Chapter 42, Problem 11P

(a) In an HCl molecule, take the Cl atom to be the isotope ³⁵Cl. The equilibrium separation of the H and Cl atoms is 0.127 46 nm. The atomic mass of the H atom is 1.007 825 u and that of the ³⁵Cl atom is 34.968 853 u. Calculate the longest wavelength in the rotational spectrum of this molecule. (b) What If? Repeat the calculation in part (a), but take the Cl atom to be the isotope ³⁷Cl, which has atomic mass 36.965 903 u. The equilibrium separation distance is the same as in part (a). (c) Naturally occurring chlorine contains approximately three parts of ³⁵Cl to one part of ³⁷Cl. Because of the two different Cl masses, each line in the microwave rotational spectrum of HCl is split into a doublet as shown in Figure P42.11. Calculate the separation in wavelength between the doublet lines for the longest wavelength.

To determine

The longest wavelength in the rotational spectrum of HCl molecule when $\text{C}\text{l}$ isotope is used.

Answer to Problem 11P

The longest wavelength in the rotational spectrum of $\text{HCl}$ molecule when $\text{C}\text{l}$ isotope is used is $472\text{μm}$.

Explanation of Solution

Write the equation for the smallest energy difference between the rotational energy levels.

    $\Delta E_{\min }=\frac{{\hslash }^2}{I}$                                                                                                           (I)

Here, $\Delta E_{\min }$ is the smallest energy difference between the rotational energy levels, $\hslash$ is the reduced Plank’s constant and $I$ is the moment of inertia.

Write the equation for the wavelength.

    $\lambda =\frac{hc}{\Delta E_{\min }}$                                                                                                       (II)

Here, $h$ is the Plank’s constant and $c$ is the speed of light.

Substitute equation (I) in the equation (II) and substitute $\frac{h}{2\pi }$ for $\hslash$ to find $\lambda$.

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

Write the equation for the reduced mass.

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

Here, $\mu$ is the reduced mass, $m_{\text{H}}$ is the atomic mass of hydrogen and $m_{\text{Cl}}$ is the atomic mass of chlorine.

Write the equation for moment of inertia.

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

Here, $I$ is the moment of inertia and $r$ is the separation distance of the atoms between hydrogen and chlorine.

Conclusion:

Substitute equation (V) in equation (III).

    $\lambda =\frac{4{\pi }^2\mu r^{2}c}{h}$

Substitute $0.12746\text{nm}$ for $r$, $2.997925\times {10}^8\text{m/s}$ for $c$ and $6.626075\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ to find $\lambda$ in above equation.

$\begin{array}{c}\lambda =\frac{4{\pi }^2\mu {\left(\left(0.12746\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2\left(2.997925\times {10}^8\text{m/s}\right)}{6.626075\times {10}^{-34}\text{J}\cdot \text{s}}\\ =\left(2.901830\times {10}^{23}\text{m/kg}\right)\mu \end{array}$ (VI)

Substitute $1.007825\text{u}$ for $m_{\text{H}}$ and $34.968853\text{u}$ for $m_{\text{Cl}}$ in equation (IV) to find ${\mu }_{35}$.

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

Here, ${\mu }_{35}$ is the reduced mass for $\text{C}\text{l}$.

Substitute $1.626653\times {10}^{-27}\text{kg}$ for $\mu$ in equation (VI) to find ${\lambda }_{35}$.

    $\begin{array}{c}{\lambda }_{35}=\left(2.901830\times {10}^{23}\text{m/kg}\right)\left(1.626653\times {10}^{-27}\text{kg}\right)\\ =\left(472\times {10}^{-6}\text{m}\right)\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =472\text{μm}\end{array}$

Thus, the longest wavelength in the rotational spectrum of HCl molecule when $\text{C}\text{l}$ isotope is used is $472\text{μm}$.

To determine

The longest wavelength in the rotational spectrum of HCl molecule when $\text{C}\text{l}$ isotope is used.

Answer to Problem 11P

The longest wavelength in the rotational spectrum of HCl molecule when $\text{C}\text{l}$ isotope is used is $473\text{μm}$.

Explanation of Solution

From the equation (IV) for the reduced mass.

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

From the equation (VI) for wavelength.

    $\lambda =\left(2.901830\times {10}^{23}\text{m/kg}\right)\mu$

Conclusion:

Substitute $1.007825\text{u}$ for $m_{\text{H}}$ and $36.965903\text{u}$ for $m_{\text{Cl}}$ in equation (IV) to find ${\mu }_{37}$.

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

Here, ${\mu }_{35}$ is the reduced mass for $\text{C}\text{l}$.

Substitute $1.629118\times {10}^{-27}\text{kg}$ for $\mu$ in equation (VI) to find ${\lambda }_{37}$.

    $\begin{array}{c}{\lambda }_{37}=\left(2.901830\times {10}^{23}\text{m/kg}\right)\left(1.629118\times {10}^{-27}\text{kg}\right)\\ =\left(473\times {10}^{-6}\text{m}\right)\left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =473\text{μm}\end{array}$

Thus, the longest wavelength in the rotational spectrum of HCl molecule when $\text{C}\text{l}$ isotope is used is $473\text{μm}$.

To determine

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

Answer to Problem 11P

The separation in wavelength between the doublet lines for the longest wavelength is $0.715\text{μm}$.

Explanation of Solution

Write the equation for the separation in wavelength between the doublet lines for the longest wavelength.

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

Here, $d$ is the separation in wavelength between the doublet lines for the longest wavelength.

Conclusion:

Substitute $472.7424\text{μm}$ for ${\lambda }_{37}$ and $472.0270\text{μm}$ for ${\lambda }_{35}$ in the above equation find $d$.

    $\begin{array}{c}d=472.7424\text{μm}-472.0270\text{μm}\\ =0.715\text{μm}\end{array}$

Thus, the separation in wavelength between the doublet lines for the longest wavelength is $0.715\text{μm}$.