Chapter 11, Problem 11B.11P

Interpretation Introduction

Interpretation:

An expression for the value of $J$ for the most populated rotational energy level of a diatomic rotor with degeneracy of $2J+1$ has to be developed.  The expression has to be evaluated for $\text{ICl}$.  An expression for the value of $J$ for the most populated rotational energy level of a diatomic rotor with degeneracy of ${\left(2J+1\right)}^2$ has to be developed.  The expression has to be evaluated for ${\text{CH}}_4$.

Concept introduction:

A three-dimensional rotating molecule is termed as a rotor.  If the bond length of the molecule does not change during the rotation, the molecule is termed as a rigid rotor.  The population of the states depends upon the temperature.  As the temperature increases, the population of the higher states increases.

Answer to Problem 11B.11P

An expression for the value of $J$ for the most populated rotational energy level of a diatomic rotor with degeneracy of $2J+1$ is shown below.

    $J=\sqrt{\frac{kT}{2hc\tilde{B}}}-\frac{1}{2}$

The value of the expression for $\text{ICl}$ is $\underset{\_}{29.656}$.

An expression for the value of $J$ for the most populated rotational energy level of a diatomic rotor with degeneracy of ${\left(2J+1\right)}^2$ is shown below.

    $J=\sqrt{\frac{kT}{hc\tilde{B}}}-\frac{1}{2}$

The value of the expression for ${\text{CH}}_4$ is $\underset{\_}{5.796}$.

Explanation of Solution

The energy of a rigid rotor ($E_J$) is calculated by the expression given below.

    $E_J=hc\tilde{B}J\left(J+1\right)$        (1)

Where,

• $h$ is the Planck’s constant. $\left(6.6\times {10}^{-34}\text{J}\text{s}\right)$.
• $c$ is the speed of light. $\left(3\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)$
• $\tilde{B}$ is the rotational constant.
• $J$ is the rotational quantum number.

The degeneracy ($g_J$) is calculated by the formula given below.

    $g_J=2J+1$        (2)

The expression for population, $N_J$ of a level $J$ is given below.

    $N_J=N{g}_J{e}^{-\frac{E_J}{kT}}$        (3)

Where,

• $N$ is the total number of molecules in the sample.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

Substitute equation (1) and (2) in (3).

    $N_J=N\left(2J+1\right)e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}$        (4)

For maximum value of $N_J$ (maximum population), the relation given below will hold true.

    $\frac{\text{d}{N}_J}{\text{d}J}=0$        (5)

Substitute the value of $N_J$ from equation (4) in (5).

    $\begin{array}{c}\frac{\text{d}}{\text{d}J}\left(N\left(2J+1\right)e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\right)=0\\ N\left(\left(2J+1\right)\frac{\text{d}}{\text{d}J}{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}+e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\frac{\text{d}}{\text{d}J}\left(2J+1\right)\right)=0\\ N\left(\left(2J+1\right)e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\left(-\frac{hc\tilde{B}}{kT}\right)\left(2J+1\right)+2e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\right)=0\\ N{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\left({\left(2J+1\right)}^2\left(-\frac{hc\tilde{B}}{kT}\right)+2\right)=0\end{array}$

The equation given above was solved further as shown below.

    $\begin{array}{c}{\left(2J+1\right)}^2\left(-\frac{hc\tilde{B}}{kT}\right)+2=0\\ {\left(2J+1\right)}^2=\frac{2kT}{hc\tilde{B}}\\ 2J+1=\sqrt{\frac{2kT}{hc\tilde{B}}}\\ J=\sqrt{\frac{kT}{2hc\tilde{B}}}-\frac{1}{2}\end{array}$        (6)

The value of $\tilde{B}$ for $\text{ICl}$ is $0.1142{\text{cm}}^{-1}$.

The value of $T$ $\text{25}\text{°C}$.

The conversion of Celsius to Kelvin is done as shown below.

  ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{25}\text{°C}$ to Kelvin is done as shown below.

  $\begin{array}{c}\text{25}\text{°C}=25+2\text{73 K}\\ =298\text{K}\end{array}$

The value of $k$ is $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$.

The value of $h$ is $6.626\times {10}^{-34}\text{J}\text{s}$.

The value of $c$ is $3\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

Substitute the value of $\tilde{B}$, $T$, $k$, $h$, and $c$ in equation (6).

    $\begin{array}{c}J=\sqrt{\frac{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 298\text{K}}{2\times 6.6\times {10}^{-34}\text{J}\text{s}\times 3\times {10}^{10}\text{cm}{\text{s}}^{-1}\text{K}\times 0.1142{\text{cm}}^{-1}}}-\frac{1}{2}\\ =\sqrt{909.36}-\frac{1}{2}\\ =30.156-\frac{1}{2}\\ =\underset{\_}{29.656}\end{array}$

Therefore, the value of maximum populated level of $\text{ICl}$ is $\underset{\_}{29.656}$.

The degeneracy ($g_J$) is calculated by the formula given below.

    $g_J={\left(2J+1\right)}^2$        (7)

Substitute equation (1) and (7) in (3).

    $N_J=N{\left(2J+1\right)}^2{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}$        (8)

For maximum value of $N_J$ (maximum population), the relation given below will hold true.

    $\frac{\text{d}{N}_J}{\text{d}J}=0$        (5)

Substitute the value of $N_J$ from equation (8) in (5).

    $\begin{array}{c}\frac{\text{d}}{\text{d}J}\left(N{\left(2J+1\right)}^2{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\right)=0\\ N\left(e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\frac{\text{d}}{\text{d}J}{\left(2J+1\right)}^2+{\left(2J+1\right)}^2\frac{\text{d}}{\text{d}J}{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\right)=0\\ N\left(2\left(2J+1\right)2e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}+{\left(2J+1\right)}^2{e}^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\left(-\frac{hc\tilde{B}}{kT}\right)\left(2J+1\right)\right)=0\\ N\left(2J+1\right)e^{-\frac{hc\tilde{B}J\left(J+1\right)}{kT}}\left(4+{\left(2J+1\right)}^2\left(-\frac{hc\tilde{B}}{kT}\right)\right)=0\end{array}$

The equation given above was solved further as shown below.

    $\begin{array}{c}4+{\left(2J+1\right)}^2\left(-\frac{hc\tilde{B}}{kT}\right)=0\\ \left(2J+1\right)=\sqrt{\frac{4kT}{hc\tilde{B}}}\\ 2J=\sqrt{\frac{4kT}{hc\tilde{B}}}-1\\ J=\sqrt{\frac{kT}{hc\tilde{B}}}-\frac{1}{2}\end{array}$        (9)

The value of $\tilde{B}$ for ${\text{CH}}_4$ is $5.24{\text{cm}}^{-1}$.

The value of $T$ $\text{25}\text{°C}$.

The conversion of Celsius to Kelvin is done as shown below.

  ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{25}\text{°C}$ to Kelvin is done as shown below.

  $\begin{array}{c}\text{25}\text{°C}=25+2\text{73 K}\\ =298\text{K}\end{array}$

The value of $k$ is $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$.

The value of $h$ is $6.626\times {10}^{-34}\text{J}\text{s}$.

The value of $c$ is $3\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

Substitute the value of $\tilde{B}$, $T$, $k$, $h$, and $c$ in equation (6).

    $\begin{array}{c}J=\sqrt{\frac{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 298\text{K}}{6.6\times {10}^{-34}\text{J}\text{s}\times 3\times {10}^{10}\text{cm}{\text{s}}^{-1}\text{K}\times 5.24{\text{cm}}^{-1}}}-\frac{1}{2}\\ =\sqrt{39.64}-\frac{1}{2}\\ =6.296-\frac{1}{2}\\ =\underset{\_}{5.796}\end{array}$

Therefore, the value of maximum populated level of ${\text{CH}}_4$ is $\underset{\_}{5.796}$.