Klein-Nishina for estimating X-ray cross section

Question

I'm looking at interaction probability for X-rays with water and DNA, and recently have starting reading up on the Klein-Nishina identities for differential cross section. When integrated over all angles, this can be expressed per electron as

$$e \sigma_{KN} = 2\pi r_{e}^2\left(\frac{1 + \alpha}{\alpha^2}\left(\frac{2( 1 + \alpha)}{1 + 2\alpha} - \frac{\ln(1 + 2\alpha)}{\alpha} \right) + \frac{\ln(1 + 2\alpha)}{\alpha} - \frac{1 + 3\alpha}{(1 + 2\alpha)^2} \right)$$

where $\alpha$ is a constant relating photon energy and electron mass by $\alpha = h\nu / m_e c^2$ and $r_e$ electron radius. From this, the cross section per atom is simply $Z_{e}e \sigma_{KN}$. This is quite reasonable, but my question is how to use this in practice to deduce probability for an interaction; if I had a stream of photons with intensity $I$ with $N$ potential electron targets, then my rate of scattering events would be $W = Z_e NIe \sigma_{KN} $ - but let's say my target material is a given volume of water or DNA with density akin to that of water. Let's also for simplicity consider the target volume as a cube of volume 1 cubic centimetre; given a constant stream of X-ray photons with intensity $I$, how would one use the cross section here to estimate the number of events in a given time, or would there be a better method? I assume the vast majority of incident photons will pass without interacting, but I would like to quantify this if possible. All advice welcomed!

Answer 1

From the formula that you used, that is the cross section for Compton scattering only. If you are looking for other types of interaction, you will have to look for other formulas. A good way to avoid this, if you only need the values, is to check databases like NIST. You can find the cross sections for Compton, positron-electron generation and photoelectric effect, just select the material.

About the way to use the cross section, I find this material useful:

http://www.springer.com/cda/content/document/cda_downloaddocument/9783642008283-c1.pdf?SGWID=0-0-45-856085-p173898906

It links the cross section to the mean free path and also to the probability of having an interaction.

Last advice. You should keep an eye on this:

http://geant4-dna.org/

Geant4 is a software that simulates the interaction of particles with matter so your scenario should be included.

Answer 2

You work out the density of electrons in your sample, using $n_e = \rho/\mu_e m_u$, where $\rho$ is density and $\mu_e$ is the number of atomic mass units ($m_u$) per electron. (e.g. $\mu_e = 18/10$ for water).

The absorption coefficient is then given by $\alpha = n_e \sigma$ and the light beam intensity at a depth $x$ in the substance will obey $I = I_0 \exp(-\alpha x)$.

The probability of a given photon interacting is $1-\exp(-\alpha x)$ and when $\alpha x\ll 1$, most of the radiation goes through the sample, and the probability of interaction becomes $\sim \alpha x$. So the number of scatterings per second is equal to the number of photons incident upon the sample per second, muliplied by $\alpha x$.

Note that your cross-section only considers scattering from the electrons and not other possible processes.