DPA Calculations
Calculation of displacements per atom (DPA) is crucial to the
characterization of radiation damage in material. The chart below is
most informative in the rate of displacement creation based on the
energy of the incident ion. Basically, below a certain energy, the ion
cannot displace an atom. Past twice that energy, it can displace an
increasing number of ions with increasing energy until reaching an
energy at which it can only displace \(\frac{E_{c}}{2E_{d}}\) atoms
per ion.
We can use this to determine the displacement rate. We start by defining
the maximum energy transfered, which is \(T_{max}=\Lambda E\), with
\[\Lambda=\frac{4M_{1}M_{2}}{M_{1}+M_{2}}\]
. Then, we can define the displacement rate as
\[\dot{\nu}=\int_{E_{d}}^{\Lambda E}\phi\left(E\right)\sigma_{d}\left(E,T\right)dE\]
. We can see that the displacement cross section \(\sigma_{d}\), so
we can define that as
\[ \begin{align}\begin{aligned}\sigma_{d}=\int_{E_{d}}^{\Lambda E}\nu\left(T\right)\sigma\left(E,T\right)dT\\This in turn requires a cross section, for which we will use the hard\end{aligned}\end{align} \]
sphere cross section
\[ \begin{align}\begin{aligned}\sigma\left(E,T\right)=\frac{\sigma_{p}}{\Lambda E}=\frac{c}{E}\\Then, all that's left to do is math\end{aligned}\end{align} \]
\[\sigma_{d}=\int_{0}^{E_{d}}\left[\left(0\right)\frac{\sigma_{p}}{\Lambda E}+\left(1\right)\frac{\sigma_{p}}{\Lambda E}+\left(\frac{T}{2E_{d}}\right)\frac{\sigma_{p}}{\Lambda E}\right]dT\]
\[ \begin{align}\begin{aligned}\sigma_{d}=\left(\frac{\sigma_{p}}{\Lambda E}\right)\left(\Lambda E_{d}-E_{d}\right)+\left(\frac{\sigma_{p}}{2\Lambda E_{d}}\right)\left(\frac{\left(\Lambda E\right)^{2}}{2}-\frac{E_{d}^{2}}{2}\right)\\From the last result, and the first definition, we can find\end{aligned}\end{align} \]
\(\dot{\nu}\) in \(\frac{\mathrm{dpa}}{\mathrm{s}}\)
Bibliography
-
J. K. Nastasi, M., Mayer, J. W., Hirvonen.
Ion-Solid Interactions: Fundamentals and Applications.
Cambridge University Press, 1996.
