# What is unit system in this reduction process?

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 :

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

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

• Link to abstract page? Mar 21, 2022 at 5:49
• @Qmechanic emm, I have not link.... Is it necessary? Mar 24, 2022 at 10:57

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 $$(Z_1^{1/2}+Z_2^{1/2})^{2/3}$$
• ohh, thank you, but the $Z_1^0.23+Z_2^0.23$ is Ziegler universal formula factor. Anyway, thank you very much. I have tried several units to get it and I'm not familiar with CGS, without trying again. Thank you!!! Mar 24, 2022 at 10:55
• it's cause by V->statV,so it's $\frac{1}{10^{-8}c}$in CSG,right? Mar 26, 2022 at 7:16