Efficiency

7.1 MeV 18F on 50 ug/cm2 CH2 target; alphas emitted isotropic in lab.

Angular distribution for alphas in centre of mass and lab, and for 15O in lab.



Energy of 15O as a function of alpha energy



Angle of 15O as a function of alpha energy.



efficiency, as a function of positions of 2 detectors:
LEDA + S2



Two LEDAs: the more distant one is positioned so that its maximum angle is 1' less than the minimum angle of the near one.



The bottom line: although the efficiency increases from 43% to 50% for the LEDA at 50 mm as the S2 behind it moves from 25 mm behind to 5 mm behind, an efficiency of 53% is obtained for one leda at 62 mm and a second at 167 mm from the target. Moreover, for a minimum distance of 100 mm (consistent with TOF PID), LEDA+S2 has an efficiency of ~6%, whereas 2LEDA has an efficiency of ~44% (at 100 mm and 272 mm).

***Update***

Here's a figure showing the efficiency of two detector configurations for 5.7 MeV 18F.



The S2 is always 5 mm behind the LEDA; for the two-leda configuration, the positions are as follows:
forward backward
leda leda
30 90
40 120
50 150
60 180
70 210
80 241
100 303
120 366
140 430
160 495
180 561
200 629
(all in mm from the target)

New non-monte-carlo simulation for TUDA

The idea: Quickly produce energy spectra for all strips of a segmented detector, for reactions that populate many excited states in the final nucleus.

Background: During the TUDA 12C+12C experiment, I wrote a monte carlo code to do just this, but to get reasonable statistics in each strip, you need to run 100000 events (about 3 minutes run time on my laptop) per excited state, which means that it's quite labor-intensive to produce spectra with many excited states and several reactions...and of course making mistakes is both inevitable and a pain. Götz Ruprecht asked why it was necessary to do a full event-by-event monte carlo simulation, instead of just finding peak locations and widths from energy loss/straggling codes; I couldn't think of a good reason why not, but I didn't have time to try out that way of doing things--until now.


Approach:

• Calculate the beam's energy loss up to half way through the target. This gives the average beam energy when the reaction takes place. For the sigma of the beam energy distribution, use the energy loss itself (because (a) this is easy and (b) it should be approximately right: the beam will lose somewhere between no energy at all and twice the average energy, and the straggling will be something like 10% of the energy loss).
  
• For a detector at a certain position (I started with a LEDA at 10 cm; I've tried moving it farther back, and am planning to add a forward S2 and a backward LEDA), find the angles corresponding to the inner and outer edges of each strip.
• For each excited energy, cycle through all strips: do the reaction using the inner and outer angles for the strip as the angle of the ejectile; find the initial energy of the ejectile; find the energy lost by the ejectile as it passes through the remaining half of the target (and again use sigma = energy loss); find its energy loss in the dead layer of the detector (this is probably negligible but it is included for completeness; sigma = 10% energy loss since there is no variability in the thickness in this case).
• Now for each strip we have θ_inner, θ_outer, E_{final inner}, E_{final outer}, and σ_E (which is the sum in quadrature of σ for the energy loss of beam in target, ejectile in target, and ejectile in dead layer) . E_{final inner} is always bigger than E_{final outer} (for forward angles), so make the energy spectra as follows: assign a channel number to E_{final inner} (so far I'm using 1 keV per channel), and for higher channel numbers calculate a gaussian distribution with amplitude 1000, mean E_{final inner}, and sigma; for channels below E_{final outer}, calculate a gaussian with mean E_{final outer}; between E_{final inner} and E_{final outer}, write the peak value of the gaussians. (Since both gaussians have the same amplitude and sigma, their peaks will have to be the same size; just in case, there are a couple of lines to calculate both peaks and choose the greater.) The result is a rectangle with softened edges. The shape represents the number of particles of a given energy hitting the strip. The flat top is not quite accurate, since (particularly for a close detector for which each strip spans a broader range of angles) there will be a variation in intensity across the strip; also, the cross section will actually change from strip to strip, which is not reflected in the code. The results, then, do not reflect the peak shapes or relative heights that can be expected, but should reflect the locations and maximum widths.
  
• For additional exitation energies, the spectra are incremented instead of being overwritten, so that a composite of all excitation energies is formed. It would be possible to do multiple reactions in a single run, but I think that it's more useful to do each reaction separately so that you can (for example) plot the alpha particle spectra in blue and the proton spectra in red (or the other way around); having separate spectra for separate reactions makes particle identification easier.

Below I compare the results of this simulation with the event-by-event simulation. The peak locations are the same; the widths are similar (although the new simulation necessarily gives an upper bound on the width and in fact is a slight overestimate because currently the strip width of LEDA is taken to be 5 mm instead of ...4.9 mm? It doesn't account for the inactive gaps between the strips, anyhow.); but the run time for the new simulation was practically instant whereas the event-by-event simulation took ~ 6 minutes. The spectra for the innermost strip of a LEDA at 10 cm is plotted in both cases. The reaction is 80 MeV 12C on a 50 ug/cm2 12C target --> proton ejectile.



And here's a comparison of two detector locations: 10 cm vs 50 cm. The farther position has narrower peaks because of the smaller angular range spanned by each strip, and of course the energies change because the angles are smaller for the farther detector.

18F(p,a): contaminants

The following figures show kinematic curves for a bunch of open reaction channels (results from Catkin) for 4.71 MeV 18(F,O,Ne) on CH2. I've considered proton scattering as well as (p,a) and (12C,p) and (12C,a). The most problematic reaction is 18Ne(p,a) because both the light and heavy products' kinematic curves lie almost on top of the 18F(p,a) curves; fortunately the Q values are quite different: +2882 keV for 18F vs. -6594 keV for 18Ne, so even if there is 18Ne present in the beam it may not produce any (p,a).

18F(p,a): reactions at different depths in the target

A slight simplification: an "event" is just an α being emitted with an angle equal to or less than the maximum angle of LEDA; not worried right now about coincidences.
The question is what c.m. energies correspond to different depths in the target; what beam energy we need to get a given c.m. energy at a given depth; and how many counts come from different depths (i.e. can we run with a thin target without sacrificing total count rate).
The cross section used is an approximation to the one found in the figure from Bardayan et al., reproduced in the expt proposal.
The figure below shows the total number of hits for a run at each of four different beam energies (only 17 s run time at the nominal beam current), from depths within each of five chunks within the 50 ug/cm2 target, plotted as a function of the average c.m. energy of those events within that chunk (with E(cm) standard deviation as x error bar).
N.B. The E(cm) given is the actual god-given number, not the number that would be reconstructed from measured energies. I can plot that too, but right now it's a sunny evening, so it will have to wait until tomorrow when it's forecast to rain. :D
Comments? questions? requests?

out-of-context quotation JOY

"What we ACTUALLY need is a force field."
"Oh! Oh! What are you DOING, woman?"
--Tom "Tuda" Davinson

"Licking Beryllium is fine."
--Lothar

"Are you smelling my fruit?"
--Kelly

"What? That's not nice? ...I just don't understand people, do I?"
--Alex

"Of course there's a plan. As soon as I know what it is, I'll tell you."
--Pierre

"What we do is a bit of a mess."
--Alison

"I'm too chilled out to care."
--Simon

18O elastic scattering

31.5 MeV 18O beam corresponds to 1.75 MeV/u.

18Ne(a,p): rutherford backscattering from entrance foil

*Monitoring the backscattered beam in the upstream LEDA: What energies should we expect, and what resolutions? How much will LEDA be shielded by the entrance to the gas cell?
*Use 26.45 MeV beam (the new approved number for populating the 2.52 MeV resonance in the middle of the 2 cm cell); 4 cm diameter gas cell entrance 6 cm upstream of entrance foil; LEDA 20 cm upstream from the entrance.
*SRIMulations were used to set the energy losses of the beam through the foil
*plot the average energy in each strip against the average angle for each strip; error bars are the standard deviation of the energies.

Update on various simulations for 18Ne(a,p)

(Unless otherwise noted, all simulations below use 29.7 MeV 18Ne on 2 um Ni entrance window, 8 cm length of 150 torr He gas, baffle in place at 6 cm, Ni exit window (default 6 um), 1000 um LEDA at 20 cm downstream, S2 immediately behind LEDA.)

1. Consider effect of different thicknesses of exit window: 6 um Ni vs 2 um Ni; also degraders in front of S2.
Here is the energy-vs-angle plot for energy deposited in the detectors (*includes* dead layer effects; also includes a 50 um Al degrader foil in front of the S2 only).



Here is a plot of the energy spectrum in Strip #7 of the LEDA, for both exit window thicknesses. The energy resolution changes only very slightly.


Effective thickness of the detectors--oops, still assuming LEDA is 500 um instead of 1000 um, so ignore the larger groups on this plot. The smaller groups are the particles that hit the S2. Their energies are still high enough and their angles low enough that they can punch through the detector--this is with the 50 um Al degrader in place.


to do: check effect of 6 um Ni degrader in front of S2.

2. Elastic scattering of protons from contaminants on exit window (assume entrance window is blocked by baffle). The protons are on the inside surface of the 2 um exit window: the beam stops in the window, so there is no scattering from the outside surface.
(sorry for the low stats) (I also tried running a 6 um exit window, but I think the protons must all stop in the window--to do: double check this with Srim range tables.)

3. Beam elastic scattering on Ni entrance window. (It was suggested to use a gold foil in front of the entrance window, but Kelly checked the relevant energy losses and they would push the resonance out of the active area of the cell.) --in progress!

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.

18Ne(α,p) simulations: downstream distance of detectors

The problem: populate a resonance in the compound system and observe outgoing protons: what distance for a downstream LEDA gives the optimal energy resolution/efficiency?
The simulation: choose a random point in the gas; calculate beam energy loss up to that point; if the energy is within +- 10 keV of 13.75 MeV (2.5 MeV c.m.?), go ahead with the reaction. (This is a crude but easy approximation to a real cross section function.) Choose a random (isotropic in the lab) direction for the outgoing protons; calculate their energy losses and angular straggling in the gas and the exit foil; calculate where they hit the plane of the LEDA; if their radial position is between 5 and 13 cm, the event is a hit. Add up the total number of hits and divide by the number of reaction events to find the efficiency of LEDA at its chosen position.
Use a gas cell pressure of 150 mbar (113 Torr); length is 8 cm; 6 μm Ni exit foil; test LEDA positions between 10 cm and 30 cm downstream from exit foil.

Here are the energies measured in each strip, for different positions. The + marks represent the average energy deposited in a strip, plotted against the average angle of the hit events (rather than the strip number). The error bars are the standard deviation of the energies in each strip.
As the detector array gets closer, the range of angles covered increases, but so does the spread in energies per strip.



Here's a more succinct way of presenting the same information. The average of all the strips' energies' standard deviations is plotted against the detector position; on the other axis is the total detector efficiency.
It's pretty much a direct trade-off between efficiency and resolution. Since there's no clear optimization, it is suggested that we put the LEDA at 20 cm, to allow adequate time-of-flight measurements.

E1106 calibration reaction

The reaction of interest is 17O (beam energy 3.45 MeV) + p (gas pressure 10 Torr) --> α + 14N, via a resonance in 18F of 183 keV lab resonance energy and width Γ=0.3 eV. The candidate calibration reaction is 18O+p-->15N + α via a 151 keV resonance in 19F with width Γ=0.3 keV.

kinematic curves for both reactions (the legend is wrong because I am a numpty: the residuals are obviously N isotopes not O isotopes)


cross sections for both reactions, as a function of position in the gas cell


Cross section as a function of position, for coincidence events


Residual (14N or 15N) hitting S2 detector: for coincidence events


alphas hitting barrel detector: for the same coincidence events

Darren's Louvain experiment revisited

Here's an annotated copy of the detector configuration paper from Darren's thesis. Dimensions of the detectors and their positions are indicated.

Orsay data: theta calibration?

In this post, I simulated the elastic scattering events that are coincidences between dets 1 and 4. Here are the results for some data.
sort routine: require exactly one good det 1 event and one good quad event. (A good event in det 1 means that both ends of one strip fire. A good event in det 4 means that both ends of one strip fire, and the corresponding strip in det 3 does not fire.) Also require that strip number 6 in det 1 is always the one to fire. (Having a single strip in det 1 simplifies the energy gain matching: the gain-match parameters have not been selected at all so it would be surprising if the energies of all strips of det 1 were even similar.) The hits in det 1 look like this (all strips, E vs p):



and the hits in det 4 look like this (all strips, E vs p):



Distribution of theta hits in det 1 strip 6: theta conversion used is theta = 15-9*position



Distribution of theta hits in det 4 all strips: theta conversion used is theta = 63-14*position



Theta for det 4 all strips vs theta for det 1 strip 6:



I was hoping that, as I messed around like this with the data, some patterns would appear and/or start making sense. But I still don't know...

• why the det 1 E vs p loci seems to slope up to the right. Does that mean that right (= large position) = small theta, or just that the gainmatching is bad?
  
• why the det 1 E vs p loci don't extend all the way across the detectors. Does this mean that left (= small position) = small theta = high intensity, or just that there's some kind of threshold/triggering problem?
  
• what all the blobs in det 4 correspond to...it looks like there are several distinct things happening, but I have no idea what any of them are.

I'm going to select one region in strip 6 of det 4 and see what patterns emerge in its data. Here's the region:



and here are the corresponding events in strip 6 of det 1: Now we're onto something!



Here are the blobs from det 4 strip 6 (on the right; the one used for gating) and det 1 strip 6 (on the left; the resultant).



For completeness, try gating on the other three evident blobs in det 4 strip 6:




For comparison/orientation, here's the ungated data for det 1 (top) and det 4 (bottom):