Author: Øystein H. Færder
Time: Summer 2010
Location: Rez, Czech Republic
Field: Data analysis in nuclear physics
Introduction
Before we start using a fissile fuel, it is important to know how feasible the fuel is. In this case, we will study the feasibility of
235
U as a fissile fuel.
The technical part of this experiment has been done before me, and it goes like this: a beam of protons strikes a lead target; neutrons are then produced as an outgoing beam by spallation; the spallation neutrons hits a sample of 235 U; the sample is then undergoing different reactions, mainly fission, but also alpha decay, beta decay and neutron decay, and therefore many new isotopes are produced from the sample. All of the isotopes emit photons due to beta decay and electron transitions in the atoms. The photons are then detected by spectrographs such that photons with different energy are registered in different channels. Finally the spectrum is import to a computer by an ADC converter.
Goal
My task is to use the spectra to find out how much which was produced of each isotope, hence also how much neutrons were produced in the spallation. These data are important because we need to know the efficiency of this fuel (number of neutrons produced per proton per foil mass), and because we want to know what kind of nuclear waste which will be produced.
Equipment
DEIMOS, a computer program used to evaluate spectra
Spectra from two different samples of
235
U, referred to as sample no. 5 and no. 6. We got 11 measurements from each sample, taken at different times and different positions at the spectrograph
The website http://nucleardata.nuclear.lu.se/nucleardata/toi/index.asp
for nuclear data, in this text referred to as “the nuclear database”
Execution
To evaluate the spectra, we use a program called DEIMOS. The first thing to do after loading in a spectrum was to check that the calibration was OK. We know that we should expect a huge peak centered about energy of 511 keV, because this is the energy of photons created in annihilation of electrons and positrons. Looking at the spectra from the first measurement of sample 5, which was taken at position no. 8 at time 51 minutes after end of irradiation, I found that peak very quickly, and it DEIMOS calculated the peak to be centered at 510.8 keV. Thus,

the annihilation peak is where it should be, with an error less than 1 keV. Hence, the calibration cannot be too bad.
I continued focusing on sample no. 5, and started measuring the areas under each peak of the spectrum from measurement 1, that is to say, the area between the peak and the continuum curve in the spectrum.
Then it was time to look at some of the peaks and see if I could figure out which isotope which the line corresponds to. I started with a peak at 846.9 keV with area 9701.4, which was a relatively high area in this measurement. I used the nuclear database to search for gammas with an energy close to this in a range of 1 keV.
Things to keep in mind while looking for a proper mother isotope: This first measurement is taken at 51 minutes after end of irradiation, at the spectrograph position with the lowest efficiency and the lowest measurement time 𝑡 𝑟𝑒𝑎𝑙
. Therefore, if the half-life, 𝜏
1/2
, of the mother isotope is shorter than 10 min, there will probably not be much left of the isotope, and we would not see any great line from the isotope. If 𝜏
1/2
is in the order of years or higher, it will probably radiate so slowly that we won’t obtain much photons from it. This is not always true, but if we locate a line where we can’t find any proper mother isotope in this range, we will skip that line and come back to it later, after evaluating the other measurements of the same sample. We will also use a “probability curve” that tells us the probability of creating a specific isotope from
235
U as a function of the atomic mass. In that way we have an idea about which isotopes we can expect there will be produced a lot of. With these criteria in mind, we should first of all check if there we can see other lines from the same isotope. Second: for small energy variations, in distances of 1-300 keV, the area of the line divided by the relative intensity 𝐼 𝑔
should almost constant if the lines are from the same isotope. For higher ranges this fraction will decrease with the energy, because the efficiency decreases with energy.
Back to the peak at 846.9 keV (with area 9701.4), I found several matches on the database following the above mentioned criteria, but in the end I concluded that it had to be a line from
134
I with energy 847.0 keV and relative intensity 95.4. This isotope has a half-life at 52.5 min, which is close to the time from end of irradiation to the time of measurement( 𝑡
0
=
51 min ), and the atomic weight is near the top of the probability curve, which makes it highly probable that we should see lines from this isotope. The isotope also has lines 884.1 keV and
1072.6 keV with rel. int. at 64.9 and 14.9, respectively, in agreement with observed peaks at
884.0 keV and 1072.5 keV with areas 6488.0 and 1310.6, respectively. This looked very promising, and we managed to locate more than 20 lines from this isotope in the spectrum, I can with very high certainty say that this isotope has been produced in the fissions of the sample during the irradiation of neutrons. The energy difference between all of these lines and the corresponding theoretical
134
I-lines were less than 1 keV, which also makes us more sure that the calibration is good. The area of each of these lines also agreed good with the relative intensity, except for a few lines which had too high area, but that may be due to blending with another line with almost same energy.

I managed to find a lot of other isotopes using this easy method, but there still remained a lot of strong lines which I had trouble finding a proper mother isotope to. There were some good mother isotope candidates for these lines too, but one problem was that these “candidates” only had one or two strong lines, and it wasn’t possible to tell which candidate which was the right one by using this simple method. I needed to obtain more information from the other spectra.
So, the next step was starting evaluating the other measurements from the same sample. Then
I could compare the area of a specific line at two different measurements and try to estimate the half-life of the mother isotope.
To estimate the half-life, we’ll do some computations: We know that the amount N of an isotope at a time t is given by
𝑁 = 𝑁
0 𝑒 −𝜆𝑡 where 𝑁
0
is the number of the isotope at 𝑡 = 0 , and 𝜆 = ln 2 /𝜏
1/2
is the decay constant. If the measurement begins at 𝑡
1
and ends at 𝑡
2
, we have
Δ𝑁 = 𝑁
1
− 𝑁
2
= 𝑁
0
(𝑒 −𝜆𝑡
1 − 𝑒 −𝜆𝑡
2 ) = 𝑁
0
(𝑒 −𝜆𝑡
1 − 𝑒 −𝜆(𝑡
1
+Δ𝑡) ) = 𝑁
0 𝑒 −𝜆𝑡
1 (1 − 𝑒 −𝜆Δ𝑡 )
From now on, I’ll change the names such that 𝑡
1
becomes 𝑡
0
, which often is referred to as the cooling time, and Δ𝑡 becomes 𝑡 𝑟𝑒𝑎𝑙
, the time of the measurement. To take the “dead time” in to consideration, we should also include the fraction 𝑡 𝑙𝑖𝑣𝑒 𝑡 𝑟𝑒𝑎𝑙
, where 𝑡 𝑙𝑖𝑣𝑒
is the live time of the measurement. But we will neglect it in this estimation, since it doesn’t make a big error. With some rough approximation, the area of a peak goes as
𝐴 ∝ Δ𝑁𝜀 𝑝
(𝐸)𝐼 𝑔 where 𝜀 𝑝
(𝐸) is the efficiency of the spectrograph position as a function of energy, and 𝐼 𝑔
is the relative intensity of the line. When comparing to different measurements, we should then have
𝐴
1
𝐴
2
≈
ΔN
1 𝜀
1
(𝐸
1
)𝐼 𝑔1
ΔN
2 𝜀
2
(𝐸
2
)𝐼 𝑔2
= 𝑒 −𝜆𝑡
1 (1 − 𝑒 −𝜆𝑡 𝑟𝑒𝑎𝑙,1 𝑒 −𝜆𝑡
2
(1 − 𝑒 −𝜆𝑡 𝑟𝑒𝑎𝑙,2
)𝜀
1
(𝐸
1
)𝜀
2
(𝐸
2
)𝐼 𝑔1
)𝐼 𝑔2
When looking at the same line (at different measurements) the fraction
𝐼 𝑔1
cancels out. I
𝐼 𝑔2 implemented this equation into a MATLAB code where I could estimate the half-life 𝜏
1/2 by solving this equation for 𝜆 graphically. I also imported a program code from Dubna to calculate the efficiency for a given position and a given energy. It was originally written for another programming language but was easy to translate into MATLAB.
First I had to try out this method on some of the lines which I had decided the mother isotope by the simple method which I used first. For the three strongest lines of the
134
I-isotope mentioned above, I got half-lives at 69.3, 69.0 and 71.6 min, which is not too far away from the actual half-life at 52.5 min. For an assumed
92
Sr-line at 1383.9 keV, I found the half-life to

be at 190 and 176.5 min by comparing measurement 1 and 2 and comparing meas. 2 and 3, respectively. And the actual half-life of
92
Sr is 162.5, so the error is still not too big. We managed to detect several new isotopes by using this method. E.g. an unknown line at 743 keV seemed to have a half-life at 340 min by comparing meas. 1 & 2 and between 900 and
1200 min by different comparisons of meas. 2 to 8. I concluded that the isotope
97
Zr with a half-life of 16.9 h (1014 min) was the main contributor to this line at long cooling times, but at short cooling times, there also got to be a more short-lived contributor (together with
97
Zr).
Some of the lines which we used this method on seemed to come from mother isotopes with half-life which could be anything between 5000 min and infinity. Looking at the database, all of these seemed to have mother isotopes with half-lives at the magnitude of thousands of years or higher. Because of the high half-life, the error is very, very huge.
After using this method for a while, I had located lines from both 131 I, 133 I and 134 I, so I expected that we should have some lines from
132
I, since this isotope have a half-life of 2.295 hours, which is within the range of half-lifes such as the lines should be obtainable. I actually found some lines which seemed to be the strongest lines of this isotope, but from my half-lifeestimation method, it seemed like the half-life was about 3-4 days. After some research, I realized that 132 I is being produced by beta-decay of 132 Te, which have a half-life of 3.204 days, and that looks like a good explanation why this lines yielded a half-life of 3-4 days.
Actually, both
131
I,
132
I,
133
I and
134
I is produced from this chain:
X
Sn
 X
Sb
 X
Te
 X
I, where X is the atomic mass. But for these masses, Sn and Sb have so short half-life that they won’t affect the yielded half-life mention-worthly much (though 131
Sb have just long enough half-life so that its lines are barely visible in the first measurement). It’s also the same with
131
Te and
133
Te.
134
Te have a half-life of 41.8 min and produces
134
I in a rate such that the yielded half-life from its lines in the 1 st
and 2 nd
measurement is about 70 min, while the real half-life is 52.5 min. These four above-mentioned isotopes of I decays further on to Xe, which is stable.
Now I had obtained a bit more knowledge about these isotopes and how they are produced, which leaded me to the final and maybe most efficient way to localize lines. This method goes like this: Looking systematically through isotopes of all masses between 80 and 150 (masses outside this range are not probable to be produced from
235
U, nor
238
U); for each mass, start with the first isotope which have a high enough half-life (minimum 20 min); from the half-life you should be able to see which measurement where you are most probable to see the strongest lines; see if you can locate the lines; estimate the half-life by the above-mentioned method; make a conclusion! You’ll do the same with the next isotopes with same mass until you come to a stable isotope, or an isotope with too high half-life. And remember, if one isotope have a much higher half-life than the next isotope in the beta-decay chain, the yielded half-life from the different measurements will be affected by that. By doing this, I was able to locate really many different isotopes. Now, let’s go to the final step in this analysis, the yield calculation!
To compute the number of produced nuclei of an isotope, we use the following equation:

𝑁 𝑦𝑖𝑒𝑙𝑑
=
𝐼 𝑔
𝑁 𝑑𝑒𝑡
(𝐸)𝜀 𝑝
(𝐸)
(𝐸)COI 𝑡 𝑟𝑒𝑎𝑙 𝑡 𝑙𝑖𝑣𝑒
1 𝑚𝐼(𝑝 + ) exp(𝜆𝑡
0
)
1 − exp(−𝜆𝑡 𝑟𝑒𝑎𝑙
) 𝜆𝑡 𝑖𝑟𝑟
1 − exp(−𝜆𝑡 𝑖𝑟𝑟
) where 𝑁 𝑦𝑖𝑒𝑙𝑑
is the number produced of the mother isotope of the line with energy E during the whole period of irradiation divided by mass of the activation detector and proton flux, from now on, only referred to as the yield of an isotope. 𝑁 𝑑𝑒𝑡
(𝐸) is the area of the line, measured from the spectra, COI is a correction factor which we here approximate to 1, m is the foil mass, 𝐼(𝑝 + ) is the proton flux, 𝑡 𝑖𝑟𝑟
is the total time of irradiation by the spallation neutrons, and the other are as mentioned earlier.
By inserting numbers into the equation from different measurements of different lines of the same isotope, I got several yields for each isotope. For different measurements of the same line, the yield seemed not to be too different, and after some yield calculations, I decided not to calculate yields from many measurements of the same line, but rather look at different lines from the measurement where the lines of the specified isotope are strongest. For the different lines (which are not blended by another line), there were some notable variations in the yield, and the variation was systematic: the yield seems to increase with the line energy. This issue can have different explanation: there could be a systematic error in the calculated efficiencies, or there could be a notable correction factor (COI) which increases with energy. The variations are though only within one order of magnitude.
A list of all located isotopes and some yields are included on a separate Excel file. At some of the yields which are different from other yields of the same isotopes, I have been able to figure out the reason, and I have wrote it down as note (look for a small red triangle in upper right corner of some of the cells).
Conclusion
I have been able to locate many isotopes produced from neutron radiation of
235 U, and I’ve got some yields for some of these isotopes, telling me how much of it that’s produced per proton per foil mass. I thereby hope the results of my analysis will be helpful for the scientists who will continue the feasibility study.