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I have a simulation program which uses cgs-units. Additionally, I defined a natural system of units:

\begin{align} \text{L} = [R], \quad \text{M} = \left[ \frac{e^2}{p^2 R^9} \right], \quad \text{T} = \left[ \frac{1}{p R^3} \right]. \end{align}

Here $R$ denotes a particle radius, $e$ is the elementary charge and $p$ describes new charged particles created due to ionization per volume and time. I now want to convert a literature value for the energy (in eV) into this natural units. First I derived the unit of the energy in this system of units:

\begin{align} E = \frac{\text{M}\,\text{L}^2}{\text{T}^2}= \frac{e^2}{p^2 R^9} R^2 p^2 R^6 = \frac{e^2}{R} \end{align}

But now I'm not sure how to convert Joule or respectively electron volt to this kind of unit. How do you derive such a conversion factor?

Regarding @J.G. answer:

I'm now distinguishing the elementary charge in CGS with an index from the one in SI. So the conversion factor for $1\text{eV}$ should be

\begin{align} \alpha&=\frac{1\text{eV}}{E} = \frac{e}{E} \frac{\text{J}}{\text{C}} = e \frac{R}{e^2_{\text{CGS}}}\, \frac{\text{J}}{\text{C}} = e \frac{4\pi \epsilon_0 R}{e^2} \, \frac{\text{J}}{\text{C}} \\ &= \frac{4\pi \cdot 8.854 \cdot 10^{-12}\,\frac{\text{As}}{\text{Vm}} \cdot 10^{-7} \,\mu\text{m}} {1.602\cdot 10^{-19}\text{C}} \, \frac{\text{J}}{\text{C}} \\ &= 69,4538 \frac{\text{J}\text{As}}{\text{C}^2\text{V}} \\ &= 69,4538 \frac{\text{J}}{\text{C} \text{V}} \\ &= 69,4538 \frac{\text{J}}{\text{C}} \frac{\text{C}}{\text{J}} \\ &= 69,4538 \end{align}

in the natural system of unit. $R$ was chosen as $0.1\,\mu\text{m}$.

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    $\begingroup$ You can't write $1\text{eV}=1e\text{J}$, because that's dimensionally inconsistent: it should be $1\text{eV}=(1e/\text{C})\text{J}=1e\text{J/C}=1e\text{V}$ (hence the name $\text{eV}$). $\endgroup$
    – J.G.
    Apr 16, 2020 at 19:22
  • $\begingroup$ Off-topic remark: the singular form of "indices" is "an index." $\endgroup$
    – rob ♦
    Apr 16, 2020 at 21:56

2 Answers 2

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In this context $e^2$ is an abbreviation for $e^2/(4\pi\varepsilon_0)$. If you work in SI units, you'll have no trouble getting $E$ in Joules. Since $e$ has the same value in Coulombs an electronvolt does in Joules, you can get the result in electronvolts by not squaring the $e$-in-Coulombs factor.

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  • $\begingroup$ Thank you for your fast response! I updated my Question with further calculations $\endgroup$
    – enco909
    Apr 16, 2020 at 19:09
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For the electric potential energy between unit charges $e$, the relation

$$ U=\frac{\alpha\hbar c}{r} $$

is correct in both SI and CGS units. So in CGS units, the squared unit charge $e^2$ has the same units as $\hbar c$, since the fine-structure constant $\alpha$ is dimensionless.

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