# Where is the missing gram unit in this expression for energy loss?

In the discussion of "Landau Straggling", the following expression comes up for the energy loss $\xi$ in Landau's original paper:

$$\xi = x\frac{2\pi N e^4\varrho\sum Z}{mv^2\sum A}$$

Here $N$ is avogadro's number, $e$ is the electron charge, $\varrho$ is the density, $m$ the electron's mass and $v$ a velocity. Moreover $x$ is a measure of length and $Z$ and $A$ are the usual atomic and mass numbers.

The point here is not to get into the physics, but rather to look at the dimensional analysis: $$\rm [\xi]=[MeV]$$ But the units of the RHS are: $$\rm[cm][statC^4][gm\cdot cm^{-3}][gm^{-1}][s^2\cdot cm^{-2}]=[MeV\cdot gm] \neq[MeV]\quad\bf!!$$ I'd ignore this as a typo, but all of the literature on the topic seems to follow the same discrepancy in units, so there must be something I'm missing.

I'd be tempted to say that $\varrho$ might be a number density, but Landau (and others) explicitly states this is mass per volume.

What am I missing? Where is the missing gram unit?

Edit: This snippet from Leroy (2009) on the topic has the same exact problem and may be simpler to understand:

![enter image description here][1]

So where is the missing gram?

The value of $n_A$, in units of $\rm cm^{-3},$ is given by $$n_A=\dfrac{N_\rho}{A}, \tag{1.39}$$ where $N$ is the Avagadro constant (see Appendix $\text{A.2}$), $\rho$ is the absorber density, in $\rm g/cm^3$, and $A$ is the atomic weight [also known as relative atomic mass].

• What is the value of $\rho$? Sep 6, 2013 at 5:18
• @LubošMotl - From what I understand of the physics, It isn't known in the text. I don't know how it "disappeared" unless it's a typo. Sep 6, 2013 at 5:22
• A link to the paper in which this expression appears could be very useful. I have a hunch that $A$ could be actual atomic mass in g/mol, but I can't confirm that without seeing the formula in context. Sep 9, 2013 at 7:53
• @DavidZ - I thought so too at first, though look at the second image I posted. Landau writes: "$\sum A$ - sum of the atomic weights" Sep 9, 2013 at 8:02

The solution to the confusion lies in the definition of $A$. It is not the relative atomic weight, which is a dimensionless number. According to the Particle Data Group, it refers to the atomic mass of the absorber, a quantity of unit g mol$^{-1}$.