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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 :

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

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

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

1 Answer 1

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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 $(Z_1^{1/2}+Z_2^{1/2})^{2/3}$

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  • $\begingroup$ 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!!! $\endgroup$ Mar 24, 2022 at 10:55
  • $\begingroup$ Emm, may i ask what is 1/300 from? $\endgroup$ Mar 26, 2022 at 7:01
  • $\begingroup$ it's cause by V->statV,so it's $\frac{1}{10^{-8}c}$in CSG,right? $\endgroup$ Mar 26, 2022 at 7:16
  • $\begingroup$ @LearningLin : Indeed, 1/300 (or, to be precise, 1/299.792458) is needed because the CGS (or, to be precise, CGS-ESU) unit of voltage (statvolt) equals approximately 300 Volts (en.wikipedia.org/wiki/Statvolt) . $\endgroup$
    – akhmeteli
    Mar 26, 2022 at 13:42
  • $\begingroup$ Yes, I understand. Thank you. $\endgroup$ Mar 27, 2022 at 14:03

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