decay kinetics

8/8/2019 Decay Kinetics

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3-1

Radioactive Decay Kinetics

Outline

Radioactive decaykinetics

Basic decay

equations Utilization of

equations

Mixtures

Equilibrium

Branching

Natural radiation

Dating



3-2

Introduction

Number of radioactive nuclei that decay in a sampledecreases with time

Exponential decrease

Independent of P, T, mass action

Conditions associated with chemical kinetics* Electron capture and internal conversion can be

affected by conditions

Specific for isotope

Irreversible

Decay of given radionuclide is random

Statistical

Evaluate behavior of group



3-3

Basic Decay Equations

Decay is 1st order

Rate proportional to amount of parentisotope

Equal to the rate of isotopedisintegration

Proportional to number of radioactivenuclei

* rate of decay=decay constant*#radioactive nuclei

Decay constant is average decayprobability per nucleus for a unit time

Represented by

P



3-4

Basic decay equations

The radioactive process is a subatomic change within

the atom

The probability of disintegration of a particular atom

of a radioactive element in a specific time interval is

independent of its past history and presentcircumstances

The probability of disintegration depends only on the

length of the time interval.

Probability of decay: p=P(t

Probability of not decaying: 1-p=1- P(t



3-5

1-p=1-P(t=probability that atom will survive (t

(1-P (t)n=probability that atom will survive n intervals of t

n(t=t, therefore (1- P (t)n =(1- P t/n)n

Since limnp(1+x/n)n=ex, (1- Pt/n)n=e-Pt, the limiting value.

Considering No atoms, the fraction remaining unchanged

after time t is N/No= e-Pt

Statistics of Radioactive Decay

N=Noe-Pt

where P is the decay constant

Statistics of Radioactive Decay



3-6

Radioactivity as Statistical

Phenomenon

Binomial Distribution for Radioactive Disintegrations

probability W(m) of obtaining m disintegrations in

time t from No original radioactive atoms

probability of atom not decaying in time t, 1-p, is

(N/No)=e-Pt, where N is number of atoms that survive

in time interval t and No is initial number of atoms

Time Intervals between Disintegrations

probability of time interval having value between t

and t+d:

m N m

o

o o p pmm N

N mW

! )(!)!(

!)( 1

dt e N

dt t P t N

o

oP

P

!)(



3-7

Average Disintegration Rate

for radioactive disintegration--if n=No and p=1-

e-Pt--average number M of atoms disintegrating

in time t is M=No(1-e

-Pt

); for small Pt, M=NoPtand disintegration R =M/t=No P, which

corresponds to -dN/dt=PN

Expected Standard Deviation

M is number of counts

Relative error = W-1

§

!

!

!!

!

nr

r

r nr

r r r W n p

q pr r n

n

r W

0 )(

!)!(

!)(

where 1- p=q

M small g ener allyi st pr act icecount i n g i n S i nce

M eee N t t t

o

!

!!

W P

W PPP

,

)(1



3-8

MeasuredActivity

In practicality, activity (A)is used instead of thenumber of atoms (N).

A= cPt, m

where c is the detection

coefficient A=AOe-Pt

Units

Curie

3.7E10 decay/s* 1 g Ra

Becquerel

1 decay/s



3-9

Half Life and decay constant

Half-life is time needed to decreasenuclides by 50%

Relationship between t1/2 and P

N/No=1/2=e-P

t

ln(1/2)=-Pt1/2

ln 2= Pt1/2

t1/2=(ln 2)/P



3-11

Exponential Decay

Average Life (X) for a radionuclide found from sum of times of existence of all

atoms divided by initial number of nuclei

1/P=1/(ln2/t1/2)=1.443t1/2=X

Average life greater than half life by factor

of 1/0.693

during time 1/P activity reduced to 1/e it¶s

initial value

PX

11

0!! ´

g!

!

t

t o

dN t N



3-12

Lifetime Total number of nuclei that decay over time

Dose

Atom at a time

Couple withHeisenberg uncertainty principle

(E (th/2T (t is X

with energy in eV

(E(4.133E-15 eV s/2TX!+

+is decay width

* Resonance energy

t1/2=1 sec, X!1.44 s, +=4.56E-16 eV



3-13

Width and energy

Need very short half-lives for

large widths

Useful in Moessbauer

spectroscopy

Absorption distribution

is centered around

EK(E

emission centered EK

(E .

overlapping part of the peaks

can be changed by changing

the temperature of the source

and/or the absorber.



3-14

Equations

Nt=Noe-Pt

N=number of nuclei, P= decay constant,

t=time

Also works for A (activity) or C (counts)

*At=Aoe-Pt, Ct=Coe

-Pt

A= PN

1/P=1/(ln2/t1/2)=1.443t1/2=X



3-15

Half-life calculation

Using Nt=N

oe-Pt

For an isotope the initial count rate was 890 Bq.After 180 minutes the count rate was found tobe 750 Bq

What is the half-life of the isotope 750=890exp(-P*180 min)

750/890=exp(-P*180 min)

ln(750/890)= -P*180 min

-0.171/180 min= -P )min-1 =P!ln2/t1/2

t1/2=ln2/9.5E-4=729.6 min



3-16

Half-life calculationA=PN

A 0.150 g sample of 248Cm has a alpha activity of 0.636 mCi.

What is the half-life of 248Cm?

Find A

* 0.636 E-3 Ci (3.7E10 Bq/Ci)=2.35E7 Bq

Find N

* .150 g x 1 mole/248 g x 6.02E23/mole= 3.64E20

atoms

P!A/N= 2.35E7 Bq/3.64E20 atoms=6.46E-14 s-1

* t1/2=ln2/P!6.46E-14 s-1=1.07E13 s

* 1.07E13 s=1.79E11 min=2.99E9 h=1.24E8 d

=3.4E5 a



3-17

Counting

A=P N Your gamma detector efficiency at 59 keV is 15.5

%. What is the expected gamma counts from 75

micromole of 241Am?

Gamma branch is 35.9 % for 241Am

C=(0.155)(0.359)P N

t1/2=432.7 a* (3.16E7 s/a)=1.37E10 s

P=ln2/1.37E10 s=5.08E-11 s-1

N=75E-6 moles *6.02E23/mole=4.52E19 atoms

C=(0.155)(0.359)5.08E-11 s-1*4.52E19 =1.28E8 Bq



3-18

Decay Scheme



3-19

Specific activity

Activity of a given amount of radionuclide

Use A=PN

Use of carrier should be included

SA

of

226

Ra 1 g 226Ra, t1/2= 1599 a

1 g * 1 mole/226 g * 6.02E23 atoms/mole =2.66E21 atom = N

t1/2=1599 a *3.16E7 s/a = 5.05E10 s P=ln2/ 5.05E10 s =1.37E-11 s-1

A= 1.37E-11 s-1 * 2.66E21=3.7E10 Bq



3-20

Specific Activity

1 g 244Cm, t1/2=18.1 a

1 g * 1 mole/244 g * 6.02E23 atoms/mole = 2.47E21

atom = N

t1/2=18.1 a *3.16E7 s/a = 5.72E8 s

P=ln2/ 5.72E8 s =1.21E-9 s-1

A= 1.21E-9 s-1 * 2.47E21=2.99E12 Bq

Generalized equation for 1 g

6.02E23/Isotope mass *2.19E-8/ t1/2 (a)

1.32E16/(Isotope mass* t1/2 (a))



3-21

Specific

Activity

Isotope t 1/2 a SA (Bq/g)

14 C 5715 1.65E+11

228 Th 1.91E+00 3.03E+13

232 Th 1.40E+10 4.06E+03

233 U 1.59E+05 3.56E+08

235 U 7.04E+08 7.98E+04

238 U 4.47E+09 1.24E+04

237 Np 2.14E+06 2.60E+07

238 Pu 8.77E+01 6.32E+11

239 Pu 2.40E+04 2.30E+09

242 Pu 3.75E+05 1.45E+08

244 Pu 8.00E+07 6.76E+05

241 Am 4.33E+02 1.27E+11

243 Am 7.37E+03 7.37E+09

244 Cm 1.81E+01 2.99E+12

248 Cm 3.48E+05 1.53E+08



3-22

1000

104

105

106

107

108

109

1010

1011

1012

1013

1014

1 100 104

106

108

1010

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

102

103

S A ( B q

/ g )

t 1/2 (a)

y = 2 /M0

Error Value

1.3169e+115.7831e+13m2

6.7326e+22Chis

0.99996R

14C

S A

( C i / g )



3-23

Specific Activity

Activity/mole

N=6.02E23

SA (Bq/mole) of 129I, t1/2=1.57E7 a

t1/2=1.57E7 a *3.16E7 s/a = 4.96E14 s P=ln2/ 4.96E14 s =1.397E-15 s-1

A= 1.397E-15 s-1 *6.02E23=8.41E8 Bq

Generalized equation SA (Bq/mole)=1.32E16/t1/2 (a)



3-24

Specific Activity

Isotope t 1/2 a

SA

(Bq/mole)

3 H 12.3 1.07E+15

14 C 5715 2.31E+12

22 Na 2.6 5.08E+15

55 Fe 2.73 4.84E+15

228 Th 1.91E+00 6.91E+15

232 Th 1.40E+10 9.43E+05

233 U 1.59E+05 8.30E+10

235 U 7.04E+08 1.88E+07

238 U 4.47E+09 2.95E+06

237 Np 2.14E+06 6.17E+09

238 Pu 8.77E+01 1.51E+14

239 Pu 2.40E+04 5.50E+11

242 Pu 3.75E+05 3.52E+10

244 Pu 8.00E+07 1.65E+08

241 Am 4.33E+02 3.05E+13

243 Am 7.37E+03 1.79E+12

244 Cm 1.81E+01 7.29E+14

248 Cm 3.48E+05 3.79E+10



3-25

105

107

109

10

11

1013

1015

1 100 104

106

108

1010

SA (Bq/mole)

S A

( B q / m

o l e )

t 1/2 (a)

y m2 / 0

Err or alue

1.9321e+121.3204e+16m2

NA3.5919e+25Chisq

NA1R



3-26

SA with carrier

1E6 Bq of 152Eu is added to 1 mmole Eu.

Specific activity of Eu (Bq/g)

Need to find g Eu

1E-3 mole *151.96 g/mole = 1.52E-1 g =1E6 Bq/1.52E-1 g =6.58E6 Bq/g

*=1E9 Bq/mole

What is SA

after 5 years t1/2=13.54 a

= 6.58E6*exp((-ln2/13.54)*5)=

*5.09E6 Bq/g



3-27

Lifetime

Atom at a time chemistry

261Rf lifetime

Find the lifetime for an atom of 261Rf

t1/2 = 65 s X=1.443t1/2

X=93 s

Determines time for experiment Method for determining half-life



3-28

Mixtures of radionuclides

Composite decay Sum of all decay particles

Not distinguished by energy

Mixtures of Independently Decaying Activities

if two radioactive species mixed together,observed total activity is sum of two separateactivities:

A=A1+A2=c1P1N1+c2P2N2

any complex decay curve may be analyzed into

its components Graphic analysis of data is possible



3-29

Can determine initial

concentration and half-lifeof each radionuclide



3-30

100

1000

104

0 5 10 15 20 25

t o t a l

T (hr )

y = m1*exp(-m2*M0)+m3*exp(-m...

Error Value

0.0006541610000m1

5.3036e-080.55452m2

0.000692062000m3

3.3669e-080.066906m4

3.7138e-07Chis

1R

P=0.554

t1/2=1.25 hr

l=0.067

t1/2=10.4 hr



3-31

Parent ± daughter decay Isotope can decay into

radioactive isotope Uranium decay series

Lower energy

Different properties

A

Z

Spin

Parity

For a decay parent -> daughter

Rate of daughterformation dependent upon

parent decay rate-

daughter decay rate



3-32

Parent - daughter

For the system 1 decays into 2

Rearranging gives

Solve and substitute for N1 using N1t=N1oe-Pt

Linear 1st order differential equation

Solve by integrating factors

Multiply by eP2t

22112

N N d t

dN PP !

d t N d t N dN 11222 PP !

d t e N d t N dN t

o1

11222

PPP

!

d t e N e N d

d t e N d t e N dN e

t

o

t

t

o

t t

)(

112

)(

11222

122

1222

)(PPP

PPPP

P

PP

!

!



3-33

Parent-daughter

Integrate from t 0->t

Multiply by e-P2t and solve for N2

)1( )(1

12

122

0 12

)(

11

0

2

122

12

2

!

!

´´

t oo

t

t t

o

t

t

e N N e N

e N e N

PPP

PP

P

PPP

PP

P

t o

t t o e N ee N t N 221

21

12

12 )()( PPP

PP

P

!

Growth of daughter from parentDecay of initial daughter



3-34

Parent daughter

Can solve equation for activity from A=P2

Find maximum daughter activity based ondN/dt=0

Solve for t

For 99mTc (t1/2=6.01 h) from 99Mo (2.75 d)

PTc=2.8 d-1, PMo=0.25 d-1

0.94 d

t

o

t t

o e Aee N A 221

21

12

212 )( PPP

PP

PP

!

t t ee 21

21

PPPP

!

)(

)ln(

12

1

2

PP

P

P

!t



3-35

Half life relationships

No daughter decay No activity of daughter

Number of daughter atoms due to parent decay

No Equilibrium

if parent is shorter-lived than daughter(P1"P2), no equilibrium attained at any time

daughter reaches maximum activity whenP1N1=P2N2

All parents decay, then decay is based ondaughter

)1( 1

12

t

o e N N P

!



3-36

Half life relationships

Transient equilibrium

Parent half life greater than 10 x daughter

half life

(P1 < P2)

Parent daughter ratio becomes constant overtime

As t goes toward infinity

0;212

2 p

t

o

t t

e N eePPP

t

oe N N 1

1

12

12

P

PP

P

} N ! N1oe

P1t N2

N1

!P1

P2 P1



3-37



3-38

Half life relationship

Secular equilibrium

Parent much longer half-life than daughter

1E4 times greater

(P1 << P2) Parent activity does not measurably

decrease in many daughter half-lives

12

1

1

2

PP

P

! N

N

2

1

1

2

P

P

! N

N

12

1122

A A

N N

!

! PP



3-39



3-40

Many Decays

Bateman solution

Only parent present at time 0

d 3

dt! P

2N2 P

3N

3

Nn ! C1eP

1t C

2eP

2t Cne

Pnt

C1 !

P1

P2.....P (n1)

(P2 P1

)(P3 P1

)...(Pn P1)N

1o

C !P1P .....P( 1)

(P1

P )(P3

P )...(P P2)N1o



3-41

Branching decay Branching Decay

partial decay constants must be considered

A has only one half life

if decay chain branches and two branches arelater rejoined, the two branches are treated as

separate chains

production of common member beyond

branch point is sum of numbers of atoms formed by the two paths

Branching ratio is based on relative constants

PiPt

§§!!

!!

N

i

i

N

i

i

t t 1 2/12/11

11;PP



3-42

Branching Decay

For a branching decay of alpha and beta

At=AE+A F

A=PN, so

* PtN =PEN+P FNPt=PE+P F

1=AEAt +A FAt 1=PEPt +P FPt

Consider 212Bi, what is the half life for each decay mode?

Alpha branch 36 %, beta branch 64 %

t1/2=60.55 min

Pt=0.0114; 0.36=PEPt; 0.36=PE0.0114; PE=0.0041

t1/2 alpha = 169 min Pt=PE+P F0.0114=0.0041+P F0.0073=P F

t1/2 beta = 95.0 min



3-43



3-44

Cross Sections

Originates from simple picture that probability for reaction

between nucleus and impinging particle is proportional to the

cross-sectional target area presented by the nucleus

doesn¶t hold for charged particles that have to

overcome Coulomb barriers or for slow neutrons

Total cross section for collision with fast particle is never

greater than twice the geometrical cross-sectional area of the

nucleus

10-24 cm2=1 barn

The probability of a nuclear process is generally

ex pressed in terms of a cross section W that has the

dimensions of an area.



3-45

iiN R W J !

iiInx R W !

For a beam of particles striking a thin target--one in which

the beam is attenuated only infinitesimally--the cross

section for a particular process is defined:

When a sam ple is embedded in a uniform flux of particles

incident on it from all direction, such as in a nuclear

reactor, the cross section is defined:

J=flux of particles/cm2/sec

N=number of nuclei contained in

sam ple

R i= # of processes of ty pe under consideration occurring in the target

per unit timeI= # of incident particles per unit

time

n= # of nuclei/cm3

x=target thickness (cm)



3-46

Production of radionuclides

N1

=N0

WJ

W=cross section

J=neutron flux

To full consider produced nuclei

N1=N0WJP 1-exp-(Pt))

t=time of irradiation

1-exp-(Pt)) gives maximum level percent

half lif e %1 50

2 75

3 87.5

4 93.75

5 96.875



3-47

Natural Radionuclides

70 naturally occurring radioactive isotopes

Mainly decay from actinides Tritium

14C

40K

70 kg ³reference man,´

4400 Bq of 40K

3600 Bq of 14C

US diet

1 pCi/day of 238U, 226Ra, and 210Po

air

~ 0.15 pCi/L of 222Rn

earth¶s crust ~ 10 ppm and ~ 4 ppm of the radioelements Th and U.

interior heat budget of the planet Earth is dominated by thecontributions from the radioactive decay of U, Th, and K



3-48

Environmental radionuclides

primordial nuclides that have survived since the timethe elements were formed

t1/2>1E9 a

Decay products of these long lived nuclides

40K., 87Rn, 238U, 235U, 232Th

cosmogenic are shorter lived nuclides formedcontinuously by the interaction of comic rays withmatter

3H., 14C, 7Be

14N(n, 1H )14C (slow n)

14N(n, 3H )12C (fast n) anthropogenic are nuclides introduced into the

environment by the activities of man

Actinides and fission products

14C and 3H



3-49

Dating

Radioactive decay as clock Based on Nt=Noe

-Pt

Solve for t

N0 and Nt are the number of radionuclides present at

times t=0 and t=t Nt from A = N

t the age of the object

Need to determine No

For decay of parent P to daughter D totalnumber of nuclei is constant

PP

t

o

o

t

N

N

N

N

t

lnln

!

!

o P t P t D ! )()(



3-50

Dating

Pt=Poe-Pt

Measuring ratio of daughter to parent atoms

no daughter atoms present at t=0

that they are all due to the parent decay none have been lost during time t

Amineral has a 206Pb/238U =0.4. What is the

age of the mineral?

t=(1/(ln2/4.5E9))ln(1+0.4)

2.2E9 years

)1ln(1

t

t

t ! P



3-51

Dating

14C dating

Based on constant formation of 14C

No longer uptakes C upon organism

death 227 Bq 14C /kgC at equilibrium

What is the age of a wooden sample with 0.15

Bq/g C?

t=(1/(ln2/5730 a))*ln(0.227/0.15)=3420 a

)ln(114

14

sampl e

eq

C C t

P!



3-53

Questions

Make excel sheets to calculate

Mass or mole to activity

Calculate specific activity

Concentration and volume to activity Determine activity for counting

Parent to progeny

Daughter and granddaughter

* i.e., 239U to 239Np to 239Pu

decay kinetics

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