Chapter 43, Problem 63CP

(a)

To determine

The vibrational frequency of ${}^{14}\text{CO}$.

Answer to Problem 63CP

The vibrational frequency of ${}^{14}\text{CO}$ is $6.15\times {10}^{13}\text{Hz}$.

Explanation of Solution

Write the expression for the reduced mass of ${}^{12}\text{CO}$.

  ${\mu }_{12}=\frac{m_{\text{C12}}{m}_{\text{O}}}{m_{\text{C12}}+m_{\text{O}}}$                                                                                                         (I)

Here, ${\mu }_{12}$ is the reduced mass of ${}^{12}\text{CO}$, $m_{\text{C12}}$ is the mass of ${}^{12}\text{C}$ and $m_{\text{O}}$ is the mass of the oxygen atom.

Write the expression for the reduced mass of ${}^{14}\text{CO}$.

  ${\mu }_{14}=\frac{m_{\text{C14}}{m}_{\text{O}}}{m_{C14}+m_{\text{O}}}$                                                                                                        (II)

Here, ${\mu }_{14}$ is the reduced mass of ${}^{14}\text{CO}$, $m_{\text{C14}}$ is the mass of ${}^{14}\text{C}$ and $m_{\text{O}}$ is the mass of the oxygen atom.

Write the expression for the vibrational frequency of ${}^{12}\text{CO}$ molecule.

  $f_{12}=\frac{1}{2\pi }\sqrt{\frac{k}{{\mu }_{12}}}$                                                                                                         (III)

Here, $f_{\text{12}}$ is the vibrational frequency of ${}^{14}\text{CO}$ molecule, and $k$ is the spring constant of the vibrating molecule.

Both ${}^{14}\text{CO}$ and ${}^{14}\text{CO}$ molecules have same interatomic potential, their spring constant is same.

Write the expression for the vibrational frequency of ${}^{14}\text{CO}$ molecule.

  $f_{14}=\frac{1}{2\pi }\sqrt{\frac{k}{{\mu }_{14}}}$                                                                                                       (IV)

Here, $f_{\text{14}}$ is the vibrational frequency of ${}^{14}\text{CO}$ molecule.

Multiply and divide inside the square root of above equation by ${\mu }_{\text{12}}$.

  $\begin{array}{c}{f}_{14}=\frac{1}{2\pi }\sqrt{\frac{k}{{\mu }_{14}}\left(\frac{{\mu }_{12}}{{\mu }_{12}}\right)}\\ =\frac{1}{2\pi }\sqrt{\frac{k}{{\mu }_{12}}}\sqrt{\left(\frac{{\mu }_{12}}{{\mu }_{14}}\right)}\end{array}$

Use equation (II) in above equation to get an expression for $f_{12}$.

  $\begin{array}{c}{f}_{14}=\frac{1}{2\pi }\sqrt{\frac{k}{{\mu }_{14}}\left(\frac{{\mu }_{12}}{{\mu }_{12}}\right)}\\ =f_{12}\sqrt{\left(\frac{{\mu }_{12}}{{\mu }_{14}}\right)}\end{array}$                                                                                               (V)

Conclusion:

In example $43.2$ it is given that frequency of photon that causes the $v=0$ to $v=1$ transition in the molecule is $6.42\times {10}^{13}\text{Hz}$.

Substitute $12\text{u}$ for $m_{\text{C12}}$ and $16\text{u}$ for $m_{\text{O}}$ in equation (I) to get ${\mu }_{\text{12}}$.

  $\begin{array}{c}{\mu }_{12}=\frac{\left(12\text{u}\right)\left(16\text{u}\right)}{\left(12\text{u}\right)+\left(16\text{u}\right)}\text{'}\\ =6.86\text{u}\end{array}$

Substitute $14\text{u}$ for $m_{\text{C12}}$ and $16\text{u}$ for $m_{\text{O}}$ in equation (I) to get ${\mu }_{\text{12}}$.

  $\begin{array}{c}{\mu }_{14}=\frac{\left(14\text{u}\right)\left(16\text{u}\right)}{\left(14\text{u}\right)+\left(16\text{u}\right)}\\ =7.47\text{u}\end{array}$

Substitute $6.42\times {10}^{13}\text{Hz}$ for $f_{12}$, $6.86\text{u}$ for ${\mu }_{12}$ and $7.47\text{u}$ for ${\mu }_{14}$ in equation (V) to get $f_{14}$.

  $\begin{array}{c}{f}_{14}=\left(6.42\times {10}^{13}\text{Hz}\right)\sqrt{\left(\frac{6.86\text{u}}{7.47\text{u}}\right)}\\ =6.15\times {10}^{13}\text{Hz}\end{array}$

Therefore, the vibrational frequency of ${}^{14}\text{CO}$ is $6.15\times {10}^{13}\text{Hz}$.

(b)

To determine

The moment of inertia of ${}^{14}\text{CO}$.

Answer to Problem 63CP

The moment of inertia of ${}^{14}\text{CO}$ is $1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$.

Explanation of Solution

Write the expression for the moment of inertia of ${}^{12}\text{CO}$.

  $I_{\text{12}}={\mu }_{12}{r}^2$                                                                                                              (VI)

Here, $I_{\text{12}}$ is the moment of inertia of the ${}^{12}\text{CO}$ molecule, and $r$ is the atomic separation.

Write the expression for the moment of inertia of ${}^{14}\text{CO}$.

  $I_{\text{14}}={\mu }_{14}{r}^2$                                                                                                              (VII)

Here, $I_{\text{14}}$ is the moment of inertia of the ${}^{14}\text{CO}$ molecule.

Multiply and divide the right hand side of above equation by ${\mu }_{\text{12}}$.

  $I_{\text{14}}=\left(\frac{{\mu }_{14}}{{\mu }_{\text{12}}}\right){\mu }_{\text{12}}{r}^2$                                                                                                  (VIII)

Use equation (VI) in equation (VIII) to get $I_{14}$.

  $I_{\text{14}}=\left(\frac{{\mu }_{14}}{{\mu }_{\text{12}}}\right)I_{\text{12}}$                                                                                                        (IX)

Conclusion:

The moment of inertia of ${}^{12}\text{CO}$ is $1.46\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$.

Substitute $1.46\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$ for $I_{12}$, $6.86\text{u}$ for ${\mu }_{12}$ and $7.47\text{u}$ for ${\mu }_{14}$ in equation (IX) to get $I_{14}$.

  $\begin{array}{c}{I}_{\text{14}}=\left(\frac{7.47\text{u}}{6.86\text{u}}\right)\left(1.46\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\right)\\ =1.56\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\end{array}$

Therefore, the moment of inertia of ${}^{14}\text{CO}$ is $1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$.

(c)

To determine

The wavelength of light that can be absorbed by ${}^{14}\text{CO}$ in the (=$\left(v=0,J=10\right)$ state that cause it end up in the $v=1$ state.

Answer to Problem 63CP

The wavelengths of light that can be absorbed by ${}^{14}\text{CO}$ in the (=$\left(v=0,J=10\right)$ state that cause it end up in the $v=1$ state are $4.96\text{μm}$ and $4.78\text{μm}$.

Explanation of Solution

The molecule lies in $\left(v=0,J=10\right)$ state. Selection rules allows only $\Delta J=\pm 1$. The possible to way to end up in the $v=1$ state is to move into $\left(v=1,J=11\right)$ or $\left(v=1,J=11\right)$ state.

Write the expression for the total energy of the ${}^{14}\text{CO}$ molecule.

  $E_{\nu ,J}=\left(v+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}J\left(J+1\right)$

Here, $E_{\nu ,J}$ is the energy of the molecule in particular vibrational and rotational state, $\nu$ is the vibrational quantum number and $J$ is the rotational quantum number.

Write energy of the molecule in $\left(v=0,J=10\right)$.

  $E_{0,10}=\left(0+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}10\left(10+1\right)$                                                                        (X)

Here, $E_{0,10}$ is the energy of the molecule in $\left(v=0,J=10\right)$.

Write energy of the molecule in $\left(v=1,J=9\right)$.

  $E_{1,9}=\left(1+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}9\left(9+1\right)$                                                                            (XI)

Here, $E_{1,9}$ is the energy of the molecule in $\left(v=1,J=9\right)$.

Write energy of the molecule in $\left(v=1,J=11\right)$.

  $E_{1,11}=\left(1+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}11\left(11+1\right)$                                                                       (XII)

Here, $E_{1,11}$ is the energy of the molecule in $\left(v=1,J=11\right)$.

Write the expression for the change in energy during transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=9\right)$.

  $\Delta E_1=E_{1,9}-E_{0,10}$                                                                                                  (XIII)

Here, $\Delta E_1$ is the change in energy during transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=9\right)$.

Write the expression for the change in energy during transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=11\right)$.

  $\Delta E_2=E_{1,11}-E_{0,10}$                                                                                                (XIV)

Here, $\Delta E_2$ is the change in energy during transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=11\right)$.

Write the expression for the energy of wavelength absorbed during a transition.

  $\Delta E=\frac{hc}{\lambda }$                                                                                                                (XV)

Use equation (X) and (XI) in equation (XIII) to get $\Delta E_1$.

  $\Delta E_1=E_{0,10}-E_{1,9}$

Conclusion:

Use equation (X) and (XI) in equation (XIII) to get $\Delta E_2$.

  $\begin{array}{c}\Delta E_1=\left[\left(1+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}9\left(9+1\right)\right]-\left[\left(0+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}10\left(10+1\right)\right]\\ =h{f}_{14}-20\frac{{\hslash }^2}{2I_{14}}\end{array}$

Substitute $\frac{hc}{\lambda }$ for $\Delta E_1$ in above equation to get wavelength emitted.

  $\begin{array}{c}\frac{hc}{\lambda }=h{f}_{14}-20\frac{{\hslash }^2}{2I_{14}}\\ \frac{c}{\lambda }=f_{14}-10\frac{2\pi {\hslash }^2}{2\pi h{I}_{14}}\\ \lambda =\frac{c}{f_{14}-10\frac{\hslash }{2\pi I_{14}}}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $6.15\times {10}^{13}\text{Hz}$ for $f_{14}$, $1.055\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ and $1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$ for $I_{14}$ in above equation to get wavelength emitted during the transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=9\right)$.

  $\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}/\text{s}}{6.15\times {10}^{13}\text{Hz}-10\frac{\left(1.055\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2\pi \left(1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\right)}}\\ =4.96\times {10}^{-6}\text{m}\times \frac{1\text{μm}}{{10}^{-6}\text{m}}\\ =4.96\text{μm}\end{array}$

Use equation (X) and (XII) in equation (XIV) to get $\Delta E_2$.

  $\begin{array}{c}\Delta E_2=\left[\left(1+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}11\left(11+1\right)\right]-\left[\left(0+\frac{1}{2}\right)h{f}_{14}+\frac{{\hslash }^2}{2I_{14}}10\left(10+1\right)\right]\\ =h{f}_{14}+22\frac{{\hslash }^2}{2I_{14}}\end{array}$

Substitute $\frac{hc}{\lambda }$ for $\Delta E_2$ in above equation to get wavelength emitted.

  $\begin{array}{c}\frac{hc}{\lambda }=h{f}_{14}+22\frac{{\hslash }^2}{2I_{14}}\\ \frac{c}{\lambda }=f_{14}+11\frac{2\pi {\hslash }^2}{2\pi h{I}_{14}}\\ \lambda =\frac{c}{f_{14}+11\frac{\hslash }{2\pi I_{14}}}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $6.15\times {10}^{13}\text{Hz}$ for $f_{14}$, $1.055\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ and $1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2$ for $I_{14}$ in above equation to get wavelength emitted during the transition from $\left(v=0,J=10\right)$ to $\left(v=1,J=11\right)$.

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}/\text{s}}{6.15\times {10}^{13}\text{Hz}+11\frac{\left(1.055\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2\pi \left(1.59\times {10}^{-46}\text{kg}\cdot {\text{m}}^2\right)}}\\ =4.78\times {10}^{-6}\text{m}\times \frac{1\text{μm}}{{10}^{-6}\text{m}}\\ =4.78\text{μm}\end{array}$

Therefore, the wavelengths of light that can be absorbed by ${}^{14}\text{CO}$ in the (=$\left(v=0,J=10\right)$ state that cause it end up in the $v=1$ state are $4.96\text{μm}$ and $4.78\text{μm}$.