<p>Just to provide a <a href="https://en.wikipedia.org/wiki/Ionization_energy#Electrostatic_explanation" rel="nofollow noreferrer">source</a>, the Ionization potential for a <strong>hydrogen-like</strong> atom is given by:</p>

<p>$$\chi_n = \frac{(13.6\:\mathrm{eV})Z^2}{n^2}$$</p>

<p>This is the energy necessary to ionize an electron in a hydrogen-like atom that "begins" at the $n^\mathrm{th}$ energy level. This equation is positive because you need to <em>input</em> energy to the electron to break it out of its potential well. Note that you may see a nearly identical equation for the energy of the electrons which will be negative. That implies the electrons themselves have negative energy because they're in the electric potential well and you need a positive amount of energy to ionize them.</p>

<p>Now, when you look at the Saha equation, you see a negative sign in the exponential term: $\exp(-\chi_i/kT)$. It was already established that $\chi_i$ is positive and so also are $k$ and $T$ simply by their nature. The negative sign comes from the derivation and meaning of this exponential term. Describing the full derivation would involve an entire lecture, but to point you along the right track, the exponential term is known as the <em>Boltzmann Factor</em> and ultimately comes from the fact that particle energies are very often distributed according to the <a href="https://en.wikipedia.org/wiki/Boltzmann_distribution" rel="nofollow noreferrer">Boltzmann distribution</a>. That negative sign is simply an artifact of the Boltzmann distribution concerning how particles like to probabilistically arrange their energies. If that sign was a positive, the universe would just blow up so be careful with your signs!</p>

CC BY-SA 3.0
2017-08-30T13:39:44.587Z
2017-08-30T13:39:44.587Z
zephyr
<p>The Saha Equation:</p>

<p>$$\frac{N_{i+1}}{N_{i}} = \frac{g_{i+1}}{g_{i}}\frac{g_e}{N_e h^3}(2 \pi m_{e} kT)^{3/2} \exp(- \chi_i / kT)$$</p>

<p>where $\chi_i$ is ionization potential. My problem is about the sign ($+$ or $-$) of $\chi_i$. I know that its formula is something like:</p>

<p>$$\chi_i = (13.6\ \mathrm{eV}) Z^2 \left(\frac{1}{n_i^2} - \frac{1}{n_j^2}\right)$$</p>

<p>and in this case $n_j$ is $\infty$. But I get confused about it's sign ($+$ or $-$). Can someone simply say what is the sign of the formula for ionization potential and the sign of the formula for the Energy we need to go from $n_i$ to $n_j$.</p>

<p>I mean: What is the complete formula for $E_m - E_n$ and also $\chi_{n}$? (with the right sign so they can be used in the first Saha Equation above)</p>

CC BY-SA 3.0
2017-08-30T09:12:27.030Z
The population number for a higher level should be less probable for equal $g$'s. Thus the minus sign is correct.
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-08-30T13:28:47.633Z
IIRC, your equation for $\chi_i$ only applies to hydrogen and  hydrogen-like atoms which only have one electron (i.e. $\mathrm{He}^+$ and $\mathrm{Li}^{2+}$)
Beta Decay
<astrophysics><stellar-astrophysics>
A simple question about Saha Equation
titansarus
# A simple question about Saha Equation

(Asked by: username_0 on 2017-08-29)

The Saha Equation:

$$\frac{N\_{i+1}}{N\_{i}} = \frac{g\_{i+1}}{g\_{i}}\frac{g\_e}{N\_e h^3}(2 \pi m\_{e} kT)^{3/2} \exp(- \chi\_i / kT)$$

where $\chi\_i$ is ionization potential. My problem is about the sign ($+$ or $-$) of $\chi\_i$. I know that its formula is something like:

$$\chi\_i = (13.6\ \mathrm{eV}) Z^2 \left(\frac{1}{n\_i^2} - \frac{1}{n\_j^2}\right)$$

and in this case $n\_j$ is $\infty$. But I get confused about it's sign ($+$ or $-$). Can someone simply say what is the sign of the formula for ionization potential and the sign of the formula for the Energy we need to go from $n\_i$ to $n\_j$.

I mean: What is the complete formula for $E\_m - E\_n$ and also $\chi\_{n}$? (with the right sign so they can be used in the first Saha Equation above)

**Tags:** astrophysics, stellar-astrophysics

> **Comments:**
> **username_1**: The population number for a higher level should be less probable for equal $g$'s. Thus the minus sign is correct.  
> **username_2**: IIRC, your equation for $\chi_i$ only applies to hydrogen and  hydrogen-like atoms which only have one electron (i.e. $\mathrm{He}^+$ and $\mathrm{Li}^{2+}$)  

---

## 1 answers

### Answer by username_3 on 2017-08-30. Score: 2

Just to provide a [source](https://en.wikipedia.org/wiki/Ionization_energy#Electrostatic_explanation), the Ionization potential for a **hydrogen-like** atom is given by:

$$\chi\_n = \frac{(13.6\:\mathrm{eV})Z^2}{n^2}$$

This is the energy necessary to ionize an electron in a hydrogen-like atom that "begins" at the $n^\mathrm{th}$ energy level. This equation is positive because you need to *input* energy to the electron to break it out of its potential well. Note that you may see a nearly identical equation for the energy of the electrons which will be negative. That implies the electrons themselves have negative energy because they're in the electric potential well and you need a positive amount of energy to ionize them.

Now, when you look at the Saha equation, you see a negative sign in the exponential term: $\exp(-\chi\_i/kT)$. It was already established that $\chi\_i$ is positive and so also are $k$ and $T$ simply by their nature. The negative sign comes from the derivation and meaning of this exponential term. Describing the full derivation would involve an entire lecture, but to point you along the right track, the exponential term is known as the *Boltzmann Factor* and ultimately comes from the fact that particle energies are very often distributed according to the [Boltzmann distribution](https://en.wikipedia.org/wiki/Boltzmann_distribution). That negative sign is simply an artifact of the Boltzmann distribution concerning how particles like to probabilistically arrange their energies. If that sign was a positive, the universe would just blow up so be careful with your signs!


---

