Background studies for the

56Fe target for the CLAS multihadron expt (updated 11 Dec 98)

The CLAS multihadron experiment (e2 run group) is planning to use a 56Fe target. 56Fe gives off 6.4 keV x-rays that produce hits in the wire chambers. Here I calculate the shielding needed to run with a 56Fe target at full luminosity. Procedure:

1. Calculate total flux of 56Fe 6.4 kev x-rays
2. Calculate  target self-shielding
3. calculate DC random hits
4. calculate needed attenuation and Pb shielding that can do this

New Summary (12/11/98): We need to attenuate the 6.4 keV x-ray flux by about 1000 to reduce the wire chamber random hits from 56Fe x-rays to an acceptable level (0.5% Region I DC hit occupancy per event).

This requires about 16 mg/cm^2 of Pb or 21 mg/cm^2 of Mo.

Corrections (10/26/98) [with thanks to Bob Owens]:

1. x-ray energy: The x-rays are 6.4 keV x-rays from K-shell knockout followed by L-shell de-excitation. The lower energy implies higher absorption cross-sections.
2. x-ray flourescence yield: only 25% of ejected K-shell electrons lead to x-rays, the rest lead to Auger electron emission.
3. Target self-absorption. A 6.4 keV x-ray cannot give secondary x-rays above 1 keV. His experience with a 30 mg iron target indicates that only 20% of the x-rays emerge from the target. My calculation for our 200 mg target indicates an effective thickness of only 5 mg! (Thus a 30 mg target and a 200 mg target have the same effective thickness for x-ray production.)
4. We might want to use materials other than Pb as our shield.
5. The free Moller cross-section overestimates the cross-section near threshold.  (I could not quantify this and thus did not include it in the calculation.  Consider it another safety factor.) 

These corrections reduce the shielding needed (from 10⁴ to 10³).

1) Flux calculation

Assume 6.4 keV x-rays come from moller scattering of incident beam on the two 56Fe k-shell electrons followed by de-excitation of the atom.

The Moller total cross-section is (from GEANT manual PHYS-330):

E_cutoff is half the lowest energy Moller e- of interest. Here E_cutoff = 0.5 * 6.4 keV.
 

sigma_tot is independent of beam energy (as it should be)

sigma_tot is determined almost entirely by the lowest energy Moller e- (as expected, a QED cross-section is dominated by the smallest angle scattering).

This calculation also agrees approximately with a differential Moller cross-section calculated by Andi Klein.

Thus the Moller scattering total lab cross-section is sigma_tot = 1.6*10⁴ fm² = 1.6*10^-22 cm².

The fluorescence yield (x-rays per Moller) is about 25%.
This gives sigma_tot^x-ray = 4.0*10^-23 cm²
At a luminosity of 10^34 / cm^2 s, this gives X-ray flux =  4*10¹¹ / s

This is (assuming the x-rays are isotropic) 3*10^10 x-rays / s sr. Rather a lot. Fortunately there is a lot of target self-shielding.

2) Target Self-shielding:

All x-rays exiting the target have to traverse part of the target.  They attenuate by e^-R/l where R is the length of target to traverse and l is the mean free path (for iron it is 13.6 mg/cm²).  Thus t(theta) = integral₀^theta e^-R/l dx = (13.6 mg/cm²) cos(theta) where R = x/cos(theta).
In the forward direction, the minimum angle is 8^o, in the back angle it is 37^o (180^o - 143^o).  The average target thickness is 5 mg/cm².
This reduces the average x-ray flux (assuming a 200 mg/cm² target) to

X-ray flux = 10¹⁰ /s = 8*10⁸ /s-sr

We also need to look at a typical angle (theta = 50^o) : t(50^o) = 13.6 * 0.64 = 8.7 mg/cm² X-ray flux(50^o) = 14*10⁸ /s-sr

3) Hits in the drift chambers:

a) We need to calculate the absorption lengths of various materials.

These are nicely presented at NIST. Local copies of the absorption lengths for Nitrogen, Argon, Iron and Lead. The interpolated attenuation lengths from the NIST tables are shown below for 6.4 keV x-rays.
 

The mean free path is the 1 / (attenuation length * density).

b) Now we need the Region I drift chamber geometry. The wire length is proportional to sin(theta), the target thickness is proportional to cos(theta), so greatest number of x-rays per wire will occur at about 45^o.

Region I has six sectors, each about 60 cm from the beam with a cell diameter of 1.4 cm.  Thus, each wire subtends a solid angle of 
Solid angle (50 degree region I cell): (2*pi/6) sin(50^o) 1.4 cm / 60 cm = 18 msr

Flux through active area = 14*10⁸ /s-sr * 18 msr = 2.5*10⁷ xrays/cell-s

c) The thickness of the air before the wire chamber is approximately 1 mfp.  The thickness of one wire cell is approximately 0.6 mfp.
(1 - 1/e) of x-rays are absorbed in the air before the chamber.
(1-e^-0.6) of x-rays are absorbed in the first wire cell (and cause all the trouble)

The drift chamber TDC gate is approximately 1 usec so the unshielded DC occupancy will be 4.2 hits/cell for each event.

The maximum acceptable occupancy is 5%. We should design in a safety factor of 10.

The shielding needs to reduce the photon flux by at least 10^3.

4) Shielding

Let's reduce the x-ray flux by 10³ so we have a safety margin of 10. This means we need log(10³) = 6.9 attenuation lengths of shielding. This can be reduced slightly since the particles exiting the scattering chamber will traverse the shielding at angle = theta + 20^o.  Thus, at the worst case of 50^o, we need 6.9*sin(70^o) = 6.5 attenuation lengths. To have a reasonable cross-section, the shielding material should be heavier than the target. Look at Pb and Mo. Pb: 10³ attenuation = 6.5 mean free paths (Pb) = 16 mg/cm^2 = 14 microns Mo: 10³ attenuation = 6.5 mean free paths (Mo) = 21 mg/cm^2 = 21 microns

Now we need to check multiple scattering. For a nominal 1 GeV e- leaving the 200 mg/cm^2 iron target we have (assuming the interaction accurs in the middle of the target and the particles leave at a small angle):
 

Thus, the multiple scattering from an iron target with a radiation shield will be 1.5 mr and without a shield will be 1.3. By contrast, the multiple scattering from the 4He target will be 0.9 mr without a shield and 1.3 mr with the shield.

Last modified: Fri 11 Dec, 1998 19:04:30 MET