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Transformation Kinetics

Transformation Kinetics. Half Life : is defined as time required for any given radionuclide to decrease to one half of its original quantity and characteristic of particular radionuclide. Example. P-32 T 1/2 = 14.3 days

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Transformation Kinetics

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  1. Transformation Kinetics Half Life: is defined as time required for any given radionuclide to decrease to one half of its original quantity and characteristic of particular radionuclide. Example. P-32 T1/2 = 14.3 days When the activity of P-32 is measured daily over a period of about 3 months and the percentage of the initial activity is plotted as a function of time. The curve shown in Figure 4-10 is obtained. . The data show that one half of P-32 is gone in 14.3 days.

  2. T1/2

  3. Half-life time Th: It is the time needed, on the average, for half of the atoms in a radioactive source to disintegrate. • In the next half-life time, one half of the remaining atoms will decay. Example:Consider a radioactive Gold 198Au with Th of 2.7 days. If one started with 100 x 106 atoms, after 2.7 days there will be 50 x 106 remaining, after 5.4 days 25 x 106, after 8.1 days 12.5 x 106. • The figure to the left shows the decay of 198 Au in the example. • This type of decay is called exponential decay. • The same data shows a straight line if plotted on a semi-log paper (where y-axis is a log scale), see the figure to the right.

  4. From the definition of the half life, it follows that the fraction of a radio nuuclide remaining after n half lives by the relationship. A/A0 = 1/2n Where A0 is the original quantity of activity and A is the activity left after n half-lives. .n: the number of half lives.

  5. Problem Cobalt-60, a gamma emitting isotope of cobalt whose half life is 5.3 years, is used as a radiation source for radiographing pipe welds, Because of the decrease in radioactivity with increasing time, the exposure time for a radiograph will be increased annually. Calculate the correction factor to be applied to the exposure time in order to account for the decrease in the strength of the sources. Solution Since A0/A= 2n By taking the logarithm of each side of the equation, we have Log A0/A = n Log 2 Where n, the number of Co-60 half-lives in 1 year , is 1/5.3 = 0.189 Log A0/A= 0.189 X 0.301 = 0.0569 A0/A = inverse log 0.0569 = 1.14 The ratio of the initial quantity of cobalt to the quantity remaining after 1 year is 1.14. the exposure time after 1 year, therefore, must be increased by 14%.

  6. Problem What percentage of a given amount of radium will decay during a period of 1000 years Solution The fraction remaining after 1000 years is given by A/A0= e –λt= e-4.02x 10-4 x 103 y .=0.67 = 0.67% The percentage that decayed during the 1000 year period , therefore , is 100-67= 33%

  7. The quantitative relationship between half-life T and decay rate constant λ may be found by setting A/A0 = ½ and solving for t. in this case, of course , the time is the half life T A/A0 = ½ = e-λt T= Ln 0.5/ λ Physical half life : T= 0.693/ λ decay

  8. Problem Given that the transformation rate constant for Ra-226 is 4.38 x 10-4 1/yr . Calculate the half life for radium. Solution T= 0.693/ λ = 0.693/ 4.38 x 10-4 1/yr = 1.6 X 103years

  9. The Curie: • It is the unit of activity named after Madame Curie. • It refers to the number of disintegration per second from 1 g of Radium which is equal to 3.7 x 1010 dis/s. Thus: • 1 Ci = 3.7 x 1010 dis/s • 1 mCi = 3.7 x 107 dis/s • 1 μCi = 3.7 x 104 dis/s The Becquerel: It is that quantity of radioactive material in which one atom is transformed per second. • 1 Bq = 1 tps (one transformation per second) or 1 dis/s • Thus, 1 Ci = 3.7 x 1010 Bq • The Becquerel is the Standard International (SI) unit for Activity. • It is a very small unit, thus we need to use the terms: • 1 KBq = 103 Bq, also, 1MBq = 106 Bq and 1GBq = 109 Bq

  10. Specific Activity: is defined as the number of becquerels (curies) per unit mass or volume SA = λN= λ X 6.02 X 1023 /A Bq/g 0.693 X 6.02 X 1023 _______________________ = A x T 4.18 X 1023 = ______________ A X T There are 3.7 x 1010 tps of Ra-226 The specific activity for Ra-226 is 3.7 x 1010 Bg/g The ratio of the specific Activity of any radionuclide SAi to that of Ra-226 is SAi/3.7 x 1010 Bq/g. = (4.18 x 1023 /Ai x Ti ) ____________________ (4.18 x 1023 /ARa x TRa) SAi = 3.7 X 1010X ARa x TRa ______________ Bq/g Ai x i

  11. Problem Calculate the specific activities of C-14 and S-35 given that their half lives are 5730 years and 87 days respectively. SA (C-14 ) = 3.7 X 1010 X 226 X 1600 ____________ Bq/g 14 x 5700 year = 1.7 x 1011 Bq/g (4.5 Ci/g) (S-32 ) = 3.7 X 1010 X 226 X 1600 year x 365 d/yr ____________ Bq/g 35 x 87 days = 1.6 x 1015 Bq/g (4.3 x 104 Ci/g)

  12. Homework 1; 1-Discuss the behaviors of the nucleons in the different radioactivity transformation mechanisms…? 2- How long will it take for each of the following radioisotopes to decrease to 0.0001% of its initial activity? (a) 99Mo (b) 99mTc (c) 131I (d) 125I (e) 60Co Half life =66 hr-Mo-99 Half life =6 hr -99mTc Half life =8 days –I31I Half life = 60 days –-I-125 Half life =5.3 years- Co-60

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