# Interaction photons-matter and dimensional analysis

I know that when photons pass through matter, the law that describes the intensity in function of the thickness is:

$$I(x)=I_0 e^{-\mu x}$$

where $\mu = \rho \frac{N_a}{A} \sigma$ and $\rho$=density of the matter, $N_a$= Avogadro constant, A= mass number of the matter and $\sigma$=cross section.

My question is about the units of measurement of $\mu$. I know that $[\mu]=[l^{-1}]$ but if I do the dimensional analysis, I obtain:

$$[\mu]=\frac{[m]}{[l^3]} \frac{atoms}{[mol]}\frac{[mol]}{[m]}[l^2]$$

and so I have:

$$[\mu]=[l^{-1}]\cdot atoms$$

Could you explain me why there is "atoms"?

Both you and freude are correct.

The units of Avagadro's constant are atoms / mole, but atoms is just a number and is dimensionless. That's why we write Avagadro's constant as mol$^{-1}$, and why the atoms units disappear from your final equation.

Response to comment:

You can write your expression for $\mu$ in various ways. In your expression:

$$\mu = \rho \frac{N_a}{A} \sigma$$

I'd guess you mean $A$ to be the molar mass (in kg). In that case the cross section $\sigma$ is the cross section per atom. That's why the number of atoms cancels out.

Suppose we take $\sigma_M$ to be the molar cross section, $\sigma_M = N_a\sigma$. Then $M/A$ is the number of moles present. We can then define $\rho_M$ to be the molar density, i.e. moles per cubic metre, and the equation becomes:

$$\mu = \rho_M \sigma_M$$

and the number of atoms disappears from the equation.

• thanks, but I'd like to understand the physical meaning of that "atoms".. could you help me? – sunrise May 1 '13 at 9:49
• @sunrise: I've edited my answer to respond to your comment. Hopefully this helps rather than making it even more confusing! – John Rennie May 1 '13 at 10:32

Avogadro constant is measured in units $mol^{-1}$ according to http://en.wikipedia.org/wiki/Avogadro_constant

What are units $atoms$? What do they measure? I have never heard about it.

• You're right. It isn't a unit of measurement. In fact, I haven't written it in square brackets.. I'd like to understand the physical meaning of the presence of "atoms".. – sunrise May 1 '13 at 9:48
• 'I'd like to understand the physical meaning of the presence of "atoms"' A lot of people seem to be making a fetish out of "physical meaning" on this site recently. "Atoms" means atoms. Who'd of thunk it? – dmckee --- ex-moderator kitten May 1 '13 at 14:15

The other answers already tell you what you need to know, but I'd like to add that this situation arises very often. One daily interaction with it is in car engines. They run at a certain amount of rpm, rotations per minute. Scientifically, these units would be $1/\mathrm{minute}$ or probably $1/\mathrm{second}$ because we hardly ever use minutes when measuring time.

The rotations are simply a dimensionless number, so the dimensional analysis wouldn't mention them. That's also why your $\mathrm{atoms}$ have no physical significance: they shouldn't be in your analysis since they're not units. They're simply the result of a counting operation: there are $1, 2, \ldots N_A$ atoms [in one mole]. Any such result of a counting operation is dimensionless.