# Saha-Boltzmann Equation

I am following Ahmed, Ahmed, Rafiqe & Baig A comparative study of Cu–Ni Alloy using LIBS, LA-TOF, EDX, and XRF paper on Laser Induced Breakdown spectroscopy.

It mentions two versions of Saha-Boltzmann equation and I can't understand the difference/similarity in the two equations. One equation is,

$$n_e\frac{C^{z+1}}{C^z}=\frac{\left(2m_ekT\right)^{3/2}}{h^3}\,\frac{2U_{z+1}}{U_z}\,\exp\left[-\frac{E_\text{ion}}{kT}\right]\,\text{cm}^{-3}\tag{6}$$ Equation 6 gives the ratio of the concentration of two charge states $$Z$$ and $$Z+1$$ of the same element ($$C^{Z+1}/C^Z$$) (Unnikrishnan et al 2012; Andrea et al 2015), from here we can easily calculate the value of $$C^{Z+1}$$ by substituting the value of $$C^Z$$ obtained from Eq. (5).

The other equation is,

Due to the insufficient number of observed lines in NiII and CuII, it was not possible to draw the Boltzmann plot for the single ionized species, separately. Thus, the Saha-Boltzmann equation was used for estimating the values of $$FC^{Ni\,I}$$ (or $$n^{NiII}$$) and $$FC^{CuII}$$ (or $$n^{CiII}$$) as below. $$\frac{n^{\alpha,z+1}}{n^{\alpha,z}}=\frac{6.04\times10^{21}\left(T_\text{eV}\right)^{3/2}\left(P_{\alpha,z+1}/P_{\alpha,z}\right)\exp\left[-\chi_{\alpha,z}/k_BT\right]}{n_e}\tag{11}$$

I tried looking into the mentioned references but that did not help much. Also, I have already calculated if the stated constants are equivalent.

• The factor $2 (2 \ m_{e} \ k_{B})^{3/2} \ h^{-3}$ is roughly $1.084 \times 10^{27}$ to make the equation allow for temperature in units of eV (as in the 2nd example above, i.e., your Equation 11). Typically the number densities are expressed in units of $cm^{-3}$ which means multiplying my above factor by $10^{-6}$. If there is a charge state of +6 floating around, the numerical factor goes to roughly $6.505 \times 10^{21}$. Note that I used double precision and 2014 CODATA values, they may have rounded things or I may be on the wrong track. – honeste_vivere Apr 23 at 22:47
• So the temperature factor is separate as the later equation expresses it in eV so that is equivalent to $(k_BT)^{1.5}$. The remaining factors are $2(2m_e)^{1.5}h^{-1.5}$. And when I calculated it, this not equivalent to the given number. – Shaz Apr 24 at 4:13
• So it is $h^3$. – Shaz Apr 24 at 8:43
• Replace the Boltzmann constant with the fundamental charge to get the conversion to eV, leaving $T^{3/2}$ in place with the understanding that the input temperature will be in eV. – honeste_vivere Apr 24 at 12:51