Rad Data Handbook 20.. - Voss Associates

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Neutron Half-Value Layers in centimeters Energy in MeV 1 5 10 15 Polyethylene 3.7 6.1 7.7 8.8 W ater 4.3 6.9 8.8 10.1 Concrete 6.8 11 14 16 Damp soil 8.8 14.3 18.2 20.8 example: How many half-value layers of polyethylene are needed to attenuate a 100 mRem/hr 5 MeV neutron source to 5 mRem/hr? How thick does the polyehylene need to be? I = I 0 x 0.5 n I = 5 mRem/hr I 0 = 100 mRem/hr n = the number of half-value layers I/I 0 = 0.5 n 5/100 = 0.05 = 0.5 n ln 0.05 = n x ln 0.5 ln 0.05/ln 0.5 = n -2.996/-0.693 = n 4.32 = n It will take 4.32 half-value layers of polyethylene to reduce attenuate the neutron source. 4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm Neutron Half-Value Layers in centimeters Energy in MeV 1 5 10 15 Polyethylene 3.7 6.1 7.7 8.8 W ater 4.3 6.9 8.8 10.1 Concrete 6.8 11 14 16 Damp soil 8.8 14.3 18.2 20.8 example: How many half-value layers of polyethylene are needed to attenuate a 100 mRem/hr 5 MeV neutron source to 5 mRem/hr? How thick does the polyehylene need to be? I = I 0 x 0.5 n I = 5 mRem/hr I 0 = 100 mRem/hr n = the number of half-value layers I/I 0 = 0.5 n 5/100 = 0.05 = 0.5 n ln 0.05 = n x ln 0.5 ln 0.05/ln 0.5 = n -2.996/-0.693 = n 4.32 = n It will take 4.32 half-value layers of polyethylene to reduce attenuate the neutron source. 4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm 60 60 Neutron Half-Value Layers in centimeters Energy in MeV 1 5 10 15 Polyethylene 3.7 6.1 7.7 8.8 W ater 4.3 6.9 8.8 10.1 Concrete 6.8 11 14 16 Damp soil 8.8 14.3 18.2 20.8 example: How many half-value layers of polyethylene are needed to attenuate a 100 mRem/hr 5 MeV neutron source to 5 mRem/hr? How thick does the polyehylene need to be? I = I 0 x 0.5 n I = 5 mRem/hr I 0 = 100 mRem/hr n = the number of half-value layers I/I 0 = 0.5 n 5/100 = 0.05 = 0.5 n ln 0.05 = n x ln 0.5 ln 0.05/ln 0.5 = n -2.996/-0.693 = n 4.32 = n It will take 4.32 half-value layers of polyethylene to reduce attenuate the neutron source. 4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm Neutron Half-Value Layers in centimeters Energy in MeV 1 5 10 15 Polyethylene 3.7 6.1 7.7 8.8 W ater 4.3 6.9 8.8 10.1 Concrete 6.8 11 14 16 Damp soil 8.8 14.3 18.2 20.8 example: How many half-value layers of polyethylene are needed to attenuate a 100 mRem/hr 5 MeV neutron source to 5 mRem/hr? How thick does the polyehylene need to be? I = I 0 x 0.5 n I = 5 mRem/hr I 0 = 100 mRem/hr n = the number of half-value layers I/I 0 = 0.5 n 5/100 = 0.05 = 0.5 n ln 0.05 = n x ln 0.5 ln 0.05/ln 0.5 = n -2.996/-0.693 = n 4.32 = n It will take 4.32 half-value layers of polyethylene to reduce attenuate the neutron source. 4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm 60 60
Exposure Rate in an Air-Filled Ion Chamber X = I / m[1 / (2.58E-4 C / kg)-R] X = exposure rate (R / sec) I = current (amperes) m = mass of air in chamber (kg) % Resolution of a Gamma Spec System % R = FWHM / peak energy x 100 = % resolution FWHM = peak energy width at full width half-max height peak energy = photopeak energy of interest True Count Rate Based on the Resolving Time of a Gas-Filled Detector R C = R 0 / (1 - R0Y) = true count rate R 0 = observed count rate Y = resolving time Specific Gamma-Ray Constant (Ã) for Source Activity (A) Ã = öE ã(ì en / ñ) aire / W 2 Ã = specific gamma constant (R-cm / hr-A) ö = 2 photon fluence rate (ã / cm -hr) E ã = gamma photon energy (MeV) (ì en / ñ) = density thickness of air (g / cm ) e = electron charge (Coulombs) W = average amount of energy to produce an ion pair in air (eV) Dose Rate (D) to Air from a Point Beta Source D = 2 300 A / d = rad /hr A = source activity in curies d = distance from source in feet Exposure Rate in an Air-Filled Ion Chamber X = I / m[1 / (2.58E-4 C / kg)-R] X = exposure rate (R / sec) I = current (amperes) m = mass of air in chamber (kg) % Resolution of a Gamma Spec System % R = FWHM / peak energy x 100 = % resolution FWHM = peak energy width at full width half-max height peak energy = photopeak energy of interest True Count Rate Based on the Resolving Time of a Gas-Filled Detector R C = R 0 / (1 - R0Y) = true count rate R 0 = observed count rate Y = resolving time Specific Gamma-Ray Constant (Ã) for Source Activity (A) Ã = öE ã(ì en / ñ) aire / W 2 Ã = specific gamma constant (R-cm / hr-A) ö = 2 photon fluence rate (ã / cm -hr) E ã = gamma photon energy (MeV) 2 (ì en / ñ) = density thickness of air (g / cm ) e = electron charge (Coulombs) W = average amount of energy to produce an ion pair in air (eV) Dose Rate (D) to Air from a Point Beta Source D = 2 300 A / d = rad /hr A = source activity in curies d = distance from source in feet 61 61 Exposure Rate in an Air-Filled Ion Chamber X = I / m[1 / (2.58E-4 C / kg)-R] X = exposure rate (R / sec) I = current (amperes) m = mass of air in chamber (kg) % Resolution of a Gamma Spec System % R = FWHM / peak energy x 100 = % resolution FWHM = peak energy width at full width half-max height peak energy = photopeak energy of interest True Count Rate Based on the Resolving Time of a Gas-Filled Detector R C = R 0 / (1 - R0Y) = true count rate R 0 = observed count rate Y = resolving time Specific Gamma-Ray Constant (Ã) for Source Activity (A) Ã = öE ã(ì en / ñ) aire / W 2 Ã = specific gamma constant (R-cm / hr-A) ö = 2 photon fluence rate (ã / cm -hr) E ã = gamma photon energy (MeV) (ì en / ñ) = density thickness of air (g / cm ) e = electron charge (Coulombs) W = average amount of energy to produce an ion pair in air (eV) Dose Rate (D) to Air from a Point Beta Source D = 2 300 A / d = rad /hr A = source activity in curies d = distance from source in feet Exposure Rate in an Air-Filled Ion Chamber X = I / m[1 / (2.58E-4 C / kg)-R] X = exposure rate (R / sec) I = current (amperes) m = mass of air in chamber (kg) % Resolution of a Gamma Spec System % R = FWHM / peak energy x 100 = % resolution FWHM = peak energy width at full width half-max height peak energy = photopeak energy of interest True Count Rate Based on the Resolving Time of a Gas-Filled Detector R C = R 0 / (1 - R0Y) = true count rate R 0 = observed count rate Y = resolving time Specific Gamma-Ray Constant (Ã) for Source Activity (A) Ã = öE ã(ì en / ñ) aire / W 2 Ã = specific gamma constant (R-cm / hr-A) ö = 2 photon fluence rate (ã / cm -hr) E ã = gamma photon energy (MeV) 2 (ì en / ñ) = density thickness of air (g / cm ) e = electron charge (Coulombs) W = average amount of energy to produce an ion pair in air (eV) Dose Rate (D) to Air from a Point Beta Source D = 2 300 A / d = rad /hr A = source activity in curies d = distance from source in feet 61 61
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Abbreviations 115 Activity vs Expos
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RADIOLOGICAL EMERGENCY RESPONSE Wri
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HAZARD CONTROL PRIORITIES DURING ME
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ACUTE RADIATION EFFECTS 0 - 25 REM
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TABLE OF THE ELEMENTS Z Density Z D
Page 11 and 12: Relative Locations of Products of N
Page 13 and 14: RADIOACTIVE DECAY MODES OF COMMONLY
Page 15 and 16: Progeny kev and % abundance 55 Fe 5
Page 17 and 18: Progeny kev and % abundance 99 Mo 9
Page 19 and 20: Progeny kev and % abundance 208 Tl
Page 21 and 22: Progeny kev and % abundance 229 Th
Page 23 and 24: Progeny kev and % abundance 241 Am
Page 25 and 26: Progeny kev and % abundance Progeny
Page 27 and 28: Progeny kev and % abundance 222 Rn
Page 29 and 30: Neptunium Decay Chain (4n + 1) st 1
Page 31 and 32: Actinium Decay Chain (4n + 3) st 1
Page 33 and 34: Ci / g = 3.578E5 / (T 1/2 in years
Page 35 and 36: Rem/hr / Ci Sv/hr / GBq Half-Life C
Page 45 and 46: mRem/hr mSv/hr 1ì 10ì 100ì 1ì 1
Page 47 and 48: RADIATION BIOLOGY Maximum survivabl
Page 49 and 50: INTERNAL DOSIMETRY Calculating CDE
Page 51 and 52: EQUIVALENT DOSE, EFFECTIVE DOSE, an
Page 53 and 54: CALCULATING TODE AND TEDE TEDE = DD
Page 55 and 56: SHIELDING MATERIALS α a single she
Page 57 and 58: Beta Dose Rates in rad/hr per mCi M
Page 59 and 60: Combining Radiation Types to Determ
Page 61: NEUTRON SHIELD THICKNESS I = I0e -N
Page 65 and 66: Inverse Square Law 2 2 X 1 (D) 1 =
Page 67 and 68: Table 1 of DOE 5400.5 Surface Activ
Page 69 and 70: 6CEN The 6CEN equation can be used
Page 71 and 72: Ventilation Rates Ventilation rates
Page 73 and 74: AIR FLOW METER CORRECTIONS Mass Flo
Page 75 and 76: 10CFR20 DAC 10CFR835 DAC 10CFR20 AL
Page 103 and 104: External Exposure in a Cloud of Air
Page 105 and 106: Characteristic X-Rays (KeV) of the
Page 107 and 108: Z # K K L L 103 Lr 71 Lu 54.06 61.2
Page 109 and 110: COUNTING STATISTICS COUNTING STATIS
Page 111 and 112: ELEVATION VS AIR PRESSURE Barometri
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NE Omaha 983 UT Cedar City 5,623 NH
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COMPOSITION OF AIR Density of Symbo
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ABBREVIATIONS ampere A, or amp angs
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Mass 1 gram = 0.03527 ounces 1 ounc
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Radiological 1 BTU = 235 1.28E-8 g
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CONSTANTS Avogadro's number (N 0) 6
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ELECTROMAGNETIC SPECTRUM Wavelength
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5. Air-proportional alpha detectors
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32 7. The bremsstrahlung from a 1 C
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RULES OF THUMB FOR NEUTRONS 1. The
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Spontaneous Fission Neutron and Gam
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Energy & Yield of neutrons from the
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WG Pu 15 years after fabrication 23
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Isotopic Mix of Natural U 238 U 234
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MISCELLANEOUS RULES OF THUMB 1. One
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PUBLIC RADIATION DOSES Average per
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COMPARATIVE RISKS OF RADIATION EXPO
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Voss Associates provides valuable t
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Publications Radiation Data Radiati
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Rad Data Handbook 20.. - Voss Associates

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