Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

Answer to Problem 31P

The moment of inertia is $1.454\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$ and rotational energy is $0.2385\text{meV}$ .

Explanation of Solution

Given:

The equilibrium separation is $r=0.113\text{nm}$ .

Formula used:

The expression for moment of inertia is given by,

  $I=\frac{m_{\text{C}}{m}_{\text{O}}{r}^2}{m_{\text{C}}+m_{\text{O}}}$

The expression for rotational energy is given by,

  $E_{\text{0r}}=\frac{{\hslash }^2}{2I}$

Calculation:

The moment of inertiais calculated as,

  $\begin{array}{c}I=\frac{m_{\text{C}}{m}_{\text{O}}{r}^2}{m_{\text{C}}+m_{\text{O}}}\\ =\frac{\left(12\text{u}\right)\left(16\text{u}\right)}{\left(12\text{u}\right)+\left(16\text{u}\right)}{\left(\left(0.113\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2\left(1.661\times {10}^{-27}\text{kg}/\text{u}\right)\\ =1.454\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\end{array}$

The rotational energy is calculated as,

  $\begin{array}{c}{E}_{\text{0r}}=\frac{{\hslash }^2}{2I}\\ =\frac{\left(6.58\times {10}^{-16}\text{eV}\cdot \text{s}\right)\left(\frac{1.602\times {10}^{-19}\text{J}}{\text{1}\text{eV}}\right)}{2\left(1.454\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\right)}\\ =\left(\left(0.2385\times {10}^{-3}\text{eV}\right)\left(\frac{{10}^3\text{meV}}{1\text{eV}}\right)\right)\\ =0.2385\text{meV}\end{array}$

Conclusion:

Therefore, the moment of inertia is $1.454\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$ and rotational energy is $0.2385\text{meV}$ .

(b)

To determine

The energy level diagram.

Answer to Problem 31P

The energy level diagram is shown in figure 1.

Explanation of Solution

Calculation:

The energy level diagram for the rotational level from $\mathcal{l}=0\text{to}\mathcal{l}=5$ is shown below,

  

Figure 1

Conclusion:

Therefore, the energy level diagram is shown in figure 1.

(c)

To determine

The wavelength for each transition of part (b).

Answer to Problem 31P

The wavelengths for each transition from start are $2.6\times {10}^{-3}\text{m}$ , $1.3\times {10}^{-3}\text{m}$ , $0.867\times {10}^{-3}\text{m}$ , $0.650\times {10}^{-3}\text{m}$ and $0.52\times {10}^{-3}\text{m}$ .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

  ${\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}$

Calculation:

The first wavelength is calculated as,

  $\begin{array}{c}{\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}\\ {\lambda }_{1,0}=\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}{2\left(1\right)\left(\left(0.2385\text{meV}\right)\left(\frac{{10}^{-3}\text{eV}}{1\text{meV}}\right)\right)}\\ =2.6\times {10}^{-3}\text{m}\end{array}$

The second wavelength is calculated as,

  $\begin{array}{c}{\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}\\ {\lambda }_{2,1}=\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}{2\left(2\right)\left(\left(0.2385\text{meV}\right)\left(\frac{{10}^{-3}\text{eV}}{1\text{meV}}\right)\right)}\\ =1.3\times {10}^{-3}\text{m}\end{array}$

The third wavelength is calculated as,

  $\begin{array}{c}{\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}\\ {\lambda }_{3,2}=\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}{2\left(3\right)\left(\left(0.2385\text{meV}\right)\left(\frac{{10}^{-3}\text{eV}}{1\text{meV}}\right)\right)}\\ =0.867\times {10}^{-3}\text{m}\end{array}$

The fourth wavelength is calculated as,

  $\begin{array}{c}{\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}\\ {\lambda }_{4,3}=\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}{2\left(4\right)\left(\left(0.2385\text{meV}\right)\left(\frac{{10}^{-3}\text{eV}}{1\text{meV}}\right)\right)}\\ =0.650\times {10}^{-3}\text{m}\end{array}$

The fifth wavelength is calculated as,

  $\begin{array}{c}{\lambda }_{\mathcal{l},\mathcal{l}-1}=\frac{hc}{2\mathcal{l}\Delta E_{\text{0r}}}\\ {\lambda }_{5,4}=\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}{2\left(5\right)\left(\left(0.2385\text{meV}\right)\left(\frac{{10}^{-3}\text{eV}}{1\text{meV}}\right)\right)}\\ =0.52\times {10}^{-3}\text{m}\end{array}$

Conclusion:

Therefore, the wavelengths for each transition from start are $2.6\times {10}^{-3}\text{m}$ , $1.3\times {10}^{-3}\text{m}$ , $0.867\times {10}^{-3}\text{m}$ , $0.650\times {10}^{-3}\text{m}$ and $0.52\times {10}^{-3}\text{m}$ .