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Chapter 3: Problem 16

Assume that we are considering only the ionization of hydrogen, so that the electron density is equal to the positive ion density, and the Saha equation simplifies to \(n_{\mathrm{e}}^{2} | n_{0}=F(T)\) where \(F(T)\) is the right side of equation (3.10) Assuming that the total amount of hydrogen. \(n_{\mathrm{TOT}}=n_{\mathrm{e}}+n_{0},\) is known and constant, find an expression for \(n_{\mathrm{e}}^{2} / n_{\mathrm{roT}},\) the fraction of hydrogen ionized, in terms of \(n_{\text {TOT }}\) and \(F(T)\).

Short Answer

Expert verified
\(\frac{n_{\text{e}}^2}{n_{\text{TOT}}} = \frac{\left( \frac{F(T) + \sqrt{(F(T))^2 + 4n_{\text{TOT}} F(T)}}{2} \right)^2}{n_{\text{TOT}}}\).

Step by step solution

01

Define the given variables

We are given the following variables: \(n_{\text{TOT}} = n_{\text{e}} + n_0\), which represents the total amount of hydrogen, and the simplified Saha equation \(n_{\text{e}}^2 / n_0 = F(T)\).
02

Express \(n_0\) in terms of \(n_{\text{e}}\) and \(F(T)\)

From the simplified Saha equation, we have \(n_{\text{e}}^2 / n_0 = F(T)\). Solving for \(n_0\), we get \(n_0 = \frac{n_{\text{e}}^2}{F(T)}\).
03

Substitute \(n_0\) into the total hydrogen equation

Substitute \(n_0 = \frac{n_{\text{e}}^2}{F(T)}\) into \(n_{\text{TOT}} = n_{\text{e}} + n_0\). This gives us: \[n_{\text{TOT}} = n_{\text{e}} + \frac{n_{\text{e}}^2}{F(T)}\]
04

Rearrange the equation to solve for \(n_{\text{e}}\)

Rearrange the equation to yield a quadratic equation in terms of \(n_{\text{e}}\). This gives: \[n_{\text{e}}^2 + n_{\text{e}} F(T) - n_{\text{TOT}} F(T) = 0\]
05

Solve the quadratic equation using the quadratic formula

The quadratic formula is given by \(x = \frac{{-b \, \pm \, \sqrt{{b^2 - 4ac}}}}{{2a}}\). Here, \(a = 1\), \(b = F(T)\), and \(c = -n_{\text{TOT}} F(T)\). Thus, \(n_{\text{e}} = \frac{{-F(T) \, \pm \, \sqrt{{(F(T))^2 + 4n_{\text{TOT}} F(T)}}}}{{2}}\). Since electron density must be positive, we take the positive solution: \[n_{\text{e}} = \frac{{F(T) + \sqrt{{(F(T))^2 + 4n_{\text{TOT}} F(T)}}}}{2}\]
06

Find the fraction of hydrogen ionized

The fraction of hydrogen ionized is given by \(\frac{n_{\text{e}}}{n_{\text{TOT}}}\). Since we want \(\frac{n_{\text{e}}^2}{n_{\text{TOT}}}\), we square the above value of \(n_{\text{e}}\) first: \[n_{\text{e}}^2 = \left(\frac{{F(T) + \sqrt{(F(T))^2 + 4n_{\text{TOT}} F(T)}}}{2}\right)^2\] Dividing by \(n_{\text{TOT}}\), we get: \[\frac{n_{\text{e}}^2}{n_{\text{TOT}}} = \frac{\left( \frac{{F(T) + \sqrt{(F(T))^2 + 4n_{\text{TOT}} F(T)}}}{2} \right)^2}{n_{\text{TOT}}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Ionization
Hydrogen ionization is a pivotal concept in astrophysics and plasma physics. It refers to the process where a hydrogen atom loses its electron, becoming a positively charged ion. This ionization process is governed by several factors like temperature and electron density.
The Saha equation simplifies the calculation of ionization states in a plasma. Ionization is essential for understanding stellar atmospheres and various energetic processes in stars. By knowing the fraction of ionized hydrogen, we can infer the physical conditions in such environments.
Electron Density
Electron density, denoted as \(n_{\text{e}}\), is a measure of the number of electrons per unit volume in a given space. It plays a significant role in determining the ionization state in a plasma.
In our context, the electron density is equal to the positive ion density since we are considering only hydrogen ionization. Electron density is crucial because it influences how atoms and ions interact with radiation, affecting the overall behavior of the plasma. By understanding electron density, we can better predict the properties of the plasma.
Quadratic Formula
The quadratic formula is a mathematical tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). In our problem, we derived a quadratic equation for \(n_{\text{e}}\), the electron density.
The quadratic formula is given by: \[ x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = F(T)\), and \(c = -n_{\text{TOT}} F(T)\). Using this formula helps us find the electron density, which is crucial for calculating the fraction of ionized hydrogen. The positive root of the quadratic equation gives us a meaningful physical quantity - the electron density.

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