Chapter 42, Problem 13P

(a)

To determine

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond.

Answer to Problem 13P

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond is $4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

A hydrogen molecule makes a transition from ground level to $v=1$ and $J=1$ after absorbing a photon of wavelength $2.2112\text{μm}$. The molecule emit a photon of wavelength $2.4054\text{μm}$ after dropping to level $v=0$ and $J=2$.

Write the formula for energy levels.

    $E_{vJ}=\left(v+\frac{1}{2}\right)hf+\frac{{\hslash }^2}{2I}J\left(J+1\right)$                                                                                 (I)

Here, $E_{vJ}$ is the energy, $v$ is the vibrational quantum number, $h$ is the Planck’s constant, $f$ is the frequency, $J$ is the rotational quantum number, and $I$ is the moment of inertia.

Refer equation (I) and find energy of $v=0$ and $J=0$.

    $E_{00}=\frac{1}{2}hf$

Here, $E_{00}$ is the energy of level  $v=0$ and $J=0$.

Refer equation (I) and find energy of $v=1$ and $J=1$.

    $E_{11}=\frac{3}{2}hf+\frac{{\hslash }^2}{I}$

Here, $E_{11}$ is the energy of level $v=1$ and $J=1$.

Refer equation (I) and find energy of $v=0$ and $J=2$.

    $E_{02}=\frac{1}{2}hf+\frac{3{\hslash }^2}{I}$

Here, $E_{02}$ is the energy of level $v=0$ and $J=2$.

Write the formula for the energy difference between $v=0$, $J=0$ level to $v=1$, $J=1$.

    $\begin{array}{c}\Delta E_1=E_{11}-E_{00}\\ =hf+\frac{{\hslash }^2}{I}\\ =\frac{hc}{{\lambda }_1}\end{array}$                                                                                          (II)

Here, $\Delta E_1$ is the energy difference between $v=0$, $J=0$ level to $v=1$, $J=1$, ${\lambda }_1$ is the wavelength of the photon required for the transition from energy difference between $v=0$, $J=0$ level to $v=1$, $J=1$.

Write the formula for the energy difference between $v=1$, $J=1$ level to $v=0$, $J=2$.

    $\begin{array}{c}\Delta E_2=E_{11}-E_{02}\\ =hf-\frac{2{\hslash }^2}{I}\\ =\frac{hc}{{\lambda }_2}\end{array}$                                                                                    (III)

Here, $\Delta E_2$ is the energy difference between $v=1$, $J=1$ level to $v=0$, $J=2$, ${\lambda }_2$ is the wavelength required from the transition between $v=1$, $J=1$ level to $v=0$, $J=2$.

Subtract equation (III) from (II).

    $\begin{array}{c}\Delta E_1-\Delta E_2=\left(hf+\frac{{\hslash }^2}{I}\right)-\left(hf-\frac{2{\hslash }^2}{I}\right)\\ =\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2}\end{array}$

Re-write the above equation.

    $\begin{array}{c}\frac{3{\hslash }^2}{I}=hc\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)\\ \frac{3h}{4{\pi }^{2}I}=c\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)\end{array}$

Re-write the above equation to obtain $I$.

    $I=\frac{3h}{4{\pi }^{2}c}{\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)}^{-1}$

Conclusion:

Substitute $6.626075\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.2112\text{μm}$ for ${\lambda }_1$, $2.4054\text{μm}$ for ${\lambda }_2$, $2.997925\times {10}^8\text{m/s}$ for $c$.

    $\begin{array}{c}I=\frac{3\left(6.626075\times {10}^{-34}\text{J}\cdot \text{s}\right)}{4{\pi }^2\left(2.997925\times {10}^8\text{m/s}\right)}{\left(\frac{1}{2.2112\text{μm}}-\frac{1}{2.4054\text{μm}}\right)}^{-1}\\ =\frac{3\left(6.626075\times {10}^{-34}\text{J}\cdot \text{s}\right)}{4{\pi }^2\left(2.997925\times {10}^8\text{m/s}\right)}{\left(\frac{1}{2.2112\times {10}^{-6}\text{m}}-\frac{1}{2.4054\times {10}^{-6}\text{m}}\right)}^{-1}\\ =4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond is $4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}$.

(b)

To determine

The vibrational frequency of the hydrogen molecule.

Answer to Problem 13P

The vibrational frequency of the hydrogen molecule is $1.32\times {10}^{14}\text{Hz}$.

Explanation of Solution

Refer section (a) and write the formula for the energy difference between $v=0$, $J=0$ level to $v=1$, $J=1$.

    $\begin{array}{c}\Delta E_1=E_{11}-E_{00}\\ =hf+\frac{{\hslash }^2}{I}\end{array}$

Here, $\Delta E_1$ is the energy difference between $v=0$, $J=0$ level to $v=1$, $J=1$, $J=0$ level to $v=1$, $J=1$, $E_{11}$ is the energy of level $v=1$ and $J=1$, $E_{00}$ is the energy of level  $v=0$ and $J=0$, $h$ is the Planck’s constant, $f$ is the frequency, $I$ is the moment of inertia.

Re-write the above equation to get an expression for $f$.

    $\begin{array}{c}hf=\Delta E_1-\frac{{\hslash }^2}{I}\\ f=\frac{\Delta E_1}{h}-\frac{h^2}{2\pi I}\end{array}$                                                                                               (III)

Write the formula for $\Delta E_1$.

    $\Delta E_1=\frac{hc}{{\lambda }_1}$                                                                                                         (IV)

Here, ${\lambda }_1$ is the wavelength of the photon required for the transition from energy difference between $v=0$, and $c$ is the speed of light in vacuum.

Conclusion:

Substitute $6.626075\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.2112\text{μm}$ for ${\lambda }_1$, $2.997925\times {10}^8\text{m/s}$ for $c$ in equation (IV).

    $\begin{array}{c}\Delta E_1=\frac{\left(6.626075\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.997925\times {10}^8\text{m/s}\right)}{2.2112\text{μm}}\\ =\frac{\left(6.626075\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.997925\times {10}^8\text{m/s}\right)}{2.2112\times {10}^{-6}\text{m}}\\ =8.983573\times {10}^{-20}\text{J}\end{array}$

Substitute $8.983573\times {10}^{-20}\text{J}$ for $\Delta E_1$, $6.626075\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$ in equation (III).

    $\begin{array}{c}f=\frac{8.983573\times {10}^{-20}\text{J}}{\text{6}\text{.626 075}\times {10}^{-34}\text{J}\cdot \text{s}}-\frac{{\left(6.626075\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\pi \left(4.600060\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\\ =1.32\times {10}^{14}\text{Hz}\end{array}$

The vibrational frequency of the hydrogen molecule is $1.32\times {10}^{14}\text{Hz}$.

(c)

To determine

The equilibrium separation distance for the molecule.

Answer to Problem 13P

The equilibrium separation distance for the molecule is $0.\text{0741}\text{nm}$.

Explanation of Solution

Write the formula for the moment of inertia of the molecule.

    $I=\mu r^2$

Here, $I$ is the moment of inertia, $\mu$ is the reduced mass, and $r$ is the equilibrium separation distance.

Reduced mass of hydrogen molecule is half of the mass of it.

    $I=\frac{m}{2}{r}^2$

Here, $m$ is the mass of the molecule.

Re-write the above equation to get an expression for $r$.

    $r=\sqrt{\frac{2I}{m}}$

Conclusion:

Substitute $4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$, $1.007825\text{u}$ for $m$.

    $\begin{array}{c}r=\sqrt{\frac{2\left(4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\right)}{1.007825\text{u}}}\\ =\sqrt{\frac{2\left(4.60\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\right)}{1.007825\text{u}}\left(1.67\times {10}^{-27}\text{kg}\right)}\\ =0.\text{0741}\text{nm}\end{array}$

The equilibrium separation distance for the molecule is $0.\text{0741}\text{nm}$.