left(0+1\right)\right)\\ =0\text{eV}\end{array}$

Substitute $1$ for $J$ in the above equation to find $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\left(18.4\times {10}^{-6}\text{ eV}\right)\left(1\left(1+1\right)\right)\\ =36.9\times {10}^{-6}\text{eV}\left(\frac{1\text{μeV}}{{10}^{-6}\text{eV}}\right)\\ =36.9\text{μeV}\end{array}$

Substitute $2$ for $J$ in the above equation to find $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\left(18.4\times {10}^{-6}\text{ eV}\right)\left(2\left(2+1\right)\right)\\ =111\times {10}^{-6}\text{eV}\left(\frac{1\text{μeV}}{{10}^{-6}\text{eV}}\right)\\ =111\text{μeV}\end{array}$

Substitute $3$ for $J$ in the above equation to find $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\left(18.4\times {10}^{-6}\text{ eV}\right)\left(3\left(3+1\right)\right)\\ =221\times {10}^{-6}\text{eV}\left(\frac{1\text{μeV}}{{10}^{-6}\text{eV}}\right)\\ =221\text{μeV}\end{array}$

Substitute $4$ for $J$ in the above equation to find $E_{\text{rot}}$.

  $\begin{array}{c}{E}_{\text{rot}}=\left(18.4\times {10}^{-6}\text{ eV}\right)\left(4\left(4+1\right)\right)\\ =369\times {10}^{-6}\text{eV}\left(\frac{1\text{μeV}}{{10}^{-6}\text{eV}}\right)\\ =369\text{μeV}\end{array}$

Therefore, the possible rotational energies about the axis are $0\text{μeV,}36.9\text{μeV, 111}\text{μeV, 221}\text{μeV, and 369}\text{μeV}$.