Chapter 8, Problem 8.10P

(a)

To determine

To compare the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ for the temperatures of $5000\text{K}$, $15,000\text{K}$, and $25,000\text{K}$.

Answer to Problem 8.10P

For all the given temperatures, $\left(N_{\text{II}}/N_{\text{I}}\right)\gg \left(N_{\text{III}}/N_{\text{II}}\right)$.

Explanation of Solution

In the case under consideration, the subscript I stands for neutral helium, II stands for singly ionized helium, and III sands for the completely ionized helium atom.

Given that the ionization energies ${\chi }_{\text{I}}=24.6\text{eV}$, ${\chi }_{\text{II}}=54.4\text{eV}$ and the partition functions are $Z_{\text{I}}=1$, $Z_{\text{II}}=2$, and $Z_{\text{III}}=1$. The electron pressure is given as $P_e=20\text{N}\cdot {\text{m}}^{-2}$.

Write the expression for the Saha equation in terms of the electron pressure.

  $\frac{N_{i+1}}{N_i}=\frac{2kT{Z}_{i+1}}{P_e{Z}_i}{\left(\frac{2\pi m_{e}kT}{h^2}\right)}^{3/2}{e}^{-{\chi }_i/kT}$        (I)

Here, $k$ is the Boltzmann constant, $P_e$ is the electron pressure, $T$ is the absolute temperature, $m_e$ is the mass of electron, and $h$ is the Planck’s constant.

Rewrite equation (I) for determining the required ratios.

  $\frac{N_{\text{II}}}{N_{\text{I}}}=\frac{2kT{Z}_{\text{II}}}{P_e{Z}_{\text{I}}}{\left(\frac{2\pi m_{e}kT}{h^2}\right)}^{3/2}{e}^{-{\chi }_{\text{I}}/kT}$        (II)

  $\frac{N_{\text{III}}}{N_{\text{II}}}=\frac{2kT{Z}_{\text{III}}}{P_e{Z}_{\text{II}}}{\left(\frac{2\pi m_{e}kT}{h^2}\right)}^{3/2}{e}^{-{\chi }_{\text{II}}/kT}$        (III)

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}$ for $k$, $5000\text{K}$ for $T$, $1$ for $Z_{\text{I}}$, $2$ for $Z_{\text{II}}$, $20\text{N}\cdot {\text{m}}^{-2}$ for $P_e$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, and $24.6\text{eV}$ for ${\chi }_{\text{I}}$ in equation (II) to find the ratio $N_{\text{II}}/N_{\text{I}}$ corresponding to temperature $5000\text{K}$.

$\begin{array}{c}\frac{N_{\text{II}}}{N_{\text{I}}}=\frac{2\left(1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}\right)\left(5000\text{K}\right)\left(2\right)}{\left(20\text{N}\cdot {\text{m}}^{-2}\right)\left(1\right)}{\left(\frac{2\pi \left(9.1\times {10}^{-31}\text{kg}\right)\left(1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}\right)\left(5000\text{K}\right)}{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}\right)}^{3/2}{e}^{\frac{-\left(24.6\text{eV}\right)}{\left(1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}\right)\left(5000\text{K}\right)}}\\ =\left(1.38\times {10}^{-20}\right)\left(8.518266682\times {10}^{26}\right)e^{\frac{-\left(24.6\text{eV}\times \frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{\left(1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}\right)\left(5000\text{K}\right)}}\\ =1.9\times {10}^{-18}\end{array}$

Similarly the values of the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ can be calculated for different temperatures and the results are tabulated below.

$T\left(\text{K}\right)$	$N_{\text{II}}/N_{\text{I}}$	$N_{\text{III}}/N_{\text{II}}$
$5000$	$1.9\times {10}^{-18}$	$4.3\times {10}^{-49}$
$15,000$	$9.9\times {10}^{-1}$	$2.4\times {10}^{-11}$
$25,000$	$7.2\times {10}^3$	$1.8\times {10}^{-3}$

From the table it is clear that the ratio $\left(N_{\text{II}}/N_{\text{I}}\right)\gg \left(N_{\text{III}}/N_{\text{II}}\right)$ for all the given temperatures.

Therefore, for all the given temperatures, $\left(N_{\text{II}}/N_{\text{I}}\right)\gg \left(N_{\text{III}}/N_{\text{II}}\right)$.

(b)

To determine

To show that $N_{\text{II}}/N_{\text{total}}=\frac{N_{\text{II}}}{\left(N_{\text{I}}+N_{\text{II}}+N_{\text{III}}\right)}$ can be expressed in terms of the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$.

Answer to Problem 8.10P

The ratio $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}}{\left(N_{\text{I}}+N_{\text{II}}+N_{\text{III}}\right)}$ can be expressed in terms of the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ as $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}/N_{\text{I}}}{\left(1+\left(N_{\text{II}}/N_{\text{I}}\right)+\left(N_{\text{III}}/N_{\text{II}}\right)\left(N_{\text{II}}/N_{\text{I}}\right)\right)}$.

Explanation of Solution

Given ratio expression is;

  $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}}{\left(N_{\text{I}}+N_{\text{II}}+N_{\text{III}}\right)}$ (IV)

Take $N_{\text{I}}$ as a common factor from the denominator in equation (IV).

  $\begin{array}{c}\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}}{N_{\text{I}}\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}+\frac{N_{\text{III}}}{N_{\text{I}}}\right)}\\ \frac{N_{\text{II}}}{N_{\text{total}}}=\frac{\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)}{\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}+\frac{N_{\text{III}}}{N_{\text{I}}}\right)}\end{array}$        (V)

Multiply and divide the third term in the denominator in equation (V) with $N_{\text{II}}$.

  $\begin{array}{c}\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)}{\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}+\frac{N_{\text{III}}}{N_{\text{I}}}\times \frac{N_{\text{II}}}{N_{\text{II}}}\right)}\\ \frac{N_{\text{II}}}{N_{\text{total}}}=\frac{\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)}{\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}+\left(\frac{N_{\text{III}}}{N_{\text{II}}}\right)\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)\right)}\end{array}$        (VI)

Equation (VI) is the form of the equation (IV) written in terms of the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$.

Conclusion:

Therefore, the ratio $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}}{\left(N_{\text{I}}+N_{\text{II}}+N_{\text{III}}\right)}$ can be expressed in terms of the ratios $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ as $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{N_{\text{II}}/N_{\text{I}}}{\left(1+\left(N_{\text{II}}/N_{\text{I}}\right)+\left(N_{\text{III}}/N_{\text{II}}\right)\left(N_{\text{II}}/N_{\text{I}}\right)\right)}$.

(c)

To determine

To plot the graph of $N_{\text{II}}/N_{\text{total}}$.

Answer to Problem 8.10P

The plot the graph of $N_{\text{II}}/N_{\text{total}}$ is shown in Figure 1.

Explanation of Solution

The expression for $N_{\text{II}}/N_{\text{total}}$ written in terms of $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ is given in equation (VI).

  $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)}{\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}+\left(\frac{N_{\text{III}}}{N_{\text{II}}}\right)\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)\right)}$

From the obtained values of $N_{\text{II}}/N_{\text{I}}$ and $N_{\text{III}}/N_{\text{II}}$ from part (a), the term $\left(\frac{N_{\text{III}}}{N_{\text{II}}}\right)\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)$ can be neglected since they does not contribute significant change to the value of $N_{\text{II}}/N_{\text{total}}$ for the temperature range $5000\text{K}$ to $25,000\text{K}$. Thus, equation (VI) becomes;

  $\frac{N_{\text{II}}}{N_{\text{total}}}=\frac{\left(\frac{N_{\text{II}}}{N_{\text{I}}}\right)}{\left(1+\frac{N_{\text{II}}}{N_{\text{I}}}\right)}$        (VII)

With the given values of electron pressure, ionization energies and partition functions for the neutral helium atom and singly ionized helium atom, the ratio $N_{\text{II}}/N_{\text{I}}$ can be determined for a range of temperatures in between $5000\text{K}$ to $25,000\text{K}$ as shown in part (a). These values can be plugged in to equation (VII) and hence the plot of $N_{\text{II}}/N_{\text{total}}$ will be obtained.

Conclusion:

The graph of $N_{\text{II}}/N_{\text{total}}$ is shown in Figure 1 below.

From the graph it can be noted that when the temperature corresponding to the partial ionization of the neutral helium is about $15,000\text{K}$ or $1.5\times {10}^4\text{K}$.