Chapter 43, Problem 19P

(a)

To determine

The longest wavelength in the rotational spectrum of the HCl molecule.

Answer to Problem 19P

The longest wavelength in the rotational spectrum of the HCl molecule is $472\text{μm}$.

Explanation of Solution

The longest wavelength for a rotational spectrum of a molecule will be obtained if the transition occurs between $J=0$ and $J=1$ state.

Write expression for the energy of rotational levels.

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

Here, $E_J$ is the energy of rotational levels, $\hslash$ is the reduced Planck’s constant, $J$ is the quantum number of rotational energy levels and $I$ is the moment of inertia of molecule.

Write the expression for the energy difference between $J=0$ and $J=1$ states.

  $\Delta E_{\min }=E_1-E_0$                                                                                                         (II)

Here, $\Delta E_{\text{min}}$ is the energy of the minimum energy transition, $E_1$ is the energy of $J=1$ level and $E_0$ is the energy of $J=0$ state.

Write the expression for the change in energy levels in terms of wavelength of photon emitted.

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

Here, $\Delta E$ is the change in energy levels, $c$ is the speed of light in vacuum and $\lambda$ is the wavelength of photon emitted.

Rearrange above two equation to get $\lambda$.

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

Write the expression for the reduced mass of $\text{HCl}$ molecule.

  $\mu =\frac{m_{\text{H}}{m}_{\text{Cl}}}{m_{\text{H}}+m_{\text{Cl}}}$                                                                                                           (V)

Here, $\mu$ is the reduced mass $\text{HCl}$ molecule, $m_{\text{H}}$ is the mass of Hydrogen atom and $m_{\text{Cl}}$ is the mass of Chlorine atom.

Write the expression for the moment of inertia of $\text{HCl}$ molecule.

  $I=\mu r^2$                                                                                                                    (VI)

Here, $r$ is the equilibrium separation of the H and Cl atom.

Conclusion:

The mass of H atom is $1.007825\text{u}$, mass of ${}^{35}\text{Cl}$ atom is $34.968853\text{u}$ and equilibrium separation of atoms is $0.12746\text{nm}$.

Substitute $0$ for $J$ in equation (I) to get $E_0$.

  $\begin{array}{c}{E}_0=\frac{{\hslash }^2}{2I}0\left(0+1\right)\\ =0\end{array}$

Substitute $1$ for $J$ in equation (I) to get $E_1$.

  $\begin{array}{c}{E}_1=\frac{{\hslash }^2}{2I}1\left(1+1\right)\\ =\frac{{\hslash }^2}{I}\end{array}$

Substitute $0$ for $E_0$ and $\frac{{\hslash }^2}{I}$ for $E_1$ in (I) to get $\Delta E_{\text{min}}$.

  $\begin{array}{c}\Delta E_{\min }=\frac{{\hslash }^2}{I}-0\\ =\frac{{\hslash }^2}{I}\end{array}$

Substitute $\frac{{\hslash }^2}{I}$ for $\Delta E$ in (IV) to get $\lambda$.

  $\begin{array}{c}\lambda =\frac{hc}{\frac{{\left(\frac{h}{2\pi }\right)}^2}{I}}\\ =\frac{4{\pi }^{2}Ic}{h}\end{array}$                                                                                                           (VII)

Substitute $1.007825\text{u}$ for $m_{\text{H}}$ and $34.968853\text{u}$ for $m_{\text{Cl}}$ equation (V) to get reduced mass of HCL with Cl atom to be isotope of ${}^{35}\text{Cl}$.

  $\begin{array}{c}{\mu }_{\text{35}}=\frac{\left(1.007825\text{u}\right)\left(34.968853\text{u}\right)}{\left(1.007825\text{u}\right)+\left(34.968853\text{u}\right)}\times \left(\frac{1.660540\times {10}^{-27}\text{kg}}{\text{u}}\right)\\ =1.626653\times {10}^{-27}\text{kg}\end{array}$

Here, ${\mu }_{\text{35}}$ is the reduced mass of HCL with Cl atom to be isotope of ${}^{35}\text{Cl}$.

Substitute $1.626653\times {10}^{-27}\text{kg}$ for $\mu$ and $0.12746\text{nm}$ for $r$ in equation (VI) to get $I$.

  $\begin{array}{c}I=\left(1.626653\times {10}^{-27}\text{kg}\right){\left(0.12746\text{nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}^2\\ =2.6426688\times {10}^{-47}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $6.626075\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/{\text{s}}^2$ for $c$ and $2.6426688\times {10}^{-47}\text{kg}\cdot {\text{m}}^2$ for $I$ in equation (VII) to get $\lambda$.

  $\begin{array}{c}\lambda =\frac{4{\pi }^2\left(2.6426688\times {10}^{-47}\text{kg}\cdot {\text{m}}^2\right)\left(3\times {10}^8\text{m}/{\text{s}}^2\right)}{6.626075\times {10}^{-34}\text{J}\cdot \text{s}}\\ =4.718749601\times {10}^{-4}\text{m}\times \frac{1\text{m}}{{10}^6\text{μm}}\\ =4.72\times {10}^{-4}\text{m}\times \frac{1\text{m}}{{10}^6\text{μm}}\\ =472\text{μm}\end{array}$

Therefore, the longest wavelength in the rotational spectrum of the HCl molecule is $472\text{μm}$.

(b)

To determine

The longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope ${}^{37}\text{Cl}$.

Answer to Problem 19P

The longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope ${}^{37}\text{Cl}$ is $473\text{μm}$.

Explanation of Solution

Use equation (V) to get reduced mass of HCl, equation (VI) to get moment of inertia of HCl molecule and (VII) to get $\lambda$.

Conclusion:

Substitute $1.007825\text{u}$ for $m_{\text{H}}$ and $36.965903\text{u}$ for $m_{\text{Cl}}$ equation (V) to get reduced mass of HCL with Cl atom to be isotope of ${}^{35}\text{Cl}$.

  $\begin{array}{c}{\mu }_{\text{35}}=\frac{\left(1.007825\text{u}\right)\left(36.965903\text{u}\right)}{\left(1.007825\text{u}\right)+\left(36.965903\text{u}\right)}\times \left(\frac{1.660540\times {10}^{-27}\text{kg}}{\text{u}}\right)\\ =1.629118\times {10}^{-27}\text{kg}\end{array}$

Here, ${\mu }_{\text{37}}$ is the reduced mass of HCL with Cl atom to be isotope of ${}^{37}\text{Cl}$.

Substitute $1.629118\times {10}^{-27}\text{kg}$ for $\mu$ and $0.12746\text{nm}$ for $r$ in equation (VI) to get $I$.

  $\begin{array}{c}I=\left(1.629118\times {10}^{-27}\text{kg}\right){\left(0.12746\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}^2\\ =2.6466735\times {10}^{-47}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/{\text{s}}^2$ for $c$ and $2.6466735\times {10}^{-47}\text{kg}\cdot {\text{m}}^2$ for $I$ in equation (VII) to get $\lambda$.

  $\begin{array}{c}\lambda =\frac{4{\pi }^2\left(2.6466735\times {10}^{-47}\text{kg}\cdot {\text{m}}^2\right)\left(3\times {10}^8\text{m}/{\text{s}}^2\right)}{6.626\times {10}^{-34}\text{J}\cdot \text{s}}\\ =4.726063702\times {10}^{-4}\text{m}\times \frac{1\text{m}}{{10}^6\text{μm}}\\ =4.73\times {10}^{-4}\text{m}\times \frac{1\text{m}}{{10}^6\text{μm}}\\ =473\text{μm}\end{array}$

Therefore, the longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope ${}^{37}\text{Cl}$ is $473\text{μm}$.

(c)

To determine

The separation in wavelength between the doublet lines for the longest wavelength.

Answer to Problem 19P

The separation in wavelength between the doublet lines for the longest wavelength is $\text{0}\text{.731}\text{μm}$.

Explanation of Solution

Write the expression for the separation of wavelength.

  $\Delta \lambda ={\lambda }_{37}-{\lambda }_{35}$

Here, ${\lambda }_{35}$ is the $\lambda$ is the wavelength of photon emitted in ${\text{H}}^{35}\text{Cl}$ and ${\lambda }_{\text{37}}$ is the wavelength of photon emitted in ${\text{H}}^{37}\text{Cl}$.

Conclusion:

Substitute $4.726063702\times {10}^{-4}\text{ m}$ for ${\lambda }_{37}$ and $4.718749601\times {10}^{-4}\text{ m}$ for ${\lambda }_{35}$ in above equation to get $\Delta \lambda$.

  $\begin{array}{c}\Delta \lambda =4.726063702\times {10}^{-4}\text{ m}-4.718749601\times {10}^{-4}\text{ m}\\ =7.31\times {10}^{-7}\text{m}\times \frac{{10}^6\text{μm}}{1\text{m}}\\ =0.\text{731}\text{μm}\end{array}$

Therefore, the separation in wavelength between the doublet lines for the longest wavelength is $\text{0}\text{.731}\text{μm}$.