Calculating Intensity of Radiation with Various Half-value Layers :Calculating Intensity of Radiation with Various Half-value Layers Tad Pierson – NMT1002


Definition: Half-Value Layer (HVL) :Definition: Half-Value Layer (HVL) A half-value layer (HVL) is defined as “the thickness of absorber necessary to diminish the intensity of the radiation to half its initial strength” (Christian, 2004, p. 54)


Simple HVL Calculations :Simple HVL Calculations Radioactive source is producing 100 mR/hr HVL of lead is 10 mm To what will 30 mm of lead reduce the exposure rate? 30 mm/10 mm = 3 HVL 3 HVL = 12.5% 100 mR/hr * 0.125 = 12.5 mR/hr Radioactive source is producing 452 mR/hr HVL of lead is 6.2 mm To what will 24.8 mm of lead reduce the exposure rate? 24.8 mm/6.2 mm = 4 HVL 4 HVL = 6.25% 452 mR/hr 8 0.0625 = 28.25 mR/hr


Other ways to calculate intensity :Other ways to calculate intensity General Attenuation Equation I = Ioe-mx/HVL I = intensity after attenuation Ioe-mx/HVL = the initial intensity of the radiation to the power of the Linear Attenuation Coefficient m (value = 0.693) (e = 2.718) General Attenuation Equation - Short Form I = (Io) (0.5)N I = intensity after attenuation N = Thickness/HVL


The half-value layer of lead for Tc99m is 0.3 mm. A dose is producing 11 R/hr, what is the exposure rate if the dose is in a lead syringe shield with a thickness of 2.5 mm? Wells (1999) :The half-value layer of lead for Tc99m is 0.3 mm. A dose is producing 11 R/hr, what is the exposure rate if the dose is in a lead syringe shield with a thickness of 2.5 mm? Wells (1999) General Attenuation Equation I = Ioe-mx/HVL I = (11 R/hr)e-(0.693)(2.5 mm/0.3 mm) I = (11 R/hr)2.718-5.775 I = (11 R/hr)(0.0031) I = 0.034 R/hr Short Form I = (11 R/hr)(0.5)(2.5 mm/0.3 mm) I = (11 R/hr)(0.5)8.3 I = (11 R/hr)(0.0032) I = 0.035 R/hr


Exposure rate = 9340 mR/hr; HVL = 0.34 mm; Shield = 3.8 mm :Exposure rate = 9340 mR/hr; HVL = 0.34 mm; Shield = 3.8 mm General Attenuation Equation I = Ioe-mx/HVL I = (9340 R/hr)e-(0.693)(3.8 mm/0.34 mm) I = (9340 mR/hr)(2.718)-7.745 I = (9340 mR/hr)(0.00043) I = 4.02 mR/hr Short Form I = (9340 R/hr)(0.5)(3.8 mm/0.34 mm) I = (9340 mR/hr)(0.5)11.2 I = (9340 mR/hr)(0.00043) I = 4.02 R/hr


Exposure rate = 5241 mR/hr; HVL = 0.74 mm; Shield = 0.58 mm :Exposure rate = 5241 mR/hr; HVL = 0.74 mm; Shield = 0.58 mm General Attenuation Equation I = Ioe-mx/HVL I = (5241 mR/hr)e-(0.693)(0.58 mm/0.74mm) I = (5241 mR/hr)(2.718)-0.543 I = (5241 mR/hr)(0.581) I = 3045 mR/hr Short Form I = (5241 mR/hr)(0.5)(0.58 mm/0.74 mm) I = (5241 mR/hr)(0.5)0.784 I = (5241 mR/hr)(0.581) I = 3045 mR/hr


Works Cited :Works Cited Christian, P. E., Bernier, D. R., & Langan, J. K. (Eds.). (2004). Nuclear medicine and PET. St. Louis, MO: Mosby. Iowa State University, (2001). Calculating intensity with the inverse square law. Retrieved May 14, 2006, from Calculating intensity with the inverse square law Web site: http://www.ndt-ed.org/EducationResources/CommunityCollege/RadiationSafety/safe_use/distance.htm Iowa State University, (2001). Half-Value Layer (Shielding). Retrieved May 14, 2006, from Half-Value Layer (Shielding) Web site: http://www.ndt-ed.org/EducationResources/CommunityCollege/RadiationSafety/safe_use/shielding.htm Iowa State University, (2001). Interaction between penetrating radiation. Retrieved May 7, 2006, from Interaction Between Penetrating Radiation Web site: http://www.ndt-ed.org/EducationResources/CommunityCollege/Radiography/Physics/radmatinteraction.htm Wells, P. (1999). Practical mathematics in nuclear medicine technology. Reston, VA: Society of Nuclear Medicine.