Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 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

2 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  Students will learn about decay constants, activity, units, half-life, how to use the radioactive decay equation, and mean life Introduction

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

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

5 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)

6 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

7 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

8 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

9 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

10 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

11 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

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

13 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

14 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.

15 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

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

17 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

18 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

19 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 -

20 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 =

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

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

23 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

24 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

25 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 -

26 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

27 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

28 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

29 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

30 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

31 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

32 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

33 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

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

35 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)  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)  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)


Download ppt "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."

Similar presentations


Ads by Google