How many photons can we expect on SPC?

XFEL

This is a brief document on estimating the number of photons that we can expect during the beamtime. We make a few assumptions:

• non-thermal electrons live up to 1 fs (estimated relaxation time)
• The x-ray beam has an energy of 100 μJ, 30% of which will hit the target
• The target is of size 1μm $\times$ 1μm $\times$ 10 μm
• We base on KLL Auger recombination to K shell for Cu

How many Auger electrons will we get?

The target absorbs ~ 60% of the photons hit on it (~ 1 absoprtion depth). Thus the number of K shell coreholes:
$\frac{30 \rm {\mu J}}{9 {\rm keV}} \times 60\% \approx 1.25\times 10^{10}$

roughly 70% of these events will cause KLL Auger electrons, thus we may expect $8.75\times 10^{9}$ electrons.

consequently the number of coreholes will be $n_{\rm h} = 3.75\times 10^{9}$.

How to estimate number of photons?

Easy. You start from the cross section model:
$N_{\rm ph}=n_{\rm h}\sigma_{\rm RR}vtN_{\rm Auger}$
Note that now we have $N_{\rm h} = 3.75\times 10^{9}$, and that $N_{\rm Auger }=8.75\times 10^{9}$. The volumn of the target: $V = 1\times 10^{-17} {\rm m}^3$ yeilds $n_{\rm h} = 3.75 \times 10^{26} /{\rm m}^3$
$v$ comes from the electron energy $\sqrt{2E_{\rm e}/m_{\rm e}}$: x-ray photons are of 9 keV, and the ${\rm K_\alpha}$ energy is 8 keV, thus KLL Auger electrons are of 7 keV. Thus $v\approx5\times 10^7 {\rm m/s}$
We just take $t=1\times 10^{-15} \rm s$

What about the cross section?

The corss section can be obtained by microscopic reversibility:

$\sigma_{\rm RR} = \sigma_{\rm PI}\times \frac{g_{\rm l}}{2g_{\rm u}}\times \frac{\hbar\omega}{m_{\rm e}c^2E_{\rm e}}$.

Whatever the configuration you have, \frac{g_{\rm l}}{2g_{\rm u}} is alway 1/2 (for the upper atomic energy level, you only have 1 electron at K shell, while for lower you have two).

$\hbar\omega = 16 {\rm keV}$, $E_{\rm e}=7 \rm keV$, $m_{\rm e}c^2=0.511 {\rm MeV}$

There's an old paper (ATOMIC DATA TABLES, 5, 51-111 (1973)) that claims the PI cross section from 9 keV to be some $3\times 10^4 ~{\rm barns}$ ($1 {\rm barn}=10^{-28}{\rm m}^2$)

Putting everything together we have $\sigma_{\rm RR}\approx5\times10^{-26} {\rm m^2}$

Finally, the emission signal?

Yes, that will be
$3.75 \times 10^{26} /{\rm m}^3\times5\times10^{-26} {\rm m^2}\times5\times 10^7 {\rm m/s}\times1\times 10^{-15} \rm s\times 8.75\times 10^{9}$
That's roughly 8,000 photons. Assuming the detector is 300 mm away form the target, we will still get about
$8000\times\frac{38.4\times35.2}{4\pi(300)^2}\approx10$ photons.

Will this extimation be smaller or larger?

To be honest I don't know, that depends on the collisional time. Here we assumed 1fs, but it could be better.