Radioactive Decay
BSEN-625ADVANCES IN FOOD ENGINEERING
Activity
The rate of decay of a radionuclideIt is the number of atoms that decay per unit timeUnits – Bacquarel (Bq): one desintegation/second1 Bq = 1 s-1
Curie-(Ci): activity of 1 g of Ra226
Ci: 1 Ci = 3.7x1010 Bq
Exponential decay
A/Ao
The activity of a pure radionuclide decreases exponentially with time
t
Exponential radioactive decay law
If N = # of atoms of a radionuclide in a sample @ a given time:
teNoNor
tNoN
NotNcNotNoNCI
λ
λ
λ
−=
−=
+−=+−=
==
ln
lnln0ln
;0,:..
ctN
dtNdN
NdtdNA
NdtdN
+−=
−
=−=
−=
λ
λ
λ
λ
ln
Half-life, T
λλ
λ
λ
λ
693.02ln
2ln21ln
21
==
−=
=−
=
=
−
−
T
T
e
eAoA
T
t
Decay constant
Time required for the activity of a radionuclide drop by a factor of one-half
Exponential decay in term of T
Tt
Tt
AoA
AoA
eAoA
NoN
Tt
Tt
693.02lnln
21 /
/63.0
−=−=
=
== −te
AoA λ63.0−=
2T
A/Ao1
T
0.125
0.5
0.25
3T
T693.0
=λ
t
Example
Calculate the activity of a 30-MBq source of Na-24 after 2.5 d. What is its decay constantSolution
T = half-life =15 h (appendix D)
MBqeAhdhdtMBqAo
hT
88.13060/245.2,30
0462.015693.0693.0
)600462.0(
1
==
=×==
===
×−
−λ
Mean life, τ
The average of all the individual lifetimes that atoms in a sample of the radionuclide experienceThe mean value of t under the exponential curve
te λ−1
A/Ao
τ t
Mean life, τ
It defines a rectangle with area equal to:
te λ−1
T
T
edte tt
>
==
=−==× ∞−∞ −∫
τλ
τ
λλτ λλ
693.01
1|11 00A/Ao
τ t
Specific Activity, SA
Activity per unit massBq/gFor a pure radionuclide the SA is determined by its decay constant, λ, or half-life T, and by its atomic weight M:
]/[1017.41002.6 2323
gBqMTM
SA ×=
×=
λ
In [s]# of atoms per gram of nuclide
Example
What is the SA of Ra226 in Bq/g
gBqgsSAMT
SA
AMappendixyT
/107.31066.3)3600243651600(226
1017.41017.4226(1600
101110
2323
×=×=
××××
=×
=
===
−−
SA (T,A)
BqCi
gCiAT
SA
10107.31
]/[2261600
×=
×=
T is expressed in years
Serial radioactivity decay
A sample in which one radionuclide produces one or more radioactive offspring in a chain
Secular equilibriumTransient equilibriumNo equilibrium
Secular equilibrium (T1>>T2)
At any time a Long-Lived parent (1) decays into a Short-Lived daughter (2), which decays to a stable nuclideT1>>T2
A1of the parent is constant (assuming short intervals of time compared to T1)At any time AT = A1 + A2
Secular equilibrium (T1>>T2)
dtudu
dNduNAuA
dtNA
dN
NAdtdN
2
22221
1
221
2
2212
;constant
λ
λλ
λ
λ
−=
−=−==
=−
−=
tt eAeAA
tNANANAc
tNNCIctNA
222012
22021
221
2021
202
2221
)1(
ln
)ln(0@..
)ln(
λλ
λλλ
λ
λλ
−− +−=
−=−−
−====+−=−
Secular equilibrium
Activity A2 relatively short-lived radionuclide as function of timeI.C: A20 =0Activity of daughter builds up to that of parent in about 7 half-livesDaughter decays at the same rate it is produced (A2=A1)Secular equilibrium is said to existTotal activity is 2(A1)
Act
iviti
es
~7T2
A1A2=A1
A2 secularequilibrium
T1>>T2
0 t
Secular equilibrium
2211 NN λλ =In terms of numbers of atoms
A chain of n short-lived radionuclides can all be in secular equilibrium with a long-lived parentThe activity of each member of the chain = activity of parentTotal activity = (n+1)(A of original parent)
General Case
If there is no restriction on the relative magnitudes of T1 and T2:
m)equilibriusecular a describes (also !!!0
)(
0..
2211
2012
12
1012
20
22212
21
NNAand
eeNN
NCI
NNdtdN
tt
λλλλ
λλλ
λλ
λλ
==>>
−−
=
==
−=
−−
Transient equilibrium (T1>T2)
N20 = 0T1>T2
A2 of the daughter initially build-up steadilyWith time, e-λ2t becomes negligible, since λ2>λ1
Transient equilibrium (T1>T2)
12
122
12
101222
12
1012
12
1012
)(
)(
0t
)(
1
1
21
λλλ
λλλλλ
λλλ
λλλ
λ
λ
λλ
−=
−=
−=
>>
−−
=
−
−
−−
AA
eNN
eNN
eeNN
t
t
tt
Activities as function of time
After initially increasing, the daughter activity A2 goes thru a maximum and decreases at the same rate as the parent activityThus, transient equilibrium existThe total activity also reaches a maximum, early than the daughterThe time transient equilibrium is reached depends on T1 & T2
activ
ities
Transientequilibrium
A1 + A2
A1
A2
A10
t
T1 > T2
0
No Equilibrium (T1 < T2)
When a daughter (N20 = 0) has a longer T2 than the parent T1 its activity build ups a maximum and then declinesThe parent eventually decays away (T1is shorter)Thus, only the daughter is leftNo equilibrium occurs
No Equilibrium (T1 < T2)
Activities as function of time when T2 > T1 and N20 = 0Non equilibrium occursOnly the daughter activity remains
A1 + A2A1
A2
A10
t
T2 > T1
activ
ities
0
Example
Starting with a 10 GBq (1010 Bq) sample of pure Sr90 at time t = 0, how long will it take for the total activity (Sr90 + Y90) to build up 17.5 GBq?
Solution
Appendix D38Sr90 β- decays with a T = 29.12 y into 39Y90, which β- decays into stable 40Zr90 with T = 64 hT1 >> T2Secular equilibrium is reached in about 7T2 = 7x64= 448hAt the end of this time, the Sr90 activity A1 has not diminished appreciablyThe Y90 activity A2 has increased to the level A2=A1=10 GBqTotal activity AT = 20 GBq
Solution
Time at which Y90 reaches 7.5 GBqThe answer will be less than 448 h
hte
GBqAGBqAhT
AeAeAA
t
tt
128)1(105.7
5.7,10;0108.0/693.0
0)1(
0108.021
122
20
201222
=−=
====
=+−=
−
−
−−
λ
λλ
Example
How many gram of Y90 are in secular equilibrium with 1 mg of Sr90?
Solution
The amount of Y90 will be that having the same activity as 1mg of Sr90
The SA of Sr90 of (T1 = 29.12y) is:
ggCi
Cim
gCi
dy
hdh
yAA
CigCigA
gCiSA
µ251.0/105.5
138.0
/105.590226
3651
24164
1600SA
m)equilibriusecular (138.0/13810
/13890226
12.291600
5
52
21
31
=×
=
×=×××
=
==×=
=×=
−
Example
A sample contains 1 mCi of Os191 at time t = 0. The isotope decays by β- emission into metastable Ir191m which then decay by γ emission into Ir191 .
IrIrOs m 19177
19177
19176 →→
β−
15.4d 4.94s
γ
Example
(a) how many grams of Os191 are present at t = 0?(b) how many mCi of Ir191m are present at t = 25 d?(c) how many atoms of Ir191m decay between t = 100s and t = 102s?(c) how many atoms of Ir191m decay between t = 30d and t = 40d?
Solution
Secular equilibrium is reached at 7X4.9 = 34 sThus, A1 = A2 at the equilibriumHowever, during the time considered at (b) and (d) A2 will have decayed appreciably (transient equilibrium)
Solution
(a) Grams of Os191
(b) At t = 25d
ggCi
Cim
gCiSA
84
3
41
1023.2/1049.4
10
/1049.4191226
4.153651600
−−
×=×
=
×=××
=
mCieAA 325.01 4.15/25696.021 =×== ×−
Solution
(c) Between 100s and 102 s secular equilibrium exists with the osmium source essentially still at its original activity:
717
172
104.7107.32satoms#s 2next theuring107.31
100@
×=××=
×==
=
−
−
sd
smCiAst
Solution
(d) Between 30 and 40s A1 and A2 do not stay constantTransient equilibrium exists, so the # of atoms of Parent and Daughter that decay are equal
7
8
40
30
4030
0450.07
4.15/693.07
1073.7)259.0165.0(1022.8
|0450.0
107.3107.3
×=
−×−=
−×
=× ∫ −− tt ee
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