# 3/2003 Rev 1 I.2.7 – slide 1 of 35 Session I.2.7 Part I Review of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session.

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3/2003 Rev 1 I.2.7 – slide 1 of 35 Session I.2.7 Part I Review of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 7Radioactive Decay IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

3/2003 Rev 1 I.2.7 – slide 2 of 35 Radioactive decay is the process by which unstable atoms transform themselves into new chemical elements Radioactive decay is the process by which unstable atoms transform themselves into new chemical elements Students will learn about decay constants, activity, units, half-life, how to use the radioactive decay equation, and mean life Students will learn about decay constants, activity, units, half-life, how to use the radioactive decay equation, and mean life Introduction

3/2003 Rev 1 I.2.7 – slide 3 of 35 Content Activity Activity Units Units Decay Constant Decay Constant Half-Life Half-Life Law of Radioactive Decay Law of Radioactive Decay Mean Life Mean Life

3/2003 Rev 1 I.2.7 – slide 4 of 35 Overview Radioactive decay principles and pertinent terms will be discussed Radioactive decay principles and pertinent terms will be discussed Units to measure radioactive decay will be defined Units to measure radioactive decay will be defined

3/2003 Rev 1 I.2.7 – slide 5 of 35 1 Bq = 1 disintegration per second Activity The amount of a radionuclide present SI unit is the becquerel (Bq)

3/2003 Rev 1 I.2.7 – slide 6 of 35 Multiples & Prefixes (Activity) MultiplePrefixAbbreviation 1-------Bq 1,000,000Mega (M)MBq 1,000,000,000Giga (G)GBq 1,000,000,000,000Tera (T)TBq 1 x 10 15 Peta (P)PBq

3/2003 Rev 1 I.2.7 – slide 7 of 35 Units Curie (Ci) = 3.7 x 10 10 dps Becquerel (Bq) = 1 dps 1 Ci = 3.7 x 10 10 Bq 1 Ci = 3.7 x 10 10 Bq

3/2003 Rev 1 I.2.7 – slide 8 of 35 Non-SI Units QuantityOld UnitSI UnitConversion Activitycurie (Ci)becquerel (Bq)1 Ci=3.7 x 10 10 Bq Absorbed Doseradgray (Gy)1 rad = 0.01 Gy Equivalent Doseremsievert (Sv)1 rem = 0.01 Sv

3/2003 Rev 1 I.2.7 – slide 9 of 35 The Decay Constant is denoted by The Decay Constant is denoted by NOTE: has units of Typically or sec -1 or per second Decay Constant 1time 1sec

3/2003 Rev 1 I.2.7 – slide 10 of 35 A = N A = N Where N is number of atoms in a sample and A is the activity of the sample. A has units of disintegrations per second (dps or Bq). Activity Activity

3/2003 Rev 1 I.2.7 – slide 11 of 35 The relationship between half-life and decay constant is: Half-Life and Decay Constant T ½ = 0.693

3/2003 Rev 1 I.2.7 – slide 12 of 35 Half-Life

3/2003 Rev 1 I.2.7 – slide 13 of 35 Half-Life RadionuclideHalf-Life Phosphorus-3214.3 days Iridium-19274 days Cobalt-605.25 years Caesium-13730 years Carbon-145760 years Uranium-2384.5 x 10 9 years

3/2003 Rev 1 I.2.7 – slide 14 of 35 Sample Problem A criticality accident occurs in an Uranium processing facility. 10 19 fissions occur over a 17-hour period. Given that the fission yield for 131 I is 0.03 and its half-life is 8 days, calculate the 131 I activity at the end of the accident. Neglect 131 I decay during the accident.

3/2003 Rev 1 I.2.7 – slide 15 of 35 Solution to Sample Problem Activity = N = x x ( 10 19 x 0.03) = 3 x 10 11 Bq 131 I 0.693 8 days 1 86,400 sec day -1

3/2003 Rev 1 I.2.7 – slide 16 of 35 Differential Equation for Radioactive Decay = - N(t) dN dt

3/2003 Rev 1 I.2.7 – slide 17 of 35 Radioactive Decay Equation N(t) = N o e - t This equation is known as the law of radioactive decay

3/2003 Rev 1 I.2.7 – slide 18 of 35 Expressing the equation in terms of activity: Radioactive Decay Equation N(t) = N o e - t A(t) = A o e - t where A(t) = activity at any time t and A o = the initial activity at time t = 0 or

3/2003 Rev 1 I.2.7 – slide 19 of 35 Radioactive Decay The amount of activity decayed away after n half-lives is given by A(t) AoAoAoAo 1 -

3/2003 Rev 1 I.2.7 – slide 20 of 35 The amount of activity A(t) remaining after n half-lives is given by Radioactive Decay A(t) AoAo 1 2n2n =

3/2003 Rev 1 I.2.7 – slide 21 of 35 Mean Life T M = 1.44 T 1/2

3/2003 Rev 1 I.2.7 – slide 22 of 35 Radioactive Decay Activity (A) Bqor disintegrations time time (t)

3/2003 Rev 1 I.2.7 – slide 23 of 35 Example The area under the curve is speed x time or (50 km/hr) x 1 hr = 50 kilometers Speed (s) kphor kilometers hour time (hours) 1 50 A Vehicle Traveling at Constant Speed

3/2003 Rev 1 I.2.7 – slide 24 of 35 Example The area under the curve is (speed x time)/2 or (50 kph x 1 hr)/2 = 25 kilometers Speed (s) kphor kilometers hour time (hours) 1 50 A Decelerating Vehicle

3/2003 Rev 1 I.2.7 – slide 25 of 35 Area Under the Decay Curve A = A o e - t 0 A dt = A o e - t dt 0 = A o e - t dt = A o e - t dt 0 = A o = A o 0 e - t -

3/2003 Rev 1 I.2.7 – slide 26 of 35 Substituting and 0 for t = A o = A o e - ( ) - - e - (0) - = A o = A o - - 1 -0 =+ AoAoAoAo 01 Area Under the Decay Curve = AoAoAoAo

3/2003 Rev 1 I.2.7 – slide 27 of 35 Half-Life However, when t = T ½, the activity decreases to ½ of the original value However, when t = T ½, the activity decreases to ½ of the original value: A t = A o e - t or AtAtAtAt AoAoAoAo = e - t AtAtAtAt AoAoAoAo = ½A o AoAoAoAo = ½ = ½ ½ = e - T ½

3/2003 Rev 1 I.2.7 – slide 28 of 35 Take the natural logarithm of both sides ln (½) = - T ½ 1 = ln (½) -T ½ Regrouping terms yields But ln (½) = - ln (2) so: 1 = - ln (2) -T ½ ln (2) ln (2) T½T½T½T½= Half-Life & Decay Constant ln (½) = ln (e ) - T ½

3/2003 Rev 1 I.2.7 – slide 29 of 35 but ln(2) = 0.693 1 = ln (2) ln (2) T½T½T½T½ Mean Life & Decay Constant = 1.44 T ½ = T m 1 = 0.693 0.693 T½T½T½T½

3/2003 Rev 1 I.2.7 – slide 30 of 35 T m = 1.44 T ½ TmTmTmTm T½T½T½T½ Activity (A) Bqor disintegration time time (t) ½A o AoAoAoAo Mean Life

3/2003 Rev 1 I.2.7 – slide 31 of 35 Activity (A) Bqor disintegration time time (t) TmTmTmTm ½A o AoAoAoAo Remember the equation A = N the total # of atoms N = A o / = A o T m Mean Life

3/2003 Rev 1 I.2.7 – slide 32 of 35 A radionuclide has a half life of 10 days. What is the mean life? Sample Problem

3/2003 Rev 1 I.2.7 – slide 33 of 35 Solution to Sample Problem Mean Life = 1.44 T 1/2 = 1.44 x 10 days = 14.4 days

3/2003 Rev 1 I.2.7 – slide 34 of 35 Summary Activity defined and units discussed Activity defined and units discussed Decay constant defined Decay constant defined Half-life defined - relationship to decay constant Half-life defined - relationship to decay constant Radioactive decay equation derived Radioactive decay equation derived Mean life derived - relationship to half-life Mean life derived - relationship to half-life

3/2003 Rev 1 I.2.7 – slide 35 of 35 Where to Get More Information Cember, H., Johnson, T. E., Introduction to Health Physics, 4 th Edition, McGraw-Hill, New York (2008) Cember, H., Johnson, T. E., Introduction to Health Physics, 4 th Edition, McGraw-Hill, New York (2008) Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012) Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012) Jelley, N. A., Fundamentals of Nuclear Physics, Cambridge University Press, Cambridge (1990) Jelley, N. A., Fundamentals of Nuclear Physics, Cambridge University Press, Cambridge (1990) Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999) New York (1999)

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