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Thesis Presentation Design and Construction of a Whole Body Counter for Non-Human Biota
- 1. Jill Belcot Radiation Science and Health Physics Jarrett Johns Nuclear Engineering Ethan Werger Nuclear Engineering Thesis Advisor Dr. Anthony Waker Construction of a Whole Body Counter for Non- Human Biota
- 2. Introduction ο The construction of a whole body counter to measure radioactivity in non-human biota ο Two units built for the examination of beta radiation and gamma radiation ο Aluminum and lead used for shielding of the unit ο Internal and external dose rates were calculated
- 3. Objective ο Build two units to detect gamma radiation and beta radiation ο Determine appropriate shielding configurations ο Calculate internal and external gamma and beta dose rates to the frog from Cs-137
- 4. Evolution of Testing Equipment Gen 1 Gen 2
- 5. Evolution of Testing Equipment Gen 4 Gen 4
- 6. Evolution of Testing Equipment Gen 5
- 7. Evolution of Testing Equipment
- 8. Evolution of Testing Equipment
- 9. Evolution of Testing Equipment
- 10. Evolution of Testing Equipment
- 11. Equipment Creation Issues ο Securing over 400 lbs. of pre WW2 fallout lead ο Creation of mold ο Welding lead casting mold ο Heating mold up to 328Β°C ο Staying on track for both the schedule and budget
- 12. Results
- 13. Gamma Interactions 1 10 100 1000 10000 0 100 200 300 400 500 600 700 800 900 1000 Counts Channel Number Background With Sheilding Series1 π΅πππππππ’ππ = πππ‘ππ πππ’ππ‘π πππ‘ππ π‘πππ = 285647 174345 = 1.628 ( πππ’ππ‘π Sec )
- 14. Gamma Interactions 1 10 100 1000 10000 100000 0 200 400 600 800 1000 Counts Channel Number Cs Source With Sheilding Series1 πΆππ’ππ‘ π ππ‘π = πππ‘ππ πππ’ππ‘π πππ‘ππ π‘πππ = 1756554 300 = 5855.18 ( πππ’ππ‘π Sec )
- 15. NaI Detector π = ππ2 30 = 0.0231 π¦πβ1 π΄ = π΄0 πβ πβπ‘ π΄ = 1 β 10β6 β πβ(0.0231β8.4166) π΄ = 0.823ππΆπ β 3.7 β 1010 π΅π πΆπ = 30 461 π΅π πππππππππ¦ = (5855.18 πππ’ππ‘π sec β 1.628 πππ’ππ‘π sec ) 30 461 ( πππ πππ‘πππππ‘πππ π ππ ) = 19.21%
- 16. Gamma β No Shield 1 10 100 1000 10000 100000 1000000 0 200 400 600 800 1000 Counts Channel Number Background Without Sheilding Series1
- 17. Gamma With Lead Shield 1 10 100 1000 10000 0 200 400 600 800 1000 Counts Channel Number Background With Sheilding Series1
- 18. Gamma 1 10 100 1000 10000 100000 1000000 0 200 400 600 800 1000 Counts Channel Number Comparison of Shielding No shielding Shielded
- 19. Beta Interactions 1 10 100 1000 0 200 400 600 800 1000 Counts Channel Number Background Without Sheilding No shielding Shielding
- 20. Beta- Cap Efficiency 1 10 100 1000 10000 0 200 400 600 800 1000 Counts Channel Number Cs 137 With and Without Cap With cap Without Cap
- 21. Beta- Cap Efficiency πππ‘ππ πππ’ππ‘ πππ‘π π€ππ‘βππ’π‘ πππ = 591425 πππ‘ππ ππππ‘ πππ‘π π€ππ‘β πππ = 409830 πππππππππ¦ = π€ππ‘β πππ π€ππ‘βππ’π‘ πππ = 409830(πππ’ππ‘π ) 591425(πππ’ππ‘π ) = 0.6929
- 22. Effects of Radiation ο Assume the frog absorbs Cesium like a human, cesium is distributed everywhere in the body
- 23. Effects of Radiation β Internal ο π1 2 ππππππ‘ππ£π πΆπ β137 = 70 πππ¦π = 6048000 π ππππππ ο πΆπ’πππππ‘ππ£π π΄ππ‘ππ£ππ‘π¦ ( π΄) π΄ = 1.44π΄0 π1 2 ππππππ‘ππ£π π΄ = (1.44)(3.7 π₯ 10β4)(6048000π ππ) π΄ = 3 222.37 πππ πππ‘πππππ‘ππππ
- 24. Effects of Ξ² Radiation - Internal ο Intake of Beta in the frog is equal to one ο The frog will absorb all of the Betas ο For Cesium-137: πΈπ½ = 0.93 π₯ 0.514 + 0.07 π₯ 1.172 πππ 3 = 0.18669 MeV
- 25. Effects of πΎ Radiation - Internal ο 93% probability that gamma energy will be emitted for Cs- 137 ο Mean energy per emission is 0.662 MeV ο 7% of the energy emitted from the gamma particle will stay in the frog. πΈ πΎ = 0.622 πππ 0.93 = 0.57846 πππ πΈ πΎ π₯ π΄ππ πππππ πΉππππ‘πππ = 0.57846 πππ 0.069 = 0.0425 πππ
- 26. Effects of Total Internal Radiation β Ξ² and πΎ ο The committed absorbed dose can then be calculated: πΆπππππ‘π‘ππ π΄ππ πππππ π·ππ π = π΄ [ π½ + πΎ πππ 1.602π₯10β13 π½ 0.01654 ππ ] πΆπππππ‘π‘ππ π΄ππ πππππ π·ππ π = 7.25 π₯ 10β9 πΊπ¦
- 27. External Beta Skin Dose ο VARSKIN is a computer code for skin contamination dosimetry Layers of the Skin Thickness (Β΅m) Mucus 75 Epidermis 110 Dermis 300
- 28. External Beta Skin Dose β VARSKIN ο π΅ππ‘π π·ππ π π ππ‘π = 7.92 π₯ 10β2 πππ βπ = 7.92 π₯ 10β8 πΊπ¦ βπ ο π΅ππ‘π π·ππ π π ππ‘π = 7.81 π₯ 10β2 πππ βπ = 7.81 π₯ 10β8 πΊπ¦ βπ
- 29. External Beta Skin Dose Hand Calculation πΈ πππ πππππ = π‘ππ‘ππ π‘βππππππ π π π½ π₯ πΈ π½ = 0.599 πππ π·ππ π πππ π·ππ πππ‘πππππ‘πππ = 0.599 πππ π₯ 1.602 π₯ 10β13 π½ 0.498 π = 1.846 π₯ 10β13 πΊπ¦ π· = π΄ π₯ 1.846 π₯ 10β13 πΊπ¦ = 2.458 π₯ 10β13 πΊπ¦ βπ
- 30. Conclusion on Dose Rates ο The frog accumulates the Cesium 137, the frog receives a higher internal dose than external dose ο There is a lack of data on effects of radiation on many organisms, including amphibians. ο Many factors affect the amount of radiation that a frog can be exposed to
- 31. Conclusion ο Testing apparatus met our design requirements and provided accurate results ο Based on our detector response, it was determined we can detect a wide range of radiation from background to lethal ο Dose calculations proved the frog will be receiving both an internal and external dose from the 1Β΅Ci source.
- 32. Special Thanks ο Thesis Adviser: Dr. Waker ο Construction of apparatus: -Tuthill Metal Works -Shurer Custom Machining -Vic Powell Welding & Crane -Bill Battle -Bill Moyer -Rick Werger -Brent Thornhill -John Hart Sr. -Dave Green
Editor's Notes
- No radiation factor was found for a frog, so the committed equivalent dose was found. First the effective half life was found by using the physical and biological half life of cesium. The effective half life of cesium is 70 days, that was then converted to seconds. We used an activity of 1 microCurie or 3.7 x 10^-4 Bq The cumulative activity was the found, which is a convenient way of expressing the number of transitions or distingrations that occur. This equaled 3 222.37 distingrations. This acted as the starting point to calculate the internal dose that the frog would receive.
- The beta radiation will stay inside the frog because the cesium is distributed everywhere throughout the frog, making the yield for beta 100% and the absorbed fraction equal to one. Therefore, the beta energy needs to be determined. For cesium 137, about 7% of its transformations decays to stable Barium 137 by emission of a beta particle from a group whose maximum energy is 1.172 MeV and in 93% of its transformations decay by emitting a 0.515 MeV beta particle. These energies and yields are multiplied and added together and divided by 3 because for betas, the max energy is one third of β¦. Finally, it was found that 0.18669 MeV is the amount of energy that was deposited in the frog from beta.
- The effects of gamma radiation is a different story. The frog will absorbed 7% of the energy emitted. This was found by doing a simple extrapolation of mass vs. absorbed fractions from the textbook Introduction to Health Physics by Herman Cember. Our frog weights about 16.54 grams or 0.01654 kg. From here the energy for the gamma was calculated, this came out to be 0.57846 MeV. This energy was then multiplied by the absorbed fraction to give the amount of energy deposited into the frog.
- The energy deposited from both the gamma and beta can be added together to give the total energy, that then needs to be converted to joules and then divided by the mass of the frog. Then all of that needs to be multiplied by the cumulative activity. The square brackets gives us the dose per distingrations and the cumulative activity gives us the total number of distingrations. Therefore multiplying these together will give us the dose, which is 7.25 x 10^-9 Gy. That is the amount of dose the frog will absorb from 1micro Curie source.
- VARSKIN was used to help calculate skin dose rates, it is set up to be used for human skin models, but we decided to use it for a frog. We did this by manipulating varskin to fit with our frog model. The area of concern is the dermis layer because that is where the stem cells are located and where the radiation will do the most damage. Therefore, with VARSKIN, you can input your total skin thickness and then input a βprotective layer thicknessβ . We did this by treating the mucus layer as the protective layer to see the dose rate to the epidermis and dermis. Then we inputted the protective layer as the mucus and epidermis combined to look at the dose rate to the dermis layer of the skin.
- The first value if from inputting the mucus layer into VARSKIN with a thickness of 75 micrometers and a total thickness of the combined epidermis and dermis layer. The second value is when the mucus layer and the protective layer thicknessβs are combined and we are studying the dose to the dermis region. There isnβt a significant difference here in the dose rates, however when we compared this to the minimum detectable activity that we found, the dose rate is considerably lower.
- The range of the beta was first calculated, and it was less than 5mm so some of the beta did not penetrate through the skin. Also, in this case we looked at the total thickness of the skin, rather than individual layers, therefore the total thickness of the frogs skin is 485 micro meters. Then the thickness was divided by the range multiplied by the energy of the beta. This was done to give us the energy absorbed in the skin. We needed to find a volume, so our skin thickness was multiplied by the thickness of the frog, this gave us a volume and allowed us to find a weight by multiplying it by the density of the frogs skin (assumed to be 1). From here a dose per distingration was calculated and then multiplied by the activity to give a dose rate. The dose rate was considerably smaller than the VARSKIN calculations, this could be due to the fact that VARSKIN has pre-determined values incorporated into its system as well as we averaged our dose over the three layers rather than individually.