Chapter 3 HRTD - Nuclear Regulatory Commission

1 Chapter 3 radioactive decay Specific ActivityHRTDH uman ResourcesTraining &Development- Slide 1 -H-201 - Health Physics TechnologyRADIOACTIVE- Slide 2 -H-201 - Health Physics TechnologyDECAYO bjectives Define the terms activity, radioactive decay constant, half-life, and specify the correct units State the equation for radioactive decay and explain - Slide 3 -H-201 - Health Physics Technologyqypeach term Calculate activity (remaining or decayed away), decay constant, half-life, etc. given various terms in the radioactive decay equationA = N Activity, A, is the term used to measure the decay rate of a radionuclideActivity- Slide 4 -H-201 - Health Physics Technologyrate of a radionuclide. The activity of a sample is based on the total number of radioactive atoms, N, and the probability of each atom undergoing radioactive decay .

2 Activity has units of disintegrations per second or dpsDecay Constant, = - Slide 5 -H-201 - Health Physics Technology The decay constant, , represents the probability that a radioactive atom will decay and is dependent on the half-life of the nuclide. Units of are 1/time (1/sec, sec-1 or per second)Activity UnitsCurie (Ci) = x 1010dpsBecquerel (Bq) = 1 dps- Slide 6 -H-201 - Health Physics TechnologyBecquerel (Bq) = 1 dps1 Ci = x 1010 BqHalf-Life- Slide 7 -H-201 - Health Physics TechnologyT = Half-Life- Slide 8 -H-201 - Health Physics Technology Half-Life- Slide 9 -H-201 - Health Physics TechnologyActivity ProblemA criticality accident occurs in a Japanese uranium processing facility. 1019fissions of U-235 occur over a 17-hour period.

3 Given that the U-235 fission yield for I-131 is and the half-life of I-131 is 8 days, calculate the I131 activity at the end of the accident Neglect- Slide 10 -H-201 - Health Physics Technologythe I-131 activity at the end of the accident. Neglect I-131 decay during the = NCalculating NA fission yield of means that for every 100 fissions of U-235, three I-131 atoms are Slide 11 -H-201 - Health Physics TechnologyN = 1019x = 3 x 1017I-131 atomsSolutionActivity = N = ( days) x (1/86,400 sec/day) x (3 x 1017 atoms) = 3 x 1011atoms/sec I-131= 3 x 1011dps I-131- Slide 12 -H-201 - Health Physics TechnologyConverting to traditional units:3 x 1011/ ( x 1010dps/Ci) = Ci I-131= - NdNdtDecay Equation- Slide 13 -H-201 - Health Physics TechnologydtN(t) = N0e - tDecay Equation- Slide 14 -H-201 - Health Physics TechnologyRadioactive decay - Slide 15 -H-201 - Health Physics TechnologyMultiply both sides by ,Activity Equation- Slide 16 -H-201 - Health Physics Technology N(t) = N0 e- tRecall A = NActivity Equation- Slide 17 -H-201 - Health Physics TechnologyA(t) = A0 e- tThe fraction of activity A remaining after n half-lives is given by:A1 radioactive decay - Slide 18 -H-201 - Health Physics TechnologyAA012n=A = Aoe (- t)orA = Ao( ) nThese two equations are identical!

4 Here s how:- Slide 19 -H-201 - Health Physics TechnologyA= Aoe(- t)but = ln(2)/T1/2so thatA= Aoe{ -ln(2)/T1/2* t }but -ln(2) = ln(1/2) and t can be measured in the number Example- Slide 20 -H-201 - Health Physics Technologyn, of half-lives that have passed (t = nT1/2) Putting these values in our equation, we get:= Aoe{ ln(1/2)/T1/2* nT1/2}= Aoe { nln( )}= Aoe{ ln[( )n] } = Ao( )nSince the exponential of a logarithm eln(A)is just the value A The fraction of activity decayed away after n half-lives is given by: radioactive decay (/)- Slide 21 -H-201 - Health Physics Technology1 -(A/A0)ProblemSuppose you have 106atoms of F-18 that were created in a water target at a cyclotron facility. How many F-18 atoms remain after the target sits and decays for 220 minutes?

5 - Slide 22 -H-201 - Health Physics TechnologyRecall that A(t) = A0 e- tand in this case, the half-life of F-18 is ~ 110 minutes, so A(t) = A0 e- t= 106atoms * e min * 220 min= atomsSolutionSuppose you have 106atoms of F-18 that were created in a water target at a cyclotron facility. How many F-18 atoms remain after the target sits and decays for 220 minutes?- Slide 23 -H-201 - Health Physics TechnologyRecall that A(t) = A0 e- tand in this case, the half-life of F-18 is ~ 110 minutes, so A(t) = A0 e- t= 106atoms * e min * 220 min= atomsSolutionAnother way of solving this would be to use the relationship:A = A02n- Slide 24 -H-201 - Health Physics TechnologySince two half-lives have passed (220 min), n = 2 and:A = A0= 106atoms = 106atoms = atoms2n2242 END OF- Slide 25 -H-201 - Health Physics TechnologyRADIOACTIVEDECAYSPECIFIC- Slide 26 -H-201 - Health Physics TechnologyACTIVITYO bjectives Define the term specific activity Explain each term given the equation for specific activity- Slide 27 -H-201 - Health Physics Technologyactivity Calculate the specific activity of various radioisotopesSpecific ActivitySpecific Activity is the activity per unit mass- Slide 28 -H-201 - Health Physics Technologyper unit massTypical units.

6 Ci/kg or Bq/gThe number of atoms of a radionuclide in one gram is given x 1023atomslAtoms per Gram- Slide 29 -H-201 - Health Physics TechnologyThis gives us the number of atoms per gram of the radionuclideN = x 10moleAwgramsmoleGrams per MoleExamples of calculating number of grams in one mole of a radionuclide: In one mole of Co-60, there are 60 grams- Slide 30 -H-201 - Health Physics Technology In one mole of U-235, there are 235 grams In one mole of Na-24, there are 24 grams In one mole of P-32, there are 32 gramsThe activity in one gram is then given by: SA = N Specific Activity- Slide 31 -H-201 - Health Physics Technology= x x 1023/ Aw( )= Bqs/gram Specific Activity ( ) in curies/gram =Specific Activity= x x 1023/ Aw( )- Slide 32 -H-201 - Health Physics x 1010 atoms/sec (secs) x 1023 atomsmoleAw gramsmolecurieSpecific x 1010 atoms/sec (secs) x 1023 atomsmoleAw gramsmolecurie- Slide 33 -H-201 - Health Physics = ( x 1013) / AwT1/2 (curies/gram)

7 Where Aw= atomic weight in grams and T1/2= half-life in seconds**Recall that units on are 1/sProblemCalculate the specific activity of Pu-239, given that the half-life is 24,400 years- Slide 34 -H-201 - Health Physics TechnologyGiven that the specific activity of natural U is 7 x 10-7Ci per g, calculate the ratio of the specific activities of Pu-239 and natural Slide 35 -H-201 - Health Physics g1 g1,428,571 g60Co27226Ra88 NatUMass vs Activity- Slide 36 -H-201 - Health Physics TechnologyAmount in gramsof each isotope equaling one curieof activity2788 END OF- Slide 37 -H-201 - Health Physics TechnologySPECIFICACTIVITYEND OF- Slide 38 -H-201 - Health Physics TechnologyCHAPTER 3