chapter 3 hrtda = λn ¾activity, a, is the term used to measure the decay rate of a radionuclide...
TRANSCRIPT
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Chapter 3
Radioactive Decay
Specific Activity
HRTDHuman ResourcesTraining & Development
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RADIOACTIVE
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DECAY
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Objectives
Define the terms activity, radioactive decay constant, half-life, and specify the correct units
State the equation for radioactive decay and explain
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q y peach term
Calculate activity (remaining or decayed away), decay constant, half-life, etc. given various terms in the radioactive decay equation
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A = λN
Activity, A, is the term used to measure the decay rate of a radionuclide
Activity
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rate 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.
Activity has units of disintegrations per second or dps
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Decay Constant, λ
λ = 0.693T½
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The decay constant, λ, represents the probability that a radioactive atom will decay and is dependent on the half-life of the nuclide.
Units of λ are 1/time (1/sec, sec-1 or per second)
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Activity Units
Curie (Ci) = 3.7 x 1010 dps
Becquerel (Bq) = 1 dps
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Becquerel (Bq) = 1 dps
1 Ci = 3.7 x 1010 Bq
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Half-Life
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T½ = 0.693λ
Half-Life
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λ
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Half-Life
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Activity Problem
A criticality accident occurs in a Japanese uranium processing facility. 1019 fissions of U-235 occur over a 17-hour period. Given that the U-235 fission yield for I-131 is 0.03 and the half-life of I-131 is 8 days, calculate the I 131 activity at the end of the accident Neglect
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the I-131 activity at the end of the accident. Neglect I-131 decay during the accident.
A = λ N
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Calculating N
A fission yield of 0.03 means that for every 100 fissions of U-235, three I-131 atoms are created.
N = 1019 0 03
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N = 1019 x 0.03 = 3 x 1017 I-131 atoms
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Solution
Activity = λN
= (0.693/8 days) x (1/86,400 sec/day) x (3 x 1017 atoms)
= 3 x 1011 atoms/sec I-131
= 3 x 1011 dps I-131
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Converting to traditional units:3 x 1011 / (3.7 x 1010 dps/Ci) = 8.1 Ci I-131
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= -λNdNdt
Decay Equation
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dt
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N(t) = N0 e -λt
Decay Equation
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Radioactive Decay
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Multiply both sides by λ,
Activity Equation
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λ N(t) = λ N0 e-λt
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Recall A = λN
Activity Equation
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A(t) = A0 e-λt
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The fraction of activity A remaining after n half-lives is given by:
A 1
Radioactive Decay
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AA0
12n
=
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A = Ao e (-λt) or A = Ao (½) n
These two equations are identical! Here’s how:
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A = Ao e(-λt) but λ = ln(2)/T1/2 so that
A = Ao e{ -ln(2)/T1/2
* t }
but -ln(2) = ln(1/2) and t can be measured in the number
Example
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n, of half-lives that have passed (t = nT1/2 ) Putting these values in our equation, we get:
= Ao e{ ln(1/2)/T1/2
* nT1/2}
= Ao e { nln(½)} = Ao e{ ln[(½)n] } = Ao (½)n
Since the exponential of a logarithm eln(A) is just the value “A”
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The fraction of activity decayed away after n half-lives is given by:
Radioactive Decay
( / )
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1 - (A/A0)
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Problem
Suppose you have 106 atoms 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?
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Recall that A(t) = A0 e-λt and in this case, the half-life of F-18 is ~ 110 minutes, so
A(t) = A0 e-λt = 106 atoms * e -.693/110 min * 220 min = 2.5E5 atoms
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Solution
Suppose you have 106 atoms 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?
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Recall that A(t) = A0 e-λt and in this case, the half-life of F-18 is ~ 110 minutes, so
A(t) = A0 e-λt = 106 atoms * e -.693/110 min * 220 min = 2.5E5 atoms
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Solution
Another way of solving this would be to use the relationship:
A = A0
2n
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Since two half-lives have passed (220 min), n = 2 and:
A = A0 = 106 atoms = 106 atoms = 2.5E5 atoms2n 22 4
2
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END OF
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RADIOACTIVEDECAY
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SPECIFIC
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ACTIVITY
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Objectives
Define the term specific activity
Explain each term given the equation for specific activity
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activity
Calculate the specific activity of various radioisotopes
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Specific Activity
Specific Activity is the activity per unit mass
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per unit mass
Typical units: Ci/kg or Bq/g
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The number of atoms of a radionuclide in one gram is given by
6.02 x 1023 atomsl
Atoms per Gram
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This gives us the number of atoms per gram of the radionuclide
N = 6.02 x 10
mole
Awgramsmole
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Grams per Mole
Examples of calculating number of grams in one mole of a radionuclide:
In one mole of Co-60, there are 60 grams
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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 grams
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The activity in one gram is then given by:
SA = λN
Specific Activity
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= λ x 6.02 x 1023/ Aw (dps . gram-1)
= Bqs/gram
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Specific Activity (S.A.) in curies/gram =
Specific Activity
= λ x 6.02 x 1023/ Aw (dps . gram-1)
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3.7 x 1010 atoms/sec
0.693
T1/2
(secs)
6.02 x 1023 atoms
mole Aw grams
mole curie
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Specific Activity
3.7 x 1010 atoms/sec
0.693
T1/2
(secs)
6.02 x 1023 atoms
mole Aw grams
mole curie
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-S.A.
= (1.13 x 1013 ) / AwT1/2 (curies/gram)
where Aw = atomic weight in grams
and T1/2 = half-life in seconds*
*Recall that units on λ are 1/s
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Problem
Calculate the specific activity of Pu-239, given that the half-life is 24,400 years
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Given that the specific activity of natural U is 7 x 10-7 Ci per g, calculate the ratio of the specific activities of Pu-239 and natural U.
Problem
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0.001 g 1 g 1,428,571 g
60Co27
226Ra88
NatU
Mass vs Activity
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Amount in gramsof each isotope equaling one curieof activity
27 88
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END OF
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SPECIFICACTIVITY
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END OF
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CHAPTER 3