Chapter 13, Problem 13B.5P

(a)

Interpretation Introduction

Interpretation:

The electronic partition functions of a tellurium atom at $298\text{K}$ and $5000\text{K}$ has to be calculated with the help of given data.

Concept introduction:

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

  $q={\displaystyle \sum _i{g}_i{e}^{-{\epsilon }_i/kT}}$

Where,

• $g_i$ represents the degeneracy.
• ${\epsilon }_i$ represents the energy of $i^{th}$ microstate.
• $k$ represents the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.
• $T$ represents the temperature $\left(\text{K}\right)$.

It is also called as canonical ensemble partition function.

Answer to Problem 13B.5P

The electronic partition functions of a tellurium atom at $298\text{K}$ is $\underset{\_}{5}$.

The electronic partition functions of a tellurium atom at $5000\text{K}$ is $\underset{\_}{6.272}$.

Explanation of Solution

The electronic partition function is calculated by using the formula shown below.

    $q^{\text{E}}={\displaystyle \sum _j{g}_j{e}^{-{\epsilon }_j/kT}}$        (1)

This equation is simplified as shown below.

    $\begin{array}{l}{q}^{\text{E}}={\displaystyle \sum _j{g}_j{e}^{-{\epsilon }_j/kT}}\\ q^{\text{E}}={\displaystyle \sum _j{g}_j\exp \left(\frac{-{\epsilon }_j}{kT}\right)}\\ q^{\text{E}}={\displaystyle \sum _j{g}_j\exp \left(\frac{-hc{\stackrel{\sim }{\nu }}_j}{kT}\right)}\end{array}$

Therefore, the electronic partition function is calculated by the formula shown below.

    $q^{\text{E}}={\displaystyle \sum _j{g}_j\exp \left(\frac{-hc{\stackrel{\sim }{\nu }}_j}{kT}\right)}$        (2)

Where,

• $q^{\text{E}}$ is the electronic partition function.
• $g_j$ is the degeneracy of the level j.
• ${\epsilon }_j$ is the energy of the level j.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$.
• $c$ is the velocity of light $\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)$.
• ${\stackrel{\sim }{\nu }}_j$ is the wavenumber of the level j.
• $k$ is the Boltzmann’s constant $\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)$.
• $T$ is the temperature.

Now, first the value of $\frac{hc}{k}$ is calculated.  Substitute the values of $h$, $c$ and $k$.

    $\begin{array}{c}\frac{hc}{k}=\frac{6.626\times {10}^{-34}\text{J}\text{s}\times 2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}}{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}}\\ =\frac{1.986\times {10}^{-23}}{1.38\times {10}^{-23}}\text{cm}\text{K}\\ =1.44\text{cm}\text{K}\end{array}$

Therefore, the value of $\frac{hc}{k}$ is $1.44\text{cm}\text{K}$.

The electronic partition functions of four levels are expressed as shown below by using equation (2).

    $q^{\text{E}}=g_0{e}^{\frac{-hc{\stackrel{\sim }{\nu }}_0}{kT}}+g_1{e}^{\frac{-hc{\stackrel{\sim }{\nu }}_1}{kT}}+g_2{e}^{\frac{-hc{\stackrel{\sim }{\nu }}_2}{kT}}+g_3{e}^{\frac{-hc{\stackrel{\sim }{\nu }}_3}{kT}}$        (3)

Substitute the value of $\frac{hc}{k}$ in equation (3).

    $q^{\text{E}}=g_0{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_0}{T}}+g_1{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_1}{T}}+g_2{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_2}{T}}+g_3{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_3}{T}}$        (4)

The degeneracy and the wavenumber of ground level to third level are shown below in the Table.

Term	Degeneracy	Wavenumber $\left({\text{cm}}^{-1}\right)$
Ground	$5$	$0$
$1$	$1$	$4707$
$2$	$3$	$4751$
$3$	$5$	$10559$

(i) The electronic partition functions of a tellurium atom at $298\text{K}$:

Substitute the values of degeneracy and wavenumber from ground term to third term in equation (4) to calculate the electronic partition function at $298\text{K}$.

    $\begin{array}{c}{q}^{\text{E}}=5\times e^{\frac{-1.44\text{cm}\text{K}\times 0}{298\text{K}}}+1\times e^{\frac{-1.44\text{cm}\text{K}\times 4707}{298\text{K}}}+3\times e^{\frac{-1.44\text{cm}\text{K}\times 4751}{298\text{K}}}+5\times e^{\frac{-1.44\text{cm}\text{K}\times 10559}{298\text{K}}}\\ =5\times e^0+1\times e^{-22.745}+3\times e^{-22.95}+5\times e^{-51.0}\\ =5\times 1+1\times \left(1.324\times {10}^{-10}\right)+3\times \left(1.078\times {10}^{-10}\right)+5\times \left(7.095\times {10}^{-23}\right)\\ =5+1.324\times {10}^{-10}+3.234\times {10}^{-10}+3.5475\times {10}^{-22}\end{array}$

Simplify the above equation,

    $q^{\text{E}}=\underset{\_}{5}$

Therefore, the electronic partition functions of a tellurium atom at $298\text{K}$ is $\underset{\_}{5}$.

(ii) The electronic partition functions of a tellurium atom at $5000\text{K}$:

Substitute the values of degeneracy and wavenumber from ground term to third term in equation (4) to calculate the electronic partition function at $5000\text{K}$.

    $\begin{array}{c}{q}^{\text{E}}=5\times e^{\frac{-1.44\text{cm}\text{K}\times 0}{5000\text{K}}}+1\times e^{\frac{-1.44\text{cm}\text{K}\times 4707}{5000\text{K}}}+3\times e^{\frac{-1.44\text{cm}\text{K}\times 4751}{5000\text{K}}}+5\times e^{\frac{-1.44\text{cm}\text{K}\times 10559}{5000\text{K}}}\\ =5\times e^0+1\times e^{-1.35}+3\times e^{-1.36}+5\times e^{-3.0}\\ =5\times 1+1\times \left(0.259\right)+3\times \left(0.256\right)+5\times \left(0.049\right)\\ =5+0.259+0.768+0.245\end{array}$

Simplify the above equation,

    $q^{\text{E}}=\underset{\_}{6.272}$

Therefore, the electronic partition functions of a tellurium atom at $5000\text{K}$ is $\underset{\_}{6.272}$.

(b)

Interpretation Introduction

Interpretation:

The proportions of the tellurium atoms in the ground term and in the second term at $298\text{K}$ and $5000\text{K}$ have to be calculated.

Concept introduction:

The relative populations of energy levels are obtained with the help of the Boltzmann distribution formula.  The relative population is controlled by the difference in energy from the ground state and system temperature.  In the higher energy state, the lower population is obtained as a result of the higher magnitude in the energy difference.  Higher temperature, however, in the higher energy state leads to higher population.

Answer to Problem 13B.5P

The proportions of the tellurium atoms in the ground term and in the second term at $298\text{K}$ and $5000\text{K}$ are $\underset{\_}{1}$, $\underset{\_}{6.468\times 10^{-11}}$, $\underset{\_}{0.8}$ and $\underset{\_}{0.122}$ respectively.

Explanation of Solution

For each term, the individual values in equation (4) are referred as the populations for each level.  The relative populations for ground level and second level are calculated by the formula shown below.

    $\text{Relative}\text{population}=\frac{N_j}{q^{\text{E}}}$        (5)

The relative population at $298\text{K}$:

The proportion for ground level is calculated by the formula shown below.

    $\text{Relative}\text{population}=\frac{N_0}{q^{\text{E}}}$        (6)

The value of $N_0$ is calculated as shown below.

    $\begin{array}{c}{N}_0=g_0{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_0}{T}}\\ =5\times e^{\frac{-1.44\text{cm}\text{K}\times 0}{298\text{K}}}\\ =5\times 1\\ =5\end{array}$

The electronic partition functions of a tellurium atom at $298\text{K}$ is $5$.

Substitute the values of $N_0$ and $q^{\text{E}}$ in equation (6) to calculate the proportion of the tellurium atoms in the ground term.

    $\begin{array}{c}\text{Relative}\text{population}=\frac{5}{5}\\ =\underset{\_}{1}\end{array}$

The proportion for second level is calculated by the formula shown below.

    $\text{Relative}\text{population}=\frac{N_2}{q^{\text{E}}}$        (6)

The value of $N_2$ is calculated as shown below.

    $\begin{array}{c}{N}_2=g_0{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_2}{T}}\\ =3\times e^{\frac{-1.44\text{cm}\text{K}\times 4751}{298\text{K}}}\\ =3\times \left(1.078\times {10}^{-10}\right)\\ =3.234\times {10}^{-10}\end{array}$

The electronic partition functions of a tellurium atom at $298\text{K}$ is $5$.

Substitute the values of $N_2$ and $q^{\text{E}}$ in equation (6) to calculate the proportion of the tellurium atoms in the second term.

    $\begin{array}{c}\text{Relative}\text{population}=\frac{3.234\times {10}^{-10}}{5}\\ =\underset{\_}{6.468\times 10^{-11}}\end{array}$

The relative population at $5000\text{K}$:

The proportion for ground level is calculated by the formula shown below.

    $\text{Relative}\text{population}=\frac{N_0}{q^{\text{E}}}$        (6)

The value of $N_0$ is calculated as shown below.

    $\begin{array}{c}{N}_0=g_0{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_0}{T}}\\ =5\times e^{\frac{-1.44\text{cm}\text{K}\times 0}{5000\text{K}}}\\ =5\times 1\\ =5\end{array}$

The electronic partition functions of a tellurium atom at $5000\text{K}$ is $6.272$.

Substitute the values of $N_0$ and $q^{\text{E}}$ in equation (6) to calculate the proportion of the tellurium atoms in the ground term.

    $\begin{array}{c}\text{Relative}\text{population}=\frac{5}{6.272}\\ =\underset{\_}{0.8}\end{array}$

The proportion for second level is calculated by the formula shown below.

    $\text{Relative}\text{population}=\frac{N_2}{q^{\text{E}}}$        (6)

The value of $N_2$ is calculated as shown below.

    $\begin{array}{c}{N}_2=g_0{e}^{\frac{-1.44\text{cm}\text{K}\times {\stackrel{\sim }{\nu }}_2}{T}}\\ =3\times e^{\frac{-1.44\text{cm}\text{K}\times 4751}{5000\text{K}}}\\ =3\times \left(0.256\right)\\ =0.768\end{array}$

The electronic partition functions of a tellurium atom at $5000\text{K}$ is $6.272$.

Substitute the values of $N_2$ and $q^{\text{E}}$ in equation (6) to calculate the proportion of the tellurium atoms in the second term.

    $\begin{array}{c}\text{Relative}\text{population}=\frac{0.768}{6.272}\\ =\underset{\_}{0.122}\end{array}$

Therefore, the proportions of the tellurium atoms in the ground term and in the second term at $298\text{K}$ and $5000\text{K}$ are $\underset{\_}{1}$, $\underset{\_}{6.468\times 10^{-11}}$, $\underset{\_}{0.8}$ and $\underset{\_}{0.122}$ respectively.