Example:

You have just pressed a pellet (disk) of kaolinite to be used for a diffusion experiment to study fluid-rock interactions. You need to know the porosity of the pellet. Here's one way to proceed.

Given:

I_o = 10,000 counts per second (CPS). This is the intensity measurement with no sample in place.

I = 3,500 CPS Intensity of transmitted beam with sample in place.

Kaolinite ρ  = 2.6 g/cm³

Kaolintite µ* = 30.8 cm²/g

x = 0.02 cm (thickness of pellet)

Assume that air does not influence the absorption (which is not totally true, but we can ingore for now). Given the small value of ρ for air (i.e., ~ 0.0 g/cm³), we can simplify the calculation. The pellet is a mixture of air and kaolinite.

I = I_o exp(-µ*ρ x) _mixture

3500 = 10000 exp(-µ*ρ 0.02)

0.35 = exp(-µ*ρ 0.02)

Find natural log of each side.

-1.0498 = - (µ* ρ) 0.02

52.49 = (µ* ρ) _mixture

The value of 52.49 represents kaolinite's absorption contribution to the thickness of the pellet.

52.49 = (30.8) (2.6) (fraction of thickness)_kaolinite

If the pellet were 100% kaolinite, then (µ* ρ) _kaolinite = 80.08

A mixture of 66% kaolinite and 34% air would give the observed absorption decrease. Therefore porosity = 34%.

Alternatively, one could determine the µ* if the porosity is known.