Chapter 11, Problem 9P

Consider the HCl molecule, which consists of a hydrogen atom of mass 1 u bound to a chlorine atom of mass 35 u. The equilibrium separation between the atoms is 0.128 nm, and it requires 0.15 eV of work to increase or decrease this separation by 0.01 nm. (a) Calculate the four lowest rotational energies (in eV) that are possible, assuming the molecule rotates rigidly. (b) Find the molecule’s “spring constant” and its classical frequency of vibration. (Hint: Recall that $U=\frac{1}{2}K{x}^2$.) (c) Find the two lowest vibrational energies and the classical amplitude of oscillation corresponding to each of these energies. (d) Determine the longest wavelength radiation that the molecule can emit in a pure rotational transition and in a pure vibrational transition.

To determine

The four lowest rotational energies that are possible for the $\text{HCl}$ .

Answer to Problem 9P

The four lowest rotational energies that are possible for the $\text{HCl}$ are $\begin{array}{cc}l=0& E_{\text{rot}}=0\\ l=1& E_{\text{rot}}=2.62\times {10}^{-3}\text{ eV}\\ l=2& E_{\text{rot}}=7.86\times {10}^{-3}\text{ eV}\\ l=3& E_{\text{rot}}=1.57\times {10}^{-2}\text{ eV}\end{array}$ .

Explanation of Solution

Write the expression for the rotational energy of diatomic molecule

    $E_{rot}=\frac{{\hslash }^2}{2I}l\left(l+1\right)$        (I)

Write the expression for the moment of inertia of diatomic molecule about its center of mass

    $I=\mu R_0^2$        (II)

Write the expression for the reduced mass of the molecule, Equation 11.3

    $\mu =\frac{m_1{m}_2}{m_1+m_2}$        (III)

Here, $m_1$ and $m_2$ are the mass of the atoms, $\hslash$ is the Plank’s constant (where $\hslash =\frac{h}{2\pi }$), $I$ is moment of inertia about the center of mass, $l$ is rotational quantum number of higher, $\omega$ is the angular velocity, $\mu$ is the reduced mass and $R_0$ is the bond length.

Substitute equation (III) in (II)

$I=\left[\frac{m_1{m}_2}{m_1+m_2}\right]R_0^2$

Substitute $1\text{ u}$ for $m_H$, $35\text{ u}$ for $m_{Cl}$ and $1.28\text{ Å}$ for $R_0$ in the above equation to find the value of $I$

$\begin{array}{c}I=\left[\frac{\left(1\text{ u}\right)\left(35\text{ u}\right)}{1\text{ u}+35\text{ u}}\right]\times {\left(1.28\times {10}^{-10}\text{ m}\right)}^2\\ =\left[0.9722\text{ u}\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}/\text{u}\right]\times {\left(1.28\times {10}^{-10}\text{ m}\right)}^2\\ =\left[1.62\times {10}^{-27}\text{kg}/\text{u}\right]\times {\left(1.28\times {10}^{-10}\text{ m}\right)}^2\\ =2.65\times {10}^{-47}\text{ kg}\cdot {\text{m}}^{\text{2}}\end{array}$

From the above equation, the reduced mass is $1.62\times {10}^{-27}\text{kg}/\text{u}$ and the moment of inertia is $2.65\times {10}^{-47}\text{ kg}\cdot {\text{m}}^{\text{2}}$.

Substitute $1.054\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ and $2.65\times {10}^{-47}\text{ kg}\cdot {\text{m}}^{\text{2}}$ for $I$ in equation (I)

$\begin{array}{c}{E}_{\text{rot}}=\frac{{\left(1.054\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{2\times 2.65\times {10}^{-47}\text{ kg}\cdot {\text{m}}^{\text{2}}}l\left(l+1\right)\\ =\left(2.1\times {10}^{-22}\text{ J}\right)l\left(l+1\right)\\ =\left(1.31\times {10}^{-3}\text{ eV}\right)l\left(l+1\right)\end{array}$

For $l=0$

$E_{rot}=0$

For $l=1$

$\begin{array}{c}{E}_{\text{rot}}=\left(1.31\times {10}^{-3}\text{ eV}\right)\left(1\right)\left(1+1\right)\\ =\left(1.31\times {10}^{-3}\text{ eV}\right)\left(2\right)\\ =2.62\times {10}^{-3}\text{ eV}\end{array}$

For $l=2$

$\begin{array}{c}{E}_{\text{rot}}=\left(1.31\times {10}^{-3}\text{ eV}\right)\left(2\right)\left(2+1\right)\\ =\left(1.31\times {10}^{-3}\text{ eV}\right)\left(6\right)\\ =7.86\times {10}^{-3}\text{ eV}\end{array}$

For $l=3$

$\begin{array}{c}{E}_{\text{rot}}=\left(1.31\times {10}^{-3}\text{ eV}\right)\left(3\right)\left(3+1\right)\\ =\left(1.31\times {10}^{-3}\text{ eV}\right)\left(12\right)\\ =1.57\times {10}^{-2}\text{ eV}\end{array}$

Conclusion:

The four lowest rotational energies that are possible for the $\text{HCl}$ are $\begin{array}{cc}l=0& E_{\text{rot}}=0\\ l=1& E_{\text{rot}}=2.62\times {10}^{-3}\text{ eV}\\ l=2& E_{\text{rot}}=7.86\times {10}^{-3}\text{ eV}\\ l=3& E_{\text{rot}}=1.57\times {10}^{-2}\text{ eV}\end{array}$ .

To determine

The spring constant of the molecule and its classical frequency of vibration.

Answer to Problem 9P

The spring constant of the molecule is $480\text{N}/\text{m}$ and its classical frequency of vibration is $8.66\times {10}^{13}\text{ Hz}$.

Explanation of Solution

The elastic potential energy of the $\text{HCl}$ molecule is equal to the work done on the molecule, $\left(W=U\right)$ ,when the separation between the atom increases by $0.01\text{ nm}$.

Write the formula for the elastic potential energy

    $U=\frac{1}{2}k{x}^2$        (IV)

Write the formula for the classical frequency of vibration

    $f=\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}$        (V)

Here, $U$  is the elastic potential energy, $k$ is the spring constant, $f$ is the classical frequency of vibration, $\mu$ is the reduced mass and $x$ is the change in length.

Rearrange equation (IV) and substitute $W=U$ and solve for $k$

$k=\frac{2W}{x^2}$

Conclusion:

Substitute $0.15\text{ eV}$ for $W$ and $0.01\text{ nm}$ for $m$ in the above equation and solve for $k$

$\begin{array}{c}k=\frac{\left(0.15\text{ eV}\right)\left(1.6\times {10}^{-19}\text{J}/\text{eV}\right)}{{\left(0.01\times {10}^{-9}\text{ m}\right)}^2}\\ =480\text{N}/\text{m}\end{array}$

The spring constant of the molecule is $480\text{N}/\text{m}$.

Substitute $1.62\times {10}^{-27}\text{kg}/\text{u}$ for $\mu$ and $480\text{N}/\text{m}$ for $k$  in equation (V) to find the value of $f$

$\begin{array}{c}f=\frac{1}{2\pi }{\left[\frac{480}{1.62\times {10}^{-27}}\right]}^{1/2}\\ =8.66\times {10}^{13}\text{ Hz}\end{array}$

Thus, the spring constant of the molecule is $480\text{N}/\text{m}$ and its classical frequency of vibration is $8.66\times {10}^{13}\text{ Hz}$.

To determine

The two lowest vibrational energies and the corresponding classical amplitude of oscillation.

Answer to Problem 9P

The two lowest vibrational energies are $E_0=2.88\times {10}^{-20}\text{ J}$ and $E_1=8.63\times {10}^{-20}\text{ J}$ and the corresponding classical amplitude of oscillation are $A_0=0.0109\text{ nm}$ and $A_1=0.0189\text{ nm}$ respectively.

Explanation of Solution

Write the expression for the vibrational energy

    $E_{vib}=\left(\nu +\frac{1}{2}\right)\hslash \omega$        (VI)

Write the expression for the total energy of simple harmonic oscillator

    $E=\frac{1}{2}k{A}^2$        (VII)

Here, $E_{vib}$ is the vibrational energy of diatomic molecule, $\nu$ is the vibrational quantum number, $\omega$ is the classical frequency of vibration, $\hslash$ is Planck’s constant, $k$ is the force constant and $A$ is the amplitude of vibration.

Since $\omega =2\pi f$ equation (VI) becomes

$\begin{array}{c}{E}_{vib}=\left(\nu +\frac{1}{2}\right)\frac{h}{2\pi }2\pi f\\ =\left(\nu +\frac{1}{2}\right)hf\end{array}$        (VIII)

Substitute $0$ for $v$, $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $8.66\times {10}^{13}\text{ Hz}$ for $f$ equation (VIII) to find the value of $E_0$

$\begin{array}{c}{E}_0=\left(0+\frac{1}{2}\right)\left(6.626\times {10}^{-34}\text{ J s}\right)\left(8.66\times {10}^{13}\text{ Hz}\right)\\ =2.88\times {10}^{-20}\text{ J}\end{array}$

Substitute $1$ for $v$, $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $8.66\times {10}^{13}\text{ Hz}$ for $f$ equation (VIII) to find the value of $E_1$

$\begin{array}{c}{E}_1=\left(1+\frac{1}{2}\right)\left(6.626\times {10}^{-34}\text{ J s}\right)\left(8.66\times {10}^{13}\text{ Hz}\right)\\ =8.63\times {10}^{-20}\text{ J}\end{array}$

Equate equation (VI) and (VII) and substitute $\nu =0$

$\begin{array}{c}\left(\nu +\frac{1}{2}\right)\hslash \omega =\frac{1}{2}k{A}^2\\ k{A}^2=2\left(\nu +\frac{1}{2}\right)\hslash \omega \\ A=\sqrt{\frac{2\left(\nu +\frac{1}{2}\right)\hslash \omega }{k}}\\ A=\sqrt{\frac{2\left(\nu +\frac{1}{2}\right)\frac{h}{2\pi }\left(2\pi f\right)}{k}}\end{array}$

$\begin{array}{c}A=\sqrt{\frac{2\left(\nu +\frac{1}{2}\right)hf}{k}}\\ =\sqrt{\frac{2\left(2v+1\right)hf}{k}}\end{array}$        (IX)

Substitute $0$ for $v$, $6.626\times {10}^{-34}\text{ J s}$ for $h$, $480\text{N}/\text{m}$ for $k$ and $8.66\times {10}^{13}\text{ Hz}$ for $f$ equation (IX) to find $A_0$

$\begin{array}{c}{A}_0=\sqrt{\frac{2\left(2\left(0\right)+1\right)\left(6.626\times {10}^{-34}\text{ J s}\right)\left(8.66\times {10}^{13}\text{ Hz}\right)}{480\text{N}/\text{m}}}\\ =1.09\times {10}^{-11}\text{ m}\\ =0.109\text{Å}\\ =0.0109\text{ nm}\end{array}$

Substitute $1$ for $v$, $6.626\times {10}^{-34}\text{ J s}$ for $h$, $480\text{N}/\text{m}$ for $k$ and $8.66\times {10}^{13}\text{ Hz}$ for $f$ equation (IX) to find $A_1$

$\begin{array}{c}{A}_1=\sqrt{\frac{2\left(2\left(1\right)+1\right)\left(6.626\times {10}^{-34}\text{ J s}\right)\left(8.66\times {10}^{13}\text{ Hz}\right)}{480\text{N}/\text{m}}}\\ =1.89\times {10}^{-11}\text{ m}\\ =0.189\text{Å}\\ =0.0189\text{ nm}\end{array}$

Conclusion:

Thus, the two lowest vibrational energies are $E_0=2.88\times {10}^{-20}\text{ J}$ and $E_1=8.63\times {10}^{-20}\text{ J}$ and the corresponding classical amplitude of oscillation are $A_0=0.0109\text{ nm}$ and $A_1=0.0189\text{ nm}$ respectively.

To determine

The longest wavelength radiation that the $\text{HCl}$ molecule can emit in a pure rotational transition and in a pure vibrational transition.

Answer to Problem 9P

The longest wavelength radiation that the $\text{HCl}$ molecule can emit in a pure rotational transition is $4.73\times {10}^{-4}\text{ m}$ and in a pure vibrational transition is $3.45\times {10}^{-6}\text{ m}$.

Explanation of Solution

Write the expression for energy using Bohr’s second postulate

    $\Delta E_{\min }=\frac{hc}{{\lambda }_{\max }}$        (X)

The longest wavelength radiation that the $\text{HCl}$ molecule can emit in a pure rotational transition and in a pure vibrational transition is between $l=0$ and $l=1$ .

In pure rotation transition, between $l=0$ and $l=1$. Equation (X) becomes

$\begin{array}{c}{E}_{rot,1}-E_{rot,0}=\frac{hc}{{\lambda }_{\max }}\\ {\lambda }_{\max }=\frac{hc}{E_{rot,1}-E_{rot,0}}\end{array}$

From part (a), for  $l=0$, $E_{\text{rot}}=0$ and $l=1$ $E_{\text{rot}}=2.62\times {10}^{-3}\text{ eV}$.

Substitute $0$ for $E_{rot,0}$, $2.62\times {10}^{-3}\text{ eV}$ $E_{rot,1}$, $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of ${\lambda }_{rot,\max }$

$\begin{array}{c}{\lambda }_{rot,\max }=\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}{\left(2.62\times {10}^{-3}\text{ eV}\right)-\left(0\right)}\\ =\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}{2.62\times {10}^{-3}\text{ eV}}\\ =4.73\times {10}^{-4}\text{ m}\end{array}$

In pure vibrational transition, between $l=0$ and $l=1$. Equation (X) becomes

$\begin{array}{c}{E}_{vib,1}-E_{vib,0}=\frac{hc}{{\lambda }_{\max }}\\ {\lambda }_{\max }=\frac{hc}{E_{vib,1}-E_{vib,0}}\end{array}$

From part (c), for  $l=0$, $E_{\text{vib}}=2.88\times {10}^{-20}\text{ J}$ and $l=1$ $E_{\text{vib}}=8.63\times {10}^{-20}\text{ J}$.

Substitute $2.88\times {10}^{-20}\text{ J}$ for $E_{vib,0}$, $8.63\times {10}^{-20}\text{ J}$ $E_{vib,1}$, $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of ${\lambda }_{rot,\max }$

$\begin{array}{c}{\lambda }_{rot,\max }=\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}{\left(8.63\times {10}^{-20}\text{ J}\right)-\left(2.88\times {10}^{-20}\text{ J}\right)}\\ =3.45\times {10}^{-6}\text{ m}\end{array}$

Conclusion:

Thus, the longest wavelength radiation that the $\text{HCl}$ molecule can emit in a pure rotational transition is $4.73\times {10}^{-4}\text{ m}$ and in a pure vibrational transition is $3.45\times {10}^{-6}\text{ m}$.