# A change of units in Bethe formula

I'm reading through the following paper: Monte Carlo simulation of non relativistic electron scattering by W. Williamson and G. C. Duncan.

In the following paragraph, I want to know how to arrive at $(2)$ from $(1)$. (It's basically a change of units) :

Bethe has derived an expression which gives the kinetic energy lost by a non-relativistic electron as it traverses a path of length $ds$ in matter. We assume that the energy lost per unit path length by the electron is given by the Bethe formula: $$\frac{\mathrm{d}T}{\mathrm{d}s} = -\frac{2\pi e^4}{T} NZ \ln\left(\frac{2T(\mathrm{eV})}{11.5Z}\right)\tag{1}$$

In Eq. (I), $T$ is the kinetic energy of the electron, $e$ is the electron charge, $N$ is the number of target atoms per $cm^3$, and $Z$ is the atomic number. For calculations it is convenient to express the energy loss in the units of $keV \over\mu m$ and in terms of the atomic weight and density of the target material. In these units, Eq. $(1)$ becomes: $$\frac{\mathrm{d}T}{\mathrm{d}s} = -7.83 \left(\frac{\rho Z}{AT}\right) \ln\left(\frac{174T}{Z}\right) \left(\frac{\mathrm{keV}}{\mu\mathrm{m}}\right) \tag{2}$$ where in equation $(2)$ $\rho$ ($\mathrm{g\over cm^3}$) is the density of the target, $A$ ($\mathrm{g}$) is the atomic weight of the target, and $T$ ($\mathrm{keV}$) is the electronic kinetic energy.

The problems is with the numerical coefficient $7.83$, where the numerical values of $e$ and $\pi$ and $N=\frac{\rho N_A}{A}$ are substituted in equation $(1)$. In CGS units, $e=4.80320427 \times 10^{−10} \,\,\,(Fr)$, which gives a wrong value when substituted in $(1)$. (I found out that the correct result can be obtained, if you divide it by $1.602\times 10^{-19}$, which is electronic charge in SI).

• Of course! And why it doesn't work is my question. – JKS Jan 20 '14 at 19:35
• No, I mean you need to edit the question to describe what you tried and why it didn't work. You should take a look at the advice in our homework policy; whether or not this is actually a homework question, there is some good information there about how to write a question that shows effort. – David Z Jan 20 '14 at 19:39
• The idea isn't just to make you suffer. If you explain what you do and don't understand and what issue in particular is confusing you, it is much easier to write a focused answer which explains your confusion. Think of this not only as a courtesy to the person writing the answer, but also something that will help you get an answer faster, since people will be more likely to answer a well-explained question. – Brian Moths Jan 20 '14 at 20:09
• @DavidZ I edited the question to meet your standards. – JKS Jan 20 '14 at 20:19

Equation (1) is dimensionally correct when the electron-charge is measured in franklins, and T is measured in ergs. When you want different units, you need to use a Wallot-Stroud type substitution.

The measure of $T$ is in ergs, but keV are sought, likewise, $s$ is in cm, but needs to be in um.

You then write $\frac {T\mbox{ ergs}}{s \mbox{ cms}} = \frac{2\pi e^4}{T\mbox{ ergs}} N \mbox{ atoms}\dots..$. Since no numeric constant is needed here, but it will appear when you substitute values into it.

The basic principle here is that something like $T$ is a number of $\mbox{keV}$, but the equation expects it to be in $\mbox{ergs}$. The means of converting this, is then to imagine that $T$ becomes $T\mbox{ keV} \frac{\mbox{erg}}{\mbox{keV}}$ The second factor is a pure numeric, and one sweeps these up, along with the constants already in the equation ($2\pi e^4$), to give the numeric constant for these units. The values given here need to to become recriprocals.

$1 \mbox{ erg} = 10^{-7}/1.60211^{-19}\mbox {eV} = 6.24*10^8 \mbox{ keV}$

$1 \mbox{ cm} = 10^{-2}/10^{-6} \mbox{ microns} = 10^4 \mbox{microns}$

$1 \mbox{ atom} = 1/6.023*10^{24} \mbox{ g-moles}$

We put these values into the equation, along with the ones that are aready there, as

$T/s * (1e4/6.24e8) = (2\pi * (4.8032e-10)^4 * 6.24e8) (6.023*10^{24}) \cdots$

The numerical constant is then the RHS divided by the LHS, for those values which are no longer c.g.s. are then given by the the equation above, which directly to the value sought.

the energy unit must set to $eV$ for both side