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ACADs (08-006) Covered Keywords

Gamma Shielding Calculations. ACADs (08-006) Covered Keywords Gamma, shielding, attenuation, scatter, buildup, half value, tenth value. Description Supporting Material. Gamma Shielding Calculations. Radiation Protection III NUCP2331. SHIELDING. Gamma Radiation

chase-knox
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ACADs (08-006) Covered Keywords

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  1. Gamma Shielding Calculations ACADs (08-006) Covered Keywords Gamma, shielding, attenuation, scatter, buildup, half value, tenth value. Description Supporting Material

  2. Gamma Shielding Calculations Radiation Protection III NUCP2331

  3. SHIELDING • Gamma Radiation • Gamma rays do not give up energy continuously as charged particles do • Gamma ray intensity decreases due to absorption or scattering • Efficiency of the shield depends upon the probability of interaction and thickness of material • Attenuation-Process by which radiation is reduced in intensity when passing through some material

  4. Attenuation Coefficients • l (cm-1) Linear Absorption Coefficient is defined as the probability of interaction per unit path length of material • m (cm2/g) Mass Absorption Coefficient is defined as / where  is the linear absorption coefficient and  is density • μl = μm ρ • μm = mass attenuation coefficient • ρ = density of material in g/cc

  5. EM Formula • I = Io e (-μlx) • I = intensity with the shield, • Io = intensity without shield, • μl = linear attenuation coefficient, • x = thickness of shield

  6. Example • I0 = 1000 mR/hr • Your shield material is Lead • Shield thickness is 2 inches • Radionuclide is Cs-137 • What do you do first?

  7. Example • I0 = 1000 mR/hr • Shield is Lead • thickness 5 cm (2 in), μm = 0.115 c/g, ρ = 11.35g/cc • What is reading on other side of shield? • I = 1000 e -(0.115)(11.35)(5) • I = 1.47 mR/hr

  8. Example • I0 = 2000 mR/hr • Your shield material is Iron • Shield thickness is 4 inches • Radionuclide is Cs-137 • What do you do first?

  9. Example • I0 = 2000 mR/hr • Shield is Iron • thickness 10 cm (4 in), μm = 0.0745cc/g, ρ = 7.8 g/cc • What is reading on other side of shield? • I = 2000 e -(0.0745)(7.8)(10) • I = 5.988 mR/hr

  10. Scatter • Remember the Compton scatter • Some of the radiation interacting with the shield will create some new EM radiation, some of which may escape the shield • This will add exposure to the area on the backside of the shield , • So how does on take this into account?

  11. Build up • So if you have another parameter in your equation that changes your dose after your shield, your shield will need to be thicker to take into account eh greater dose • So that means you have 2 unknowns in your equations, not good • Need to estimate the shield thickness calculate the B (get close) and use that

  12. Build up Factors • I = BIo e (-μlx) • B= build up factor • based on μx • Dependant on energy and density of material • Caused by Compton scattering • How will your answers change?

  13. Example • I0 = 1000 mR/hr • Your shield material is Lead • Shield thickness is 2 inches • Radionuclide is Cs-137 • What do you do first? • Build up factor?

  14. Example • I0 = 1000 mR/hr • B =? μx= 6.5 energy is .662 MeV • Shield is Lead • thickness 5 cm (2 in), μm = 0.115 c/g, ρ = 11.35g/cc • What is reading on other side of shield? • I = 2.3 1000 e -(0.115)(11.35)(5) • I = 3.38 mR/hr

  15. Example • I0 = 2000 mR/hr • Your shield material is Iron • Shield thickness is 4 inches • Radionuclide is Cs-137 • What do you do first?

  16. Example • I0 = 1000 mR/hr • B μx= 6 energy is .662 MeV • Shield is Iron • thickness 10 cm (4 in), μm = 0.0745cc/g, ρ = 7.8 g/cc • What is reading on other side of shield? • I = 9.25 2000 e -(0.0745)(7.8)(10) • I = 55.3 mR/hr

  17. Half Value Layers • Half-Value Layer That thickness of material which reduces the radiation intensity to one-half of its initial value • Tenth-Value Layer That thickness of material which reduces the radiation intensity to one-tenth of its initial value Both take into account build up

  18. How many • How many half value layers would it take to reduce the radiation intensity from 500 mR/hr to less than 5 mR/hr?

  19. Total thickness • Number of HVL X the thickness of each HVL • What is total thickness of material need to reduce 500 mR/hr to less than 5 mR/hr? • Of • Iron • Concrete • Lead

  20. Problem • How many TVL does it take to decrease a reading of 350 mR/hr to under 5 mR/hr? • How many HVL are there in a TVT?

  21. Medical X ray • Several factors need to be taken into account when designing shielding for an X-ray machine • Not really for the people getting x rayed but for the people outside the X-ray room • The dose to those people are limited, as opposed to the dose to the patient • What energy the machined operates, how many patients will be x rayed, and where the X-ray room are all important to the design of the room

  22. Medical X-ray • 3 sources of radiation • Primary radiation (from the machine) • Full beam from the machine interacting with wall or floor • Secondary radiation • Scattered radiation ( from patient) • Leakage radiation ( from machine) • Dose rate at some distance (primary) • H (unshielded) =WUT/ d2

  23. Medical X-ray • Shielded/Unshielded(Dose/mA minute) = PD2/WUT • P- dose level in area needs to be r/wk • D-Distance from the machine (m) • W- workload (mA min/wk) • U-use factor how often the primary beam is faced on that barrier • T-occupancy percentage time area is in use

  24. Xray room design Secondary barrier D primary D secondary Primary barrier

  25. Questions?

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