Chapter 13, Problem 13A.1ST

Interpretation Introduction

Interpretation:

The ratio of the population of the levels with $J=2$ and $J=1$ of $\text{HCl}$ at ${25}^{\text{ο}}\text{C}$ has to be calculated.

Concept introduction:

Statistical thermodynamics is used to describe all the possible configurations in a system at given physical quantities such as pressure, temperature, and a number of particles in the system.  The important quantity in the thermodynamic is partition function that is represented as shown below.

  $q\equiv {\displaystyle \sum _i{g}_i{e}^{-\beta {\in }_i}}$

It is also known as canonical ensemble partition function.

Answer to Problem 13A.1ST

The ratio of the population of the levels with $J=2$ and $J=1$ of $\text{HCl}$ at ${25}^{\text{ο}}\text{C}$ is $\underset{\_}{1.359}$.

Explanation of Solution

The expression for the energy of the state with quantum number $J$ is shown below.

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

Where,

• $h$ is Plank’s constant.
• $c$ is the velocity of light.
• $\tilde{B}$ is rotational constant.
• $J$ is the rotational quantum number.

The energy of the state with quantum number $J=1$ is calculated by substituting $J=1$ in equation (1) as shown below.

  $\begin{array}{c}{\epsilon }_1=hc\tilde{B}\times 1\left(1+1\right)\\ =2hc\tilde{B}\end{array}$

The energy of the state with quantum number $J=2$ is calculated by substituting $J=2$ in equation (1) as shown below.

  $\begin{array}{c}{\epsilon }_2=hc\tilde{B}\times 2\left(2+1\right)\\ =6hc\tilde{B}\end{array}$

The expression for the Boltzmann population ratio is shown below.

  $\frac{N_i}{N_j}={\text{e}}^{-\beta \left({\epsilon }_i-{\epsilon }_j\right)}$        (2)

Where,

• $\frac{N_i}{N_j}$ is the ratio of the population of the upper state and lower state.
• ${\epsilon }_i$ is the energy of the state with a quantum number $i$.
• ${\epsilon }_j$ the energy of the state with a quantum number $j$.
• $\beta =1/kT$, here $k$ is Boltzmann’s constant and $T$ is the temperature.

The energy separation of the states with $J=2$ and $J=1$ is shown below.

  $\begin{array}{c}{\epsilon }_2-{\epsilon }_1=6hc\tilde{B}-2hc\tilde{B}\\ =4hc\tilde{B}\end{array}$

Substitute the values of $i=2$ and $j=1$ in equation (2) as shown below.

  $\frac{N_2}{N_1}={\text{e}}^{-\beta \left({\epsilon }_2-{\epsilon }_1\right)}$

Substitute the value of ${\epsilon }_2-{\epsilon }_1$ in the above equation.

  $\frac{N_2}{N_1}={\text{e}}^{-4hc\tilde{B}\beta }$

The upper level with rotational quantum number has five-fold degeneracy and the lower level has three-fold degeneracy.  So the relative population of the levels is calculated as shown below.

  $\frac{N_2}{N_1}=\frac{5}{3}{\text{e}}^{-4hc\tilde{B}\beta }$

Substitute $\beta =1/kT$ in the above equation as shown below.

  $\frac{N_2}{N_1}=\frac{5}{3}{\text{e}}^{-\frac{4hc\tilde{B}}{kT}}$        (3)

The value of $kT/hc$ is $207.22{\text{cm}}^{-1}$ at ${25}^{\text{ο}}\text{C}$ and the value of $\tilde{B}$ is $10.591{\text{cm}}^{-1}$.  Substitute the value of $kT/hc$ and $\tilde{B}$ in equation (3) as shown below.

  $\begin{array}{c}\frac{N_2}{N_1}=\frac{5}{3}{\text{e}}^{-\frac{4hc\tilde{B}}{kT}}\\ =\frac{5}{3}{\text{e}}^{-\frac{4\times 10.591{\text{cm}}^{-1}}{207.22{\text{cm}}^{-1}}}\\ =\frac{5}{3}{\text{e}}^{-\frac{42.364}{207.22}}\\ =\frac{5}{3}{\text{e}}^{-0.204}\end{array}$

The above equation is further solved as shown below.

  $\begin{array}{c}\frac{N_2}{N_1}=\frac{5}{3}{\text{e}}^{-0.204}\\ =\frac{5}{3}\times 0.8154\\ =\underset{\_}{1.359}\end{array}$

Therefore, the ratio of the population of the levels with $J=2$ and $J=1$ of $\text{HCl}$ at ${25}^{\text{ο}}\text{C}$ is $\underset{\_}{1.359}$.