LB PAA Nickel Investigation

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Nickel Normalization

Talk about Nickel reaction, give example of Nickel rates for 511 and 1377 keV lines. Run condition for examples below.

Ni-08-22-13

PAA_8-22-13#runlist


Using a 45 MeV linac (Jack) you get 75 uCi of Nickel per (gram Kw hr)

If using a 25 meV machine at 1 kW power, how ling should I run to keep 1 g of natural Selium from having a Se-79 Activiy less than 16 pCi?


Here is the data for the cross section of Ni-58([math]\gamma[/math],n)Ni-57 data

http://www-nds.indcentre.org.in/exfor/servlet/X4sGetSubent?reqx=119235&subID=220597006&plus=1

When comparing this data to the Se-80([math]\gamma[/math],n)Se-79 reaction, we must be careful when handling the bin sizes. The data for the Se-79 reaction is binned by intervals of 0.2 MeV, while the Ni-57 data is binned by intervals of 0.1 MeV. To create coarser bins, I must take the 3 data points from the Nickel data and get an average value, then plot that value centered at the midpoint of these two values. For example, I would take 10.1,10.2, and 10.3 MeV and average their values and plot at 10.2 MeV. Below is the plot:

Gamma np XSect.png

Here is the data for the plot above. The Zn-68 -> Cu-67 data was found in the IAEA handbook on page 160. I used data thief to find the points that were plotted.


Energy (MeV) Zn-68 -> Cu-67 (mb) Ni-58 -> Ni-57 Se-80 -> Se-79
12.0 0.0103 ** 33.8 +/- 1.8
12.2 0.0117 0.56 +/- 0.2 35.6 +/- 1.7
12.4 0.0128 1.56 +/- 0.2 40.8 +/- 1.9
12.6 0.0624 2.83 +/- 0.2 42.3 +/- 1.5
12.8 0.1496 4.23 +/- 0.2 47.4 +/- 2.2
13.0 0.2113 5.13 +/- 0.2 51.6 +/- 2.8
13.2 0.3149 5.53 +/- 0.2 54.8 +/- 2.6
13.4 0.5009 5.93 +/- 0.2 61.9 +/- 2.8
13.6 0.642 6.6 +/- 0.2 69.5 +/- 2.4
13.8 0.7151 7.46 +/- 0.2 74.4 +/- 3.3
14.0 0.8874 8.79 +/- 0.2 82.8 +/- 3.8
14.2 0.9466 9.87 +/- 0.2 94.1 +/- 3.4
14.4 0.9884 10.41 +/- 0.2 94 +/- 3.6
14.6 1.113 10.47 +/- 0.2 102 +/- 3.4
14.8 1.337 10.92 +/- 0.2 107.3 +/- 4.3
15.0 1.411 11.71 +/- 0.2 111.3 +/- 4.7
15.2 1.4574 12.76 +/- 0.2 115.6 +/- 4.7
15.4 1.6059 14.09 +/- 0.2 119.9 +/- 5
15.6 1.641 15.76 +/- 0.2 120.1 +/- 5.3
15.8 1.7318 18.13 +/- 0.2 126 +/- 7.3
16.0 1.7506 21.22 +/- 0.2 122.9 +/- 7.3
16.2 2.0141 22.89 +/- 0.2 114.6 +/- 5
16.4 2.2435 23.13 +/- 0.2 122.4 +/- 6.1
16.6 2.3172 22.84 +/- 0.2 121.2 +/- 7.7
16.8 2.6092 22.61 +/- 0.2 118.2 +/- 9.1
17.0 2.6947 22.31 +/- 0.22 116.8 +/- 8.8
17.2 2.9378 22.86 +/- 0.39 116.4 +/- 5.9
17.4 3.1341 24.23 +/- 0.46 109.5 +/- 7.4
17.6 3.3798 25.8 +/- 0.54 109.6 +/- 8.3
17.8 3.2368 27.12 +/- 0.63 102.3 +/- 8.1
18.0 4.0164 27.38 +/- 0.63 107.2 +/- 8.3
18.2 4.4739 26.5 +/- 0.62 111.9 +/- 8
18.4 4.5713 25.18 +/- 0.64 101.3 +/- 8.3
18.6 4.6206 24.16 +/- 0.64 97 +/- 7.5
18.8 5.1468 23.31 +/- 0.71 93.4 +/- 7.9
19.0 5.0914 22.38 +/- 1.10 90.5 +/- 7.1
19.2 5.6712 20.84 +/- 1.71 87.1 +/- 7.4
19.4 6.525 19.18 +/- 1.51 89.7 +/- 7.8
19.6 6.8866 19.93 +/- 1.10 90.6 +/- 6.3
19.8 6.9607 24.06 +/- 1.3 90.3 +/- 6.3
20.0 7.6705 29.25 +/- 0.9 84.9 +/- 7
20.2 8.1832 31.17 +/- 0.69 79 +/- 6.2
20.4 8.4519 29.6 +/- 0.69 75.5 +/- 7.9
20.6 8.8241 27.06 +/- 0.87 74.1 +/- 6.5
20.8 9.5163 26.29 +/- 0.82 75.4 +/- 6.4
21.0 9.9354 25.4 +/- 0.72 72.9 +/- 7.4
21.2 10.0428 23.07 +/- 0.76 67.6 +/- 5.6
21.4 10.151 21.84 +/- 0.84 69.1 +/- 8.3
21.6 10.1502 23.90 +/- 0.79 57.2 +/- 7.7
21.8 10.149 25.98 +/- 0.88 61 +/- 7.6
22.0 9.9323 25.64 +/- 1.27 58 +/- 7.6
22.2 9.9316 21.39 +/- 2.20 57.8 +/- 6.4
22.4 9.7191 17.49 +/- 1.49 65.8 +/- 9.6
22.6 9.3077 15.82 +/- 0.82 59.9 +/- 6.9
22.8 8.9141 15.85 +/- 0.76 55.7 +/- 8.4
23.0 8.7229 16.79 +/- 1.10 53.4 +/- 7.3
23.2 8.6287 17.07 +/- 0.86 57.5 +/- 7.8
23.4 8.2638 15.93 +/- 0.93 32 +/- 7.8
23.6 8.0867 12.61 +/- 0.84 43.7 +/- 7.8
23.8 7.8288 8.76 +/- 0.87 36.5 +/- 9.8
24.0 7.5787 6.96 +/- 0.78 39.7 +/- 7.2
24.2 7.8 6.45 +/- 0.86 37.3 +/- 8.5

Using a left handed Riemann Sum on each set of cross section data (The Selenium 79 reaction and the Nickel 58 reaction), the relative activity between the nickel isotopes and the selenium isotopes can be found. For the Nickel reaction I used a rectangular width equal to the bin size of the measurements, which was 0.1 MeV. Similarly for the selenium I had to use a bin size of 0.2 MeV while rounding the values down to the nearest tenth, i.e. 10.02 becomes 10.0, 10.22 becomes 10.2 and so on.

Reaction Integrated Cross Section (mb)[12->24 MeV] Abundance
Se-80(gamma,n)Se-79 5051.1 49.8%
Ni-58(gamma,n)Ni-57 1074.99 68.1%
Zn-68(gamma,p)Cu-67 293.2027 18.45 %
Se-76(gamma,n)Se-75 4598.8 9.23
Se-82(gamma,n)Se-81 5219.89 8.2
Reaction Integrated Cross Section (mb)[12.2->18.2 MeV] Abundance
Se-80(gamma,n)Se-79 2892 49.8%
Ni-58(gamma,n)Ni-57 451.63 68.1%
Zn-68(gamma,p)Cu-67 46.0847 18.45 %
Se-76(gamma,n)Se-75 2644.7 9.23%
Se-82(gamma,n)Se-81 3271.3 8.82%

[math]\frac{{68 \atop 30\; }Zn (\gamma,p){67 \atop 29\;}Cu}{{82 \atop 34\; }Se (\gamma,n){81 \atop \;Se}} = \frac{Abundance \times \int_{12.0}^{24.2}\sigma dE}{Abundance \times \int_{12.0}^{24.2}\sigma dE} = \frac{0.1845 \times 293.2027}{0.082 \times 5429.19} = 0.12 [/math]


[math]\frac{{68 \atop 30\; }Zn (\gamma,p){67 \atop 29\;}Cu}{{76 \atop 34\; }Se (\gamma,n){75 \atop \;Se}} = \frac{Abundance \times \int_{12.0}^{24.2}\sigma dE}{Abundance \times \int_{12.0}^{24.2}\sigma dE} = \frac{0.1845 \times 293.2027}{0.0923 \times 4598.8} = 0.13 [/math]


[math]\frac{{68 \atop 30\; }Zn (\gamma,p){67 \atop 29\;}Cu}{{80 \atop 34\; }Se (\gamma,n){79 \atop \;Se}} = \frac{Abundance \times \int_{12.1}^{24.2}\sigma dE}{Abundance \times \int_{12}^{24.2}\sigma dE} = \frac{0.1845 \times 293.2027}{0.498 \times 5051.1} = 0.02 [/math]

[math]\frac{{58 \atop 28\; }Ni (\gamma,n){57 \atop \;Ni}}{{82 \atop 34\; }Se (\gamma,n){81 \atop \;Se}} = \frac{Abundance \times \int_{12.0}^{24.2}\sigma dE}{Abundance \times \int_{12.0}^{24.2}\sigma dE} = \frac{0.68077 \times 1074.99}{0.082 \times 5219.89} = 1.7 [/math]

Here Segebade has the inverse of this expression, so taking the inverse we have my 0.58 compared to Segebade's 0.54

So now using 0.46g of Selenium pellets. I will calculate the expected activity for Se-82


[math]\frac{N_{Se-81}}{N_{Ni-57}} = \left ( \frac{\sigma_{Se-81}}{\sigma_{Ni-57} }\right ) \left ( \frac{\lambda_{Ni-57}}{\lambda_{Se-81}}\right ) = \left ( \frac{0.0882*3271.3}{451.63*0.681} \right ) \left ( \frac{2136.36}{18.45} \right ) [/math] = 108.6

Now using 34.5 uCi/h (Assuming 1kW and 0.46g of Selenium) we get an expected activity of 108.6*34.5uCi/h = 3.7mCi/h of Se-81. So in 2 hours we will have 7.4mCi of Se-81


[math]\frac{{58 \atop 28\; }Ni (\gamma,n){57 \atop \;Ni}}{{76 \atop 34\; }Se (\gamma,n){75 \atop \;Se}} = \frac{Abundance \times \int_{12.0}^{24.2}\sigma dE}{Abundance \times \int_{12.0}^{24.2}\sigma dE} = \frac{0.68077 \times 1074.99}{0.0923 \times 4598.8} = 1.72 [/math]

Here Segebade has the inverse of this expression, so taking the inverse we have my 0.58 compared to Segebade's 0.013


So using 0.46g of selenium, I will calculate the expected activity for Se-75


[math]\frac{N_{Se-75}}{N_{Ni-57}} = \left ( \frac{\sigma_{Se-75}}{\sigma_{Ni-57} }\right ) \left ( \frac{\lambda_{Ni-57}}{\lambda_{Se-75}}\right ) = \left ( \frac{0.0923*2644.7}{451.63*0.681} \right ) \left ( \frac{1.4833}{120} \right ) [/math] = 0.01

Now using 34.5 uCi/h (Assuming 1kW and 0.46g of selenium) we get an expected activity of 0.01*(34.5uCi/h) = 0.345 uCi/h. So in 2 hours the expected activity will be 0.69uCi



[math]\frac{{58 \atop 28\; }Ni (\gamma,n){57 \atop \;Se}}{{80 \atop 34\; }Se (\gamma,n){79 \atop \;Se}} = \frac{Abundance \times \int_{12.1}^{24.2}\sigma dE}{Abundance \times \int_{12}^{24.2}\sigma dE} = \frac{0.68077 \times 1074.99}{0.498 \times 5051.1} = 0.29 = \frac{\sigma_{Ni-57}}{\sigma_{Se-79}}[/math]

[math]\frac{N_{Se-79}}{N_{Ni-57}} = \left ( \frac{\sigma_{Se-79}}{\sigma_{Ni-57} }\right ) \left ( \frac{\lambda_{Ni-57}}{\lambda_{Se-79}}\right ) = \left ( \frac{1}{0.29} \right ) \left ( \frac{0.0041}{327000} \right ) [/math] = 0.000000043

Now using the production rate ratio above from the nickel we can determine that the production rate of Se-79 is 3.243pCi/h, this would mean that we could irradiate for 4.9 hours before reaching the threshold of 16pCi.

Segebade has a table of ratios of specific activities of certain elements compared to Ni-57. The section starts on page 167, but the ratios with selenium are found on page 176

File:Ch5a PAA SWL 1988.pdf

Since the beam energy has been changed to 18 MeV, I will recalculate the expected yield with respect to nickel


[math]\frac{{58 \atop 28\; }Ni (\gamma,n){57 \atop \;Se}}{{80 \atop 34\; }Se (\gamma,n){79 \atop \;Se}} = \frac{Abundance \times \int_{12.2}^{18.2}\sigma dE}{Abundance \times \int_{12.2}^{18.2}\sigma dE} = \frac{0.68077 \times 451.63}{0.498 \times 3090.5} = 0.19 = \frac{\sigma_{Ni-57}}{\sigma_{Se-79}}[/math]

[math]\frac{N_{Se-79}}{N_{Ni-57}} = \left ( \frac{\sigma_{Se-79}}{\sigma_{Ni-57} }\right ) \left ( \frac{\lambda_{Ni-57}}{\lambda_{Se-79}}\right ) = \left ( \frac{1}{0.19} \right ) \left ( \frac{0.0041}{327000} \right ) [/math] = 0.000000066

Now 75 microCi/h (assuming 1 kW and 1 g of natural selenium) 75(microCi/h)*0.000000066 = 0.000004949 microCi/h = 4.9 pCi/h, so we can irradiate for 3.27 hours before we hit 16 pCi.