18Ne(α,p) simulations

Use a 18Ne beam energy of 29.4 MeV and a He pressure of 200 mbar: here's the resulting distribution of beam energies in the last 2 cm of the 8 cm gas cell (entrance window 1780 ug/cm2 and exit window 5340 ug/cm^2 Ni):



Then if the reaction happens when the beam energy is within +- 100 keV of 13.75 MeV (lab), the resulting proton energies look like this, as a function of their true angles when leaving the exit window:



(The above curve is the energies for all protons that are about to hit one of the detectors, either LEDA or S2. The gap in the curve is the gap in the angular coverage for the two detectors.)

There is also a question about whether the protons will stop in the detectors. 9 MeV protons, for example, have a range of 591 um in Si (according to Srim's range tables), which means that they wouldn't stop in the 500 um LEDA that is currently in place. For each simulated proton event, therefore, I have calculated the effective thickness of the detectors, taking into account the angle at which the proton enters the detector, and compared it with the Srim range values. The results are shown here:



For a LEDA at 20 cm downstream of the exit foil, the protons all have a range smaller than the effective thickness of the detector; therefore they will all stop. The S2 is more of a problem, but I don't remember whether the S2 that's in place is really 500 um, or if it's thicker (or even thinner).

UPDATE:
1. The efficiency I reported initially is incorrect, because I was assuming isotropic distribution of protons in the lab, not the centre of mass. The actual efficiencies are as follows:

downstream distance of LEDA : percent efficiency of LEDA: total efficiency
10 cm : 15% : 21%
15 cm : 13% : 18%
20 cm : 11% : 14%

2. The optimal energy is 29.4 MeV = 1.63 MeV/u = just under the maximum energy for 18Ne(4+) (1.65 MeV/u.

3. Energy resolution figure:
a single strip of the LEDA shows a resolution of about 100 keV.


New update:
Kelly is concerned that the beam energy we need for the resonance is different than what I've been assuming. The beam energy at resonance should be 13.86 MeV instead of 13.75 MeV: this is to populate the 2.52 MeV (cm) resonance. I've re-done the simulation with a couple of changes: 29.7 MeV initial beam energy, corresponding to 1.65 MeV/u; and require the beam energy at the (randomly chosen) reaction position to be within +- 100 keV of 13.85 MeV. The beam energy at the reaction position is shown below...



...and here is the effective thickness of the detectors, as a function of particle energy. As before, punching through won't be a problem for the LEDA and a shield should take care of the S2.

18F(p,α): efficiency calculations

Revised May 14th--all-new photographic goodness.

The question: If we have one LEDA and three S2 detectors, where should we put them so as to maximize our efficiency for detecting coincidences?

Assumptions:

• The alphas are isotropic in the centre of mass
• We restrict ourselves to particles of reasonable energy: at least 300 keV deposited in the detector, to match a reasonable discriminator threshold. (This is actually a less stringent requirement than excluding all particles under 3', at least for forward angles.) (3' is the approximate scattered beam opening angle.)
• ¹⁸F beam energy 7.1 MeV (lab): no restrictions on how the reaction takes place, i.e. nothing explicit about resonances.
• ¹⁸O contaminant beam at the same energy; 50% contaminant.
  

Here's what the real energies and angles of the outgoing particles look like: assuming they hit annular detectors with thick (i.e. twice LEDA) dead layers. (Theta is the true angle of the particle, not the angle calculated by the detector.)



Here's an E vs E plot: energy of alpha vs energy of the 15O.



...and a theta vs theta plot: angle of 15O vs angle of alpha. The boxes indicate the angular ranges covered by a LEDA at 5 cm from the target and an S2 at 6 cm. This arrangement gives 50% coincidence detection efficiency; it would be even higher if we could move the detectors even closer to the target, but this is probably near the limit of convenient positions; and also past 70' the alpha energies start to get degraded by straggling through the target/deadlayer at a steep angle, and they may begin to be blocked by the target frame.



...the same thing, plotted using the program Igor to create a "binned" data set: each dot is colour-coded to represent the number of events within that particular 1'-by-1' bin. The boxes still represent the detector ranges.



Here's a plot of the summed energy (E_α + E_15O) as a function of alpha angle.


Interestingly, it may be easier to separate just the alphas from the different reactions than to separate the summed energy peaks, since the heavy ions suffer so much straggling in the target/deadlayer especially at low energies. See the following...



And here's a bonus: an image comparing E vs E for all events and for the measured coincidences.