Transcript Decay Kinetics - UNLV Radiochemistry

Radioactive Decay Kinetics
•
Outline
Radioactive decay
kinetics
 Basic decay
equations
 Utilization of
equations
 Mixtures
 Equilibrium
 Branching
 Natural radiation
 Dating
3-1
Introduction
• Number of radioactive nuclei that decay in a sample
decreases 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
3-2
Basic Decay Equations
• Decay is 1st order
 Rate proportional to amount of parent
isotope
Equal to the rate of isotope
disintegration
Proportional to number of radioactive
nuclei
* rate of decay=decay constant*#
radioactive nuclei
 Decay constant is average decay
probability per nucleus for a unit time
3-3
 Represented by l
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 present
circumstances
• The probability of disintegration depends only on the
length of the time interval.
Probability of decay: p=lDt
Probability of not decaying: 1-p=1- lDt
3-4
StatisticsofofRadioactive
Radioactive Decay
Statistics
Decay
1-p=1-lDt=probability that atom will survive Dt
(1-l Dt)n=probability that atom will survive n intervals of t
nDt=t, therefore (1- l Dt)n =(1- l t/n)n
Since limn∞(1+x/n)n=ex, (1- lt/n)n=e-lt, the limiting value.
Considering No atoms, the fraction remaining unchanged
after time t is N/No= e-lt
N=Noe-lt
where l is the decay constant
3-5
Radioactivity as Statistical
Phenomenon
• Binomial Distribution for Radioactive Disintegrations
 probability W(m) of obtaining m disintegrations in
time t from No original radioactive atoms
No!
W (m) 
p m (1  p) N o  m
( N o  m )! m!
 probability of atom not decaying in time t, 1-p, is
(N/No)=e-lt, 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
3-6
and t+d:
P(t )dt  N ole  N lt dt
o
• Average Disintegration Rate
W (r ) 
where 1-p=q
n!
p r q n r
( n  r )! r !
np   r  0 rW ( r )  r
r n
 for radioactive disintegration--if n=No and p=1e-lt--average number M of atoms disintegrating
in time t is M=No(1-e-lt); for small lt, M=Nolt
and disintegration R=M/t=No l, which
corresponds to -dN/dt=lN
• Expected Standard Deviation
s  N o (1  e  lt )e  lt  Me  lt
Since in counting practice lt is generally small , s  M
• M is number of counts
• Relative error = s-1
3-7
Measured Activity
• In practicality, activity (A)
is used instead of the
number of atoms (N).
• A= clt, m
 where c is the detection
coefficient
 A=AOe-lt
• Units
 Curie
 3.7E10 decay/s
* 1 g Ra
 Becquerel
 1 decay/s
3-8
Half Life and decay constant
Half-life is time needed to decrease
nuclides by 50%
Relationship between t1/2 and l
N/No=1/2=e-lt
ln(1/2)=-lt1/2
ln 2= lt1/2
t1/2=(ln 2)/l
3-9
Half lives
• Large variation in half-lives for different
isotopes
 Short half-lives can be measured
Evaluate activity over time
* Observation on order of half-life
 Long half-lives
Based on decay rate and sample
* Need to know total amount of nuclide
in sample
* A=ln
 A is activity, n is number of nuclei
3-10
Exponential Decay
• Average Life () for a radionuclide
 found from sum of times of existence of all
atoms divided by initial number of nuclei
1
 
No
t 
1
 t  dN  l
t 0
 1/l=1/(ln2/t1/2)=1.443t1/2=
Average life greater than half life by factor
of 1/0.693
during time 1/l activity reduced to 1/e it’s
initial value
3-11
Lifetime
• Total number of nuclei that decay over time
 Dose
 Atom at a time
• Couple with Heisenberg uncertainty principle

DE Dt≥h/2p
 Dt is 
 with energy in eV
 DE≥(4.133E-15 eV s/2p)/  G
 G is decay width
* Resonance energy
 t1/2=1 sec, 1.44 s, G=4.56E-16 eV
3-12
Width and energy
• Need very short half-lives for
large widths
• Useful in Moessbauer
spectroscopy

Absorption distribution
is centered around
Eg+DE

emission centered Eg
DE .
• overlapping part of the peaks
can be changed by changing
the temperature of the source
and/or the absorber.
3-13
Equations
• Nt=Noe-lt
 N=number of nuclei, l= decay constant,
t=time
Also works for A (activity) or C (counts)
* At=Aoe-lt, Ct=Coe-lt
• A= lN
• 1/l=1/(ln2/t1/2)=1.443t1/2=
3-14
Half-life calculation
Using Nt=Noe-lt
• For an isotope the initial count rate was 890 Bq.
After 180 minutes the count rate was found to
be 750 Bq
 What is the half-life of the isotope
750=890exp(-l*180 min)
750/890=exp(-l*180 min)
ln(750/890)= -l*180 min
-0.171/180 min= -l
9.5E4 min-1 =lln2/t1/2
t1/2=ln2/9.5E-4=729.6 min
3-15
Half-life calculation
A=lN
• 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
 lA/N= 2.35E7 Bq/3.64E20 atoms=6.46E-14 s-1
* t1/2=ln2/l0.693/6.46E-14 s-1=1.07E13 s
* 1.07E13 s=1.79E11 min=2.99E9 h=1.24E8 d
=3.4E5 a
3-16
Counting
A=lN
• 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)lN
 t1/2=432.7 a* (3.16E7 s/a)=1.37E10 s
 l=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
3-17
Decay Scheme
3-18
Specific activity
• Activity of a given amount of radionuclide
 Use A=lN
Use of carrier should be included
• SA of 226Ra
 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
l=ln2/ 5.05E10 s =1.37E-11 s-1
 A= 1.37E-11 s-1 * 2.66E21=3.7E10 Bq
3-19
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
 l=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))
3-20
Isotope
14 C
Specific
Activity
t 1/2 a
SA (Bq/g)
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
3-21
10
10
10
10
13
14
10
3
10
2
10
1
10
0
10
-1
10
-2
10
-3
10
-4
10
-5
10
-6
10
-7
y = m2 /M0
C
12
11
Value
Error
m2
5.7831e+13
1.3169e+11
Chisq
R
6.7326e+22
0.99996
NA
NA
10
10
10
10
10
10
10
9
8
SA (Ci/g)
SA (Bq/g)
10
14
7
6
5
4
1000
1
100
10
4
10
6
t 1/2 (a)
10
8
10
10
3-22
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
l=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)
3-23
Specific Activity
Isotope
SA
(Bq/mole)
t 1/2 a
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
3-24
10
15
SA (Bq/mole)
SA (Bq/mole)
10
13
10
11
10
9
y = m2 /M0
10
7
10
5
1
Value
Error
m2
1.3204e+16
1.9321e+12
Chisq
R
3.5919e+25
1
NA
NA
100
10
4
10
t 1/2 (a)
6
10
8
10
10
3-25
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
3-26
Lifetime
• Atom at a time chemistry
• 261Rf lifetime
 Find the lifetime for an atom of 261Rf
t1/2 = 65 s
=1.443t1/2
=93 s
• Determines time for experiment
• Method for determining half-life
3-27
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 separate
activities:
A=A1+A2=c1l1N1+c2l2N2
any complex decay curve may be analyzed into
its components
Graphic analysis of data is possible
3-28
Can determine initial
concentration and half-life
of each radionuclide
3-29
l=0.554
t1/2=1.25 hr
y = m1*exp(-m2*M0)+m3*exp(-m...
total Bq
10
4
Value
Error
m1
10000
0.00065416
m2
m3
0.55452
2000
5.3036e-08
0.00069206
m4
0.066906
3.3669e-08
Chisq 3.7138e-07
NA
R
1
l=0.067
t1/2=10.4 hr
NA
1000
100
0
5
10
15
T (hr)
20
25
3-30
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 daughter
formation dependent upon
parent decay ratedaughter decay rate
3-31
Parent - daughter
• For the system 1 decays into 2
dN 2
 l1 N1  l2 N 2
dt
• Rearranging gives dN2 + l2 N2dt  l1N1dt
• Solve and substitute for N1 using N1t=N1oe-lt
dN 2 + l2 N 2 dt  l1 N1o e  l1t dt
 Linear 1st order differential equation
Solve by integrating factors
• Multiply by el2t
el2t dN 2 + l2 N 2el2t dt  l1 N1o e( l2 l1 )t dt
d ( N 2el2t )  l1 N1o e( l2 l1 )t dt
3-32
Parent-daughter
• Integrate from t 0->t
t
t
( l2  l1 ) t
l
N
e
l2 t
1 1o
N
e

0 2
0 l2  l1
N 2e
l2 t
 N 2o 
l1
l2  l1
N1o (e ( l2 l1 )t  1)
• Multiply by e-l2t and solve for N2
N 2 (t ) 
l1
l2  l1
N1o (e l1t  e l2t ) + N 2o e l2t
Growth of daughter from parent
Decay of initial daughter
3-33
Parent daughter
• Can solve equation for activity from A=lN
l1l2
A2 
N1o (e l t  e l t ) + A2 o e l t
l2  l1
• Find maximum daughter activity based on
dN/dt=0
l1e l t  l2e  l t
l2
• Solve for t
ln( )
l1
t
(l2  l1 )
1
2
1
2
2
• For 99mTc (t1/2=6.01 h) from 99Mo (2.75 d)
 lTc=2.8 d-1, lMo=0.25 d-1
0.94 d
3-34
Half life relationships
• No daughter decay
 No activity of daughter
 Number of daughter atoms due to parent decay
• No Equilibrium


N 2  N1o (1  e  l1t )
if parent is shorter-lived than daughter
(l1l2), no equilibrium attained at any time
daughter reaches maximum activity when
l1N1=l2N2
 All parents decay, then decay is based on
daughter
3-35
Half life relationships
• Transient equilibrium
 Parent half life greater than 10 x daughter
half life
(l1 < l2)
• Parent daughter ratio becomes constant over
time
 As t goes toward infinity
e  l2t  e  l1t ; N 2 o e  l2t 
 0
N2 
l1
l2  l1
N1o e
 l1t
N1  N1o e
l 1t
N2
l1

N1 l2 3-36
l1
3-37
Half life relationship
• Secular equilibrium
 Parent much longer half-life than daughter
1E4 times greater
(l1 << l2)
 Parent activity does not measurably
decrease in many daughter half-lives
N2
l1
N 2 l1


N1 l2  l1
N1 l2
N 2l2  N1l1
A2  A1
3-38
3-39
Many Decays
dN3
 l 2N2  l3N3
dt
• Bateman solution
• Only parent present at time 0
Nn  C1e
C1 
 l1 t
+ C2 e
l 2t
+ Cn e
l nt
l1l 2 .....l (n1)
(l 2  l1 )(l 3  l1 )...(l n  l1 )
C2 
N1o
l1l 2 .....l(n1)
(l1  l 2 )(l3  l2 )...(ln  l2 )
N1o
3-40
Branching decay
• Branching Decay

partial decay constants must be considered
N
A has only one half life l  l ; 1 

i 1

i
t1/ 2
N
t
i
i 1 1/ 2
if decay chain branches and two branches are
later 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
 li/lt
1
3-41
Branching Decay
• For a branching decay of alpha and beta
 At=Aa+Ab
 A=lN, so
* ltN =laN+lbN; lt=la+lb
 1=Aa /At +Ab /At ; 1=la /lt +lb /lt
• Consider 212Bi, what is the half life for each decay mode?
 Alpha branch 36 %, beta branch 64 %
 t1/2=60.55 min
 lt=0.0114; 0.36=la /lt; 0.36=la /0.0114; la=0.0041
 t1/2 alpha = 169 min
 lt=la+lb; 0.0114=0.0041+lb; 0.0073=lb
 t1/2 beta = 95.0 min
3-42
3-43
Cross Sections
The probability of a nuclear process is generally
expressed in terms of a cross section s that has the
dimensions of an area.
• 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
3-44
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:
Ri  Inxs i
When a sample 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:
Ri  Ns i
Ri= # of processes of type under consideration occurring in the target
per unit time
I= # of incident particles per unit
=flux of particles/cm2/sec
time
N=number of nuclei contained in
n= # of nuclei/cm3
sample
3-45
x=target thickness (cm)
Production of radionuclides
• N1=N0s
 s=cross section
 =neutron flux
• To full consider produced nuclei
 N1=N0s/l1 (1-exp-(l1t))
t=time of irradiation
(1-exp-(l1t)) gives maximum level percent
half life
%
1
50
2
75
3
87.5
4
93.75
5
96.875
3-46
Natural Radionuclides
• 70 naturally occurring radioactive isotopes

Mainly decay from actinides

Tritium
14C

40K

• 70 kg “reference 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 the
contributions from the radioactive decay of U, Th, and K
3-47
Environmental radionuclides
• primordial nuclides that have survived since the time
the elements were formed
 t1/2>1E9 a
 Decay products of these long lived nuclides
 40K., 87Rn, 238U, 235U, 232Th
• cosmogenic are shorter lived nuclides formed
continuously by the interaction of comic rays with
matter
 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
3-48
 14C and 3H
Dating
• Radioactive decay as clock
 Based on Nt=Noe-lt
 Solve for t
Nt
No
ln
ln
No
Nt
t

l
l
• N0 and Nt are the number of radionuclides present at
times t=0 and t=t
 Nt from A = λN
• t the age of the object
 Need to determine No
 For decay of parent P to daughter D total
number of nuclei is constant
D(t ) + P(t )  Po
3-49
Dating
Dt
t  ln( 1 + )
l
Pt
1
• Pt=Poe-lt
• 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
• A mineral 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
3-50
Dating
•
14C
dating
 Based on constant formation of 14C
No longer uptakes C upon organism
14
death
Ceq
1
)
• 227 Bq 14C /kgC at equilibrium t  l ln( 14C
sample
• 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
3-51
Dating
• Determine when Oklo reactor operated
 Today 0.7 % 235U
 Reactor 3.5 % 235U
 Compare 235U/238U (Ur) ratios and use Nt=Noe-lt
e -l235t
U r (t)  U r (o) -l238t  U r (o)e (- l235t + l238t )
e
U r (t)
ln
 t (-l235 + l238 )
U r (o)
U r (t)
ln
U r (o)
t
(-l235 + l238 )
7.25E - 3
ln
3.63E - 2
t
1.94 E 9 years
(-9.85E - 10 + 1.55E  10)
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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
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