1
$\begingroup$

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.

$\endgroup$

bumped to the homepage by Community 19 hours ago

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

1
$\begingroup$

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 $\lambda$.

$\endgroup$
  • $\begingroup$ Thank you DilithiumMatrix for your answer. It seems clear now for me the μ concept. $\endgroup$ – Jafarino Dec 14 '16 at 11:23
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.