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