# What is the linear attenuation coefficient and how does it relate to interaction probability?

I have misunderstanding the linear attenuation coefficient (L.A.C) concept. As known, L.A.C is depend on absorbed medium and energy of incident radiation. Supposing, L.A.C= 100 cm-1, how can this parameter measure the probability of interaction per unit of length however the probability values are between 0 and 1.

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For a LAC $\mu$, the probability of interaction after a path length $l << 1/\mu$, is approximately: $P \approx l \cdot \mu$, e.g. if $l = 0.001 cm$ and $\mu = 100 \, \mathrm{cm}^{-1}$, then the probability of interactions is approximately: $P \approx 10\%$. This no longer applies when $l \gtrsim 1/\mu$.

More precisely, the probability of interaction (or the fraction of incident radiation which will interact) is: $$P = 1 - e^{-l \cdot \mu}$$ Or equivalently the Transmittance is, $$T = e^{-l \cdot \mu}$$

This is often expressed using the mean-free-path $\lambda = 1/\mu$, such that typically a photon will interact after a distance $\lambda$.

• Thank you DilithiumMatrix for your answer. It seems clear now for me the μ concept. – Jafarino Dec 14 '16 at 11:23

The linear attenuation coefficient $\mu$ is defined using fractional beam absorption, i.e.

$\mu(x) \equiv -\frac{1}{I} \frac{dI}{dx}$

where $I$ is the intensity of the beam. Rearranging a bit leaves you with:

$\frac{dI}{I} = -\mu(x)dx$

The right hand side can be interpreted as the probability of interaction per unit length. It is properly constrained between 0 and 1 because no change in intensity can be greater than the total intensity $I$. Integrating both sides will give the probability of absorption over some distance $l$ which DilithiumMatrix covers nicely in their answer.