LB PAA Nickel Investigation

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:

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.