What is unit system in this reduction process?

Question

I have read "The Stopping and Range of Ions in Solids", which is published in 1985 about ion implantation. There is a equation in P.53 : $$ \varepsilon=\frac{32.53M_2E_0}{Z_1Z_2(M_1+M_2)(Z_1^{0.23}+Z_2^{0.23})} $$ It's come from these : $$ \varepsilon=\frac{aE_c}{Z_1Z_2e^2} $$

$$ E_c=\frac{E_0M_2}{M_1+M_2} $$

$$\ \ \ \ \ \ \ \ a=\frac{0.8853a_0}{Z_1^{0.23}+Z_2^{0.23}} $$ $a_0=0.529\ Å$ is Bohr radius.

Above are all his book. But I can't find the unit of $e$ . And I have try Gauss(CGS) and $SI$ system. And I can't get the $32.53$ , so anyone know it?

Note :

$\mathrm{SI} : e=1.602×10^{-19} \ \mathrm{C}$

$\rm Gauss(in \ CGS): e=4.8032×10^{-10}statC$

Answer 1

So $E_0$ is ion energy in keV (according to DOI: 10.1007/978-1-4615-8103-1_3, after eq. (16)). In your second formula, everything is in CGS. So $$32.53\approx 0.8853\cdot 5.29\cdot10^{-9}\cdot 1000\cdot 4.8\cdot 10^{-10}\cdot\frac{1}{300}/(4.8\cdot 10^{-10})^2.$$ By the way, according to DOI: 10.1103/PhysRevB.15.2458 , your $Z_1^{0.23}+Z_2^{0.23}$ should be replaced by $(Z_1^{1/2}+Z_2^{1/2})^{2/3}$