Post on 16-Dec-2015
4. Radioactive Decay
radioactive transmutation and decay are synonymous expressions
4 main series4n 232Thorium4n + 2 238Uranium-Radium4n + 3 235Actinium 4n + 1 237Neptunium
4.1 Decay Series
4.2a Law and Energy of Radioactive Decay
radioactive decay law follows Poisson statistics behaves as
where: N is the number of atoms of a certain radionuclide;
-dN/dt is the disintegration rate; and
is the disintegration constant in sec-1
λNdt
dN
4.2a Law and Energy of Radioactive Decay
law of radioactive decay describes the kinetics of a reaction
Where A is the mother radionuclide;B is the daughter nuclide;X is the emitted particle; and E is the energy set free by the decay process(also known as Q-value)
ΔExBA
4.2a Law and Energy of Radioactive Decay
radioactive decay only possible when
E > 0 which can be calculated as
however decay may only arise if nuclide A surmounts an energy barrier with a threshold ES or through quantum mechanical tunneling
2XB
2 cMMMc AM E
4.2b Kinetics of Radioactivity
Half-Life
the time for any given radioisotope to decrease to 1/2 of its original quantity
range from a few microseconds to billions of years
100 -
80 -
60 -
40 -
20 -
0 -
5 10 15 20 25 30 35
years
act
ivit
y i
n
per
cen
t
t1/2 = 5 years
4.2b Kinetics of Radioactivity
each isotope has its own distinct half-life (t1/2) and in almost all cases no operation, physical or chemical, can alter the transformation rate
1st half-life 50% decay2nd half-life 75% decay3rd half-life 87.5% decay4th half-life 93.75% decay5th half-life 96.87% decay6th half-life 98.44% decay7th half-life 99.22% decay
4.2b Kinetics of Radioactivity
4.2c Probability of Disintegration
number of nuclei dN in a time interval dt will be proportional to that time interval and to the number of nuclei N that are present; or at any time t there are N nuclei dN = - Ndt
where is the proportionality constant and the -ve sign is introduced because N decreases
CtNlndtN
dNdt
N
dN
at t = 0: N = N0 therefore lnN0 = C
teoNN
teN
Nt
N
NlnNlntNln
00
0
n0 2
1
N
N
the fraction of any radioisotope remaining after n half-lives is given by
4.2c Probability of Disintegration
where No is the original quantity and N is the quantity after n half lives
1/2
t t
00
t
ln2
t 2 ln
e2e2/12/1 N
N when e
N
N
2/1
1/2 2/1t
4.2c Probability of Disintegration
if the time t is small compared with the half-life of the radionuclide ( t<<t1/2) then we can approximate
...
2
1/2t
t
2
2(ln2)
t/2t
t(ln2)λt1
2
2t)(λt1λte
4.2c Probability of Disintegration
1/2t 1.45
00
1dteNdt
N
1 t
0
Average Life of an Isotope
it is equally important to know the average life of an isotope
4.2c Probability of Disintegration
Decay Constant Problems
what is the constant 52V which has a t1/2 = 3.74 min.?
1-3 sec 1009.3sec 60
min 1
min 74.3
693.
4.2c Probability of Disintegration
what is the constant for 51Cr which has a t1/2 = 27.7 days?
1-7 sec 1089.2sec 86,400
day 1
days 7.27
693.
1-11 sec 10355.1days 365
yr1
sec 86,400
day 1
yrs1622
693.
what is the constant for 226Ra which has t1/2 = 1622 yrs
4.2c Probability of Disintegration
Decay Problem
what % of a given amount of 226Ra will decay during a period of 1000 years?
1/2 life of 226Ra = 1622 yr
14 yr1038.4
4.2c Probability of Disintegration
therefore the percentage transformed during the 1000 year period is:
100% - 64.5% = 35.5%
%5.64645.0e
yr310 1yr41038.4eeA
A
438.0
t
0
4.2c Probability of Disintegration
4.2d Activity
Curie (Ci), originally defined as the activity of 1 gm of Ra in which 3.7 1010 atoms are transformed per sec
in S.I. units activity is measured in Becquerel (Bq), where 1 Bq = 1 tps -> the quantity of radioactive material in which one atom is transformed per sec
activity of a radionuclide is given by its disintegration rate
1/2t
ln2 N
dt
dN - A
4.2d Activity
equal weights of radioisotopes do not give equivalent amounts of radioactivity
238U and its daughter 234Th have about the same no. of atoms per gm. However their half- lives are greatly different
238U = 4.5 109 yr; 234Th = 24.1 days
therefore, 234Th is transforming 6.8 1010 faster than 238U
4.2d Activity
60Co
60Ni
, 0.314 MeV
, 1.1173 MeV
, 1.332 MeV
1 Bq with 3 emissions
4.2d Activity
1 Bq with 1.18 emissions
42K
42Ca
, 2.04 MeV
18%
, 1.53 MeV
4.2d Activity
1 kilobecquerel (kBq) = 103 Bq
1 megabecquerel (MBq) = 106 Bq
1 gigabecquerel (GBq) = 109 Bq
1 terabecquerel (TBq) = 1012 Bq
1 millicurie (mCi) = 10-3 Ci
1 microcurie (μCi = 10-6 Ci
1 nanocurie (nCi) = 10-9 Ci
1 picocurie (pCi) = 10-12 Ci
1 femtocurie (fCi) = 10-15 Ci
1 Ci = 3.7 1010 Bq
4.2d Activity
since activity A is proportional to N, the number of atoms, we get
A = A0e-t
the mass m of radioactive atoms can be calculated from their number N; activity A; M
mass of nuclide; and Nav Avogadro’s number ( 6.02 X 1023)
1/2AνAνAν
tln2N
MA
λN
MA
N
MNm
4.2d Activity
Problem● how much time is required for 5 mg of 22Na
(t1/2 = 2.60 y) to reduce to 1 mg?● since the mass of a sample will be
proportional to the no. of atoms in the sample get
y6.04t
1/5e
e mg 5mg 1
ememm
t y0.693/2.60
t y0.693/2.60
t 0.693/t0
λt0
1/2
4.2d Activity
Specific Activity the relationship between mass of the material
and activity or
AS (SA) = no. of Bq's/unit mass or volume
SA) (also [Bq/g]m
AsA
4.2d Activity
Bq/g1/2tA
23104.18SA
M
23106.025
1/2t
0.693
Bq/gM
23106.025λλNSA
g/mole M
atoms/mole23106.025N
SA can also be represented in combined mathematical known terms
4.2d Activity
SA may also be derived by using the fact that there are 3.7 1010 tps in 1 gm of 226Ra
g/BqtA
tA107.3SA
tA/1018.4
tA/1018.4
g/Bq107.3
g/BqSA
i2/1i
Ra2/1Ra10
Ra23
i23
10i
Ra2/1
2/1
4.2d Activity
4.2d Activity
Problemcalculate the specific activity of 14C
(t1/2 = 5730 yrs)
g/Bq1067.1g/Bqyr 573014
yr 1622226107.3SA 1110
Problempotassium (atomic weight = 39.102 AMU)
contains: 93.10 atom % 39K, having atomic mass
38.96371 AMU
0.0118 atom % 40K, which has a mass of 40.0 AMU and is radioactive with:
t1/2 = 1.3 109 yr
6.88 atom % 41K having a mass of 40.96184 AMU
4.2d Activity
estimate the specific activity of naturally occurring potassium
specific activity refers to the activity of 1 g material
1 g of naturally occurring potassium contains: 1.18 10-4 g 40K plus non-radioactive isotopes
4.2d Activity
11
18
9
234
23
4
2/1
s 1000.3
y1047.9
0.40y103.11002.61018.1693.0
nuclei 1002.6/g 0.40g1018.1
t693.0
NA
4.2d Activity
Problemprior to use of nuclear weapons, the SA of
14C in soluble ocean carbonates was found to be 16 dis/min ·g carbon
amount of carbon in these carbonates has been estimated as 4.5 1016 kg
how many MCi of 14C did the ocean carbonates contain?
Ci M 320
Ci 102.3dis/s 103.7
Ci 1
s 60
min 1
kgC 10
min/dis 16 kgC105.4 8
10316
4.2d Activity
Problema mixture of 239Pu and 240Pu has a specific
activity of 6.0 109 dps
the half-lives of the isotopes are 2.44 104 and 6.58 103y, respectively
calculate the isotopic composition
2/1t
N 693.0NA
4.2d Activity
for 239Pu
for 240Pu
gy/1015.7g 239
1002.6
y1044.2
693.0A 16
23
4
gy/1064.2g 2401002.6
y1058.6693.0
A 1723
3
4.2d Activity
number of seconds in a year is
s 7103.15min
s 60
h
min 60
d
h 24 d 365
for 239Pu: A = 2.27 109/s g
for 240Pu: A = 8.37 109/s g
let the fraction of 239Pu = x; then the fraction 240Pu = 1 - x
4.2d Activity
(2.27 109)x+(1 – x)(8.37 109) = 6.0 109
(8.37 109) – (6.10 109) x = 6.0 109
2.37 109 = (6.1 109) x
x = 0.39 = 39% 239Pu
4.2d Activity
4.2d Activity
Problem if 3 10-9 kg of radioactive 200Au has an
activity of 58.9 Ci, what is its half-life?
no. of atoms in 3 10-9 kg of 200Au is
atoms109.04kmol
atoms106.025
kg 200
kmol 1 kg103N 15269
s
tegrationsindi1018.2
Ci 1
tion/sdisintegra 107.3 Ci 9.58 12
10
decay constant is found from
A = N
1415
112
s1041.21004.9
s1018.2
min 48s1088.2s1041.2
693.02lnt 3
142/1
finally
4.2d Activity
4.3 Radioactive Equilibria
3 nuclide2 nuclide1 nuclide
net production of nuclide 2 is given by decay rate of nuclide 1 less the decay rate of nuclide 2
22112212 NNN
dt
dN
dt
dN
0eNNdt
dN
e N N
t-01122
2
01 1
1
t
t-02
t-t-01
12
12
221 eN)e(eNN
4.3 Radioactive Equilibria
solution of first order differential equation
given that:
)e(eNN
then 0 N
t-t-01
12
12
02
21
]1[NN t)1
12
12
1
2(-e
4.3 Radioactive Equilibria
if nuclide 1 and 2 are separated at t = 0; then nuclide 2 is not produced and
)1(t)2(t
12/12/1
2/12/12
2/12/1
2
11N
)1(t/)2(t1
)1(t/)2(tN
)2(t
t
)1(t
)2(t1
)1(t
t
)2(t
t
)2(t /(1)t
2/12/1
2/1
2/12/1
1/21/2
4.3 Radioactive Equilibria
after substitution for λ:
the exponent term can be written to show the influence the ratio of
12/12/1
2/12/11
12
12 N
)1(t/)2(t1
)1(t/)2(tNN
4.3 Radioactive Equilibria
4.4 Secular Equilibrium
in secular equilibrium t1/2 (1)>> t1/2 (2) so
]e1[NN t)(-1
12
12
12
t-1
2
12
2e1NN
reduces
)1(t
)2(t
N
N
2/1
2/1
2
1
1
2
21 AA
β 64.1 h
β 2.74 m
β 28.8 y
β 33 S
90Kr 90Rb 90Sr 90Y 90Zr
4.4 Secular Equilibrium
after 10 half-lives
Practical Applicationsdetermination of long-half-life of a mother
nuclide by measuring the mass ratio of the daughter and mother nuclides providing the half-life of the daughter is known
calculation of mass ratios of radionuclidescalculation of the mass of a mother nuclide
from the measured activity of a daughter nuclide or the reverse
(1)t ln2
A
N
M m 1/2
2
av1
4.4 Secular Equilibrium
Problemhow many grams of 90Y are in secular
equilibrium with 1 mg of 90Sr
thus, the amount of 90Y having the same activity of 1 mg of 90Sr
Ci/g 14590
226
8.28
1622Srof SA 90
1
4.4 Secular Equilibrium
1 mg of sample is
Ci 0.145Ci/g 145g10A 3i
Ci/g105.50SA
90
226
day
yr
365
1
hr
day
24
164.1
1622SA
52
2
4.4 Secular Equilibrium
specific activity of 90Y is
therefore mass of 90Y is
g 264.0
g106.2Ci/g 105.50
Ci 145.0 75
4.4 Secular Equilibrium
4.5 Transient Equilibrium
in transient equilibrium the half-life of the mother is longer than the daughter
t1/2 (1)> t1/2 (2)
)2(t)1(t
)2(t
N
N
2/12/1
2/1
1
2
in secular equilibrium the mother and daughter have the same activities
in transient equilibrium the the daughter activity is always higher
)1(t
)2(t11
N
N
A
A
2/1
2/1
2
1
22
11
2
1
4.5 Transient Equilibrium
Practical Applicationsthe same applications as in secular
equilibrium except the following equation is used
)2(t)1(t2ln
A
N
Mm 2/12/1
2
Av
11
4.5 Transient Equilibrium
4.6 Half-Life of Mother Nuclide Shorter than Half-Life of Daughter
t1/2 (1)< t1/2 (2)
no radioactive equilibrium attained
fission product 141Ce has a half-life of 13.9 minutes and its daughter product 146Pr has a half-life of 24.4 mi
4.7 Similar Half-Lives and Attainment of Maximum Activity of Daughter Nuclide
an important aspect in radiochemistry and health physics is the knowledge when daughter and granddaughters’ products reach their maximum activity
by differentiating with respect to time and setting it equal to zero we get
)e(eNN t-t-01
12
12
21
4.7 Similar Half-Lives and Attainment of Maximum Activity of Daughter Nuclide
2
1
12max
121
2
tt
t
1
2
t2
t1
t2
t1
12
0112
ln1
t
tln
ee
e
ee
0eeN
dt
Nd
12
2
1
21
21
4.7 Similar Half-Lives and Attainment of Maximum Activity of Daughter Nuclide
in the following decay sequence when will the maximum activity of 135Xe occur?
Cs h 9.1
βXe
h 6.6
βI 135135135
in 11.1 hours
4.8 Branching Decay
branching decay is often seen in odd-odd nuclei or in decay series
for example, 40K decays into 40Ca by -
emission with a probability of 89.5% and into 40Ar by electron capture with a probability of 10.7%
B C
Acb
4.8 Branching Decay
AAAcAbA NNN
td
Nd
two probabilities of decay are independent and thus the decay rate is given as
4.8 Branching Decay
t)(0AA
cbeNN
integration of the equation yields
rates of production of nuclides B and C
AcC
AbB N
td
Nd and N
td
Nd
4.8 Branching Decay
CcC
BBB N
td
Nd and N
td
Nd
decay rates of nuclides B and C
4.8 Branching Decay
BBAbB NN
td
Nd
0eNNtd
Nd t)(0AbBB
B cb
net rate of production of nuclide B
using t)(0AA
cbeNN
4.8 Branching Decay
tt)(0A
cbB
bB
Bcb eeN)(
N
integrating and setting NB =0 at t = 0
tt)(0A
cbC
bC
Ccb eeN)(
N
the same holds for nuclide C
4.8 Branching Decay
cbA2/1
2ln2ln)A(t
C
c
A
C
B
b
A
B
N
N and
N
N
in secular equilibrium we get
b + C << b _
but only one half-life
4.8 Branching Decay
c(A)1/2t
(C)1/2t
ANCN
and b(A)1/2t
(B)1/2t
ANBN
cc1/2
bb1/2
ln2(A)t and
ln2(A)t
in secular equilibrium we get
placing these terms into
C
c
A
C
B
b
A
B
N
N and
N
N
4.8 Branching Decay
1eN )t(A
cb
bB
cb
N
1eNλ
N )t(A
cb
cC
cb
c
b
C
B
N
N
if the daughter nuclides are long-lived or stable (as 40K)
4.8 Branching Decay
tN
N and t
N
Nc
A
Cb
A
B
if the time t is small compared with the half- life of the mother nuclide A (t<< t1/2(A)) we get
4.8 Branching Decay
4.9 Successive Transformations
nn1n1nn NN
td
Nd
nuclide 1 nuclide 2 nuclide 3 nuclide 4 nuclide n
tn
t2
t1n
n21 ecececN
Solution of the series of differential equations with n= 1, 2, 3, 4, …n yields
4.9 Successive Transformations
with the coefficients given as:
01
1n1312
1n211 N
)())((c
01
2n2321
1n212 N
)())((c
01
n1nn2n1
1n21n N
)())((c
4.9 Successive Transformations
))((
e
))((
e
))((
e
NN
3132
t
2321
t
1312
t
01213
3
21
for example if n = 3 we get
4.9 Successive Transformations
21013 NNNN
if nuclide 3 is stable then
t
21
1t
12
2013
21 ee1NN
4.9 Successive Transformations
ecN t1n
1
01
n
11 Nc
if t1/2 of the mother nuclide is much longer than the successive ones 1 << 2, 3, n we get
4.9 Successive Transformations
and
under these conditions we get 0
(1) t
t
N
N or
N
N
1/2
(n) 1/2
1
n
2
1
1
n
4.9 Successive Transformations
and
1n AA
16. Dating by Nuclear Methods
General AspectsCosmogenic RadionuclidesTerrestrial Mother/Daughter Nuclide PairsNatural Decay SeriesRatios of Stable IsotopesRadioactive DisequilibriaFission Tracks
16.1 General Aspects
the laws of radioactive decay are the basis of chronology by nuclear methods
two kinds of dating by nuclear methods can be distinguished:
1) Measuring radioactive decay of cosmogenic radionuclides, such as 3H or 14C
2) Measuring the daughter nuclides formed by decay of primordial mother nuclides (e.g. K/Ar, Rb/Sr, U/Pb ….)
16.1 General Aspects
Rutherford was first to see the potential of determining the age of uranium minerals from the amount of helium formed by radioactive decay
this potential was realized soon after the elucidation of the natural decay series of uranium and thorium
Ernest RutherfordNobel Prize in Chemistry 1908
16.1 General Aspects
time scale of applicability for naturally occurring radionuclides depends on the half-life (t1/2)
age to be determined and t1/2 should be on roughly the same order:
0.1* t1/2 < age < 10* t1/2
16.1 General Aspects
dating on the basis of radioactive equilibrium is possible after about 10 half-lives of the longest-lived daughter nuclides
the longest lived nuclides are:
(4n+2) → 234U (t1/2 = 2.44 x 105 years)
(4n) → 228Ra (t1/2 = 5.75 years)
(4n+3) → 231Pa (t1/2 = 3.28 a 104 years)
16.1 General Aspects
stable decay products, such as 4He, 206Pb, 207Pb, 208Pb, 40Ar, and 87Sr, increase continuously with time.
if one stable atom is formed per radioactive decay of the mother nuclide, the number of stable radiogenic atoms is:
(1.1)
16.1 General Aspects
N10 is the number of atoms of the mother
nuclide at t=0.
for dating, N2 and N1 have to be determined
)1e(NN
)e1(NNNNt
12
t011
102
(16.1)
16.1 General Aspects
if several stable atoms are formed per radioactive decay of the mother nuclide, as in the case of 4He formed by radioactive decay of 238U, Th, 235U and their daughter nuclides, the number of stable radiogenic atoms is:
)1e(nNN
)e1(nN)NN(n)He(Nt
12
t011
012
(16.2)
where n is the number of 4He atoms produced in the decay series.
16.1 General Aspects
the following methods of dating by nuclear methods can be distinguished
1. measurement of cosmogenic radionuclides
2. measurement of terrestrial mother/daughter nuclide pairs
3. measurement of members of the natural decay series
16.1 General Aspects
4. measurement of isotope ratios of stable radiogenic isotopes
5. measurement of radioactive disequilibria
6. measurement of fission tracks
16.1 General Aspects
there are some problems with the methods outlined here, and these will be discussed separately in detail
one major problem with most methods is whether the system is open or closed. If it is open, then the nuclides of interest could be lost or enter the system during the time period of interest
16.2 Cosmogenic Radionuclides
cosmogenic radionuclides are produced by the interaction of cosmic rays with the components of the atmosphere, mainly in the stratosphere.
if the intensity of cosmic rays (protons and neutrons) can be assumed to be constant, then the production rate of the radionuclides is constant.
16.2 Cosmogenic Radionuclides
16.2 Cosmogenic Radionuclides
as these radionuclides take part in various natural cycles on the surface of the earth, they are incorporated in various organic and inorganic products, such as plants, sediments and glacial ice
if no exchange takes place, the activity of the radiounculides is a measure of the age.
16.2 Cosmogenic Radionuclides
tritium (T) atoms formed in the stratosphere are transformed into HTO and enter the water cycle as well as the various water reservoirs, such as surface waters, groundwaters and polar ice
large quantities of T have been released into the atmosphere due to nuclear weapons testing, causing an increase in the T:H ratio by about 1000 times
T dating is thus restricted appreciably for all but glacier and polar ice samples, in which the influence of nuclear explosions is negligible
16.2 Cosmogenic Radionuclides
Libby proved the formation of 14C by the interaction of cosmic rays with the nitrogen in the atmosphere in 1947
14C atoms are quickly oxidized in the atmosphere to CO2, which is incorporated by the process of assimilation into plants and via the food chain into animals and humans
16.2 Cosmogenic Radionuclides
death of living things signifies the end of 14C uptake.
14C activity decreases with the half-life, provided no exchange of carbon atoms with the environment takes place.
half-life of 14C is very favorable for dating of archaeological samples in the range of about 250 - 40,000 years.
16.2 Cosmogenic Radionuclides
14C dating basic assumptions
1. 14C: 12C ratio in living things is identical with that in the atmosphere
2. 14C: 12C ratio has been constant in the atmosphere during the period of time considered.
16.2 Cosmogenic Radionuclides
3. Periodic variation of the 14C : 12C ratio (~9 x 10 3y at an amplitude of ~±5%) is correlated with the variation of the magnetic field of the earth causing changes in the intensity and composition of the cosmic radiation and consequently in the production rate of 14C
16.2 Cosmogenic Radionuclides
humans have caused drastic changes in the 14C: 12C ratio since the beginning of the industrial age. fossil Fuel combustion has diluted the 14CO2
by releasing 14C-free CO2
nuclear explosions liberated neutrons in the upper atmosphere that sharply increased 14C production
these changes should not influence dating of samples more than 100 years old.
16.2 Cosmogenic Radionuclides
ratio of carbon isotopes 14C: 13C: 12C in samples of recent origin is about
1:0.9 X 1010:0.8 x 1012.
ratio cannot be measured by classical mass spectrometry because ions of the same mass are found at practically the same position.
16.2 Cosmogenic Radionuclides
accelerator mass spectrometry (AMS) has been successful at identifying some nuclides.
26Al, 32Si, 36Cl, 41Ca, and 129I have all been identified
typically dating by these nuclides is not favored for several reasons such as: low concentrations low production rates technical challenges associated with
detection
16.2 Cosmogenic Radionuclides
Radiocarbon dating, the use of long-lived radioisotopes in climate research, and new developments in accelerator mass spectrometry are the main research activities of the laboratory. Ion beams are also applied to materials analysis and modification.
16.3 Terrestrial Nuclide Pairs
where, N2 is the total number of atoms of the stable nuclide (2), N2
0 is the number of atoms of this nuclide present at t=0, and N1 (eλt-1) is the number of radiogenic atoms formed by decay of the mother nuclide
)1e(NNN t1
022 (16.3)
dating by this method requires evaluation
of the following equation:
16.3 Terrestrial Nuclide Pairs
16.3 Terrestrial Nuclide Pairs
there are two methods for sample analysis:Independent determination of N2 and N1 Simultaneous determination of N2 and N1
by mass spectrometry
properties of mother and daughter must be similar for simultaneous determination
both methods require additional determination of N2
0 , but it can be neglected in some special cases.
16.3 Terrestrial Nuclide Pairs
in the 40K/40Ar method, the mass spectrometry is complicated because of the necessary 40Ar isotope dilution
the time required for this process may introduce additional 40Ar from atmosphere, and lead to a false dating
16.3 Terrestrial Nuclide Pairs
simultaneous determination of N2 and N1 is performed by measuring the ratios with a stable non-radiogenic nuclide as reference nuclide (Nr) by using the following equation:
)1e(N
N
N
N
N
N t
r
1
r
02
r
2 (16.4)
16.3 Terrestrial Nuclide Pairs
rearranging this equations leads to the following equation for the age of the sample
r1
r02r22/1
N/N
N/NN/N1ln
2ln
tt
where t1/2 is the half-life of the radioactive mother nuclide
(16.5)
16.3 Terrestrial Nuclide Pairs
simultaneous determination of mother and daughter nuclide by MS is applied in the 87Rb/87Sr and 147Sm/134Nd methods. These methods have had applications in geochronology in the dating of minerals, magmatic rocks, and sedimentary rocks of various origins
applications of the 176Lu/176Hf and 187Re/187Os methods have no advantages over the two previous methodsmajor drawbacks are low concentrations of Lu
(<1mg/kg) and Re (~1ng/kg) found in the minerals
16.4 Natural Decay Series
16.4 Natural Decay Series
taking into account the long-lived radionuclides, radioactive equilibrium is established after about 106 y in the case of the uranium and actinium series and after about 10 y in the case of the thorium series
variations in the ratio 207Pb:206Pb indicate geological processes
since 204Pb is not radiogenic, it is commonly used as a reference nuclide
16.4 Natural Decay Series
three kinds of systems can be distinguished:
1. losing parts of the members of the decay chains or the radiogenic Pb by diffusion or recrystallization processes (i.e. open systems)
16.4 Natural Decay Series
applications of this technique are summarized in the following table
16.4 Natural Decay Series
2. the loss of members of decay chains can be neglected and in which the concentration of the mother nuclide can be taken as a measure of age (equation (16.4) applies)
)1e(N
N
N
N
N
N t
r
1
r
02
r
2
16.4 Natural Decay Series
applicable forms of equation (16.4) for case number 2.
)1e(Pb
Th
Pb
Pb
Pb
Pb
)1e(Pb
U
Pb
Pb
Pb
Pb
)1e(Pb
U
Pb
Pb
Pb
Pb
t)232(204
232
0204
208
204
208
t)235(204
235
0204
207
204
207
t)238(204
238
0204
206
204
206
(16.6)
(16.7)
(16.8)
16.4 Natural Decay Series
3. the loss of members of decay chains can be neglected, but in which the concentration of the mother nuclide cannot be taken as a measure of the age
16.4 Natural Decay Series
a practical application of equations (16.6) through (16.8) is the calculation of the age of the solar system
mass spectrometry analysis of meteorites gives isotope ratios of the Pb isotopes 206Pb:204Pb=9.4 and 207Pb:204Pb=10.3
assuming these values are the initial isotope ratios at the time of formation of the solar system, the age is found by application of equation (16.5):
Pb/U
Pb/PbPb/Pb1ln
)238(
1t
2042380
204206204206
(16.9)
16.4 Natural Decay Series
dating with 210Pb is of interest for the dating of glacier and polar ice, and climatology.
the source of 210Pb is 222Rn emitted into the airsome 222Rn is emitted from volcanos.annual amounts of 210Pb brought down with
precipitations is relatively constantthe easiest method of detection of 210Po is by α
spectrometry (detection limit ~ 10-4 Bq) after attainment of radioactive equilibrium and chemical separation
16.4 Natural Decay Series
early stages of dating by nuclear methods were by measurement of 4He formed by α decay in the natural decay series
it was difficult to ensure the prerequisites of dating by U/ 4He method, because neither 4He nor α -emitting members of the decay series can be lost or produced by any other means beside alpha decay of U
16.5 Ratios of Stable Isotopes
there are four stable isotopes of lead: 204Pb, 206Pb, 207Pb, and 208Pb.
primordial Pb is what was formed in the course of the genesis of the elements. Radiogenic Pb is the additional amounts formed by decay of 235U, 238U, and/or 232Th.
16.5 Ratios of Stable Isotopes
mineral dating is possible by taking 204Pb as a reference nuclide, and comparing the ratios of each other stable nuclide to it by mass spectrometry
if the contents of U or Th are known and losses can be neglected, eqs. (16.6, 16.7, and 16.8) can be applied.
16.5 Ratios of Stable Isotopes
measurement of the Pb/Pb ratio offers the possibility of dating without knowledge of the contents of U and Th.
basis for the Pb/Pb method is given by equations (16.6), (16.7), and (16.8) knowledge of the ratio 235U:238U as a function
of time fact that the ratio Th:U is practically constant
for minerals of the same genesis.
16.5 Ratios of Stable Isotopes
the 39Ar/40Ar method is a variant of the 40K/40Ar method.
neutron activation analysis is applied to determine the amount of K present in the sample
sample and a standard of known age are irradiated under the same conditions for about 1 day
16.5 Ratios of Stable Isotopes
Ar is produced and measured by mass spectrometry
age of the sample is calculated by the relation
1e
1e
)40(N
)39(N
)40(N
)39(Nx
s
t
t
sx
(16.10)
16.6 Radioactive Disequilibria
useful for providing information about separation processes in minerals and ores, and sediments in oceans or lakes
by measuring the decay of the separated daughter nuclide or the growth of the daughter in the phase containing the mother, the time of separation can be determined
16.6 Radioactive Disequilibria
prerequisite is that the mother and daughter nuclide exhibit different chemical behavior under the given conditions
may be caused by different solubility of mother and daughter nuclide, by different probabilities of escape or by different leaching rates due to recoil effects
examples are U/Th and U/Pa
16.6 Radioactive Disequilibria
Example with 234U/230 Th UO2
2+ ions are found in natural waters, in the form of [Uo2(CO3)3]4- ions
Th ions are completely hydrolyzed and easily sorbed on particulates, and thus settle in sediments
corals and other inhabitants form skeletons by uptake of elements dissolved in the sea
16.6 Radioactive Disequilibria
applications geochemistry for dating of crystallization
processes by measuring the ratio 238U:230Th
excess of 230Th or 231Pa found in marine sediments allows dating of these sediments and determination of the sedimentation rate
archaeology, the 234U:230Th method is applied for dating of carbonates used by humans or for dating of bones or teeth.
16.7 Fission Tracks
in this way, 234Th (daughter of 238U) and the long-lived 230Th are separated
and if the skeletons can be considered to be closed systems, the ingrowth of 230Th is a measure of the age.
16.7 Fission Tracks
fission tracks are observed in solids due to spontaneous or neutron-induced fission of heavy nuclei and can be made visible under an optical microscope.
238U is the only spontaneous fission isotope that gives dense enough tracks for dating.
16.7 Fission Tracks
the method is the same as that used with track detectors such as photographic emulsion and autoradiography, dielectric track detectors, cloud chambers, bubble chambers, and spark chambers
XO particle passing through a bubble chamber
16.7 Fission Tracks
track density (number of fission tracks/cm2) in a mineral is a function of U concentration & the age of the mineral
for the purpose of dating, a sufficient number of tracks must be counted, so the concentration of U or the age should be relatively high
238U spontaneous fission track density is first measured, and then the sample is irradiated so that the neutron-induced fission of 235U is obtained
16.7 Fission Tracks
the age t of the mineral is calculated by the following formula:
inf,n t)238,sf(
)238(
)f,n(D
)sf(D3
e252.71
ln)238(
1t
(16.11)
16.7 Fission Tracks
where λ (238) is the decay constant of 238U, 7.252 x 10-3 is the isotope ratio 235U:238U,
D(sf) and D(n,f) are the fission track densities due to spontaneous fission of 238U and due to neutron-induced fission of 235U, respectively,
σ(n,f) is the cross section of fission of 235U by thermal neutrons, and ti is the irradiation time.
16.7 Fission Tracks
for homogeneous distribution of U in the sample, the values of D(sf) and D(n,f) can be determined in different aliquots of the sample
for heterogeneous distribution of U, D(sf) and the sum D(sf) + D(n,f) must be counted in the same sample
fission tracks are also influenced by recrystallization processes in solids, and is therefore useful in determining the temperature/pressure that the mineral was exposed to over time
Consider the decay series A B C D, where the half-lives of A, B, and C are 3.45 h, 10.0 min, and 2.56 h, respectively. We first prepare some pure radionuclide A and exactly 2.75 h after this preparation we measure the activity of daughter C. What would the activity of daughter C be (in Bq) after the 2.75 h decay of pure A, if we started with 7.35 x 107 Bq’s of pure A (nuclidic mass of A = 158.9 )?
Problem
dps x107.35Bq 10 7.35 A ofActivity
.271 4.08 .201 )(h
h 2.56 min 10.0 h 3.45 t
C B A
77
1-
1/2
Bq 108.2hr
d 1000.1)atoms 1070.3)(h 271(.NA
atoms 1070.3
)1093.1()1083.9()10(2.30
]e)1006.4[(]e)1033.7[(]e)1099.3[(N
10-4.06
N 08.3)271.08.4)(08.401.2(
)08.4)(201(.N
))((C
107.33
N 0555.0)08.4271)(.08.4201(.
)08.4)(201(.N
))((C
1099.3
N 02.3)201.271)(.201.08.4(
)08.4)( 201(.N
))((C
eCeCeCN
atoms1032.1s1058.5
Bq1035.7AN
711
111c
11
12512
)75.2)(271(.12)75.2)(08.4(10)75.2)(201(.12c
12
oA
oA
CBCA
BAC
10
oA
oA
BCBA
BAB
12
oA
oA
ACAB
BAA
tC
tB
tAc
1215
7oA
CBA
A sample of 0.250 g of a pure radionuclide with a mass number of 244 was observed to have an absolute activity of 4.45 microcuries (µCi). Calculate the half-life of this radionuclide and with the aid of a chart of the nuclides tentatively identify this radionuclide.
Problem
value. this to close life-half a has Pu
y 1027.8s 1060.2s 1067.2
693.693.t
s 1067.2atoms 1017.6
dps 1065.1
N
A
dps 1065.1Ci 1
dps 107.3Ci 1045.4A
atoms 1017.6g 244
tomsa 1002.6g 250.0N
.693t NA
244
7151162/1
11620
5
510
6
2023
1/2
Calculate the activity (in mCi) of a medical 60Co source containing 1.00 mg of the isotope.
mCi 1130Ci 1
mCi 10
dps 107.3
Ci 1dps 1018.4
dps 1018.4)atoms 1000.1)(s1017.4(A
s1017.4y 131.y 2714.5
693.0
atoms 1000.1g 0.60
atoms1002.6g 1000.1N
g 101.00 mass y 2714.5t t
.693 NA
3
1010
101919
191
1923
3
3-1/2
1/2
Problem
Calculate the activity, in dps and Ci, expected for a 1.00 mg 252Cf source that is 10.0 years old. The half-life of 252Cf is 2.64 y.
Ci 1089.3dps 107.3
Ci 1dps 1044.1
e dps 1099.1A
eAA
dps 1099.1)atoms 1039.2)(s 1034.8(A
atoms 1039.2g 252
atoms 1002.6g 1000.1N
s 1034.8y 263.0y 64.2
693. NA
210
9
)s 1015.3)(s 1034.8(10
to
101819o
1823
3
191
819
Problem
Problem
(a) Calculate the mass, in grams, of the 241Am present in the smoke detector which has 1 µCi
g 1092.2atoms 1002.6
g 241atoms 1028.7
atoms 1028.7s 1008.5
dps 107.3AN
dps 107.3Ci 1
dps 107.3
Ci1
10Ci1A
s 1008.5y 00160.y 32.74
693.
t
693.
NA
723
14
14111
4
4106
1111
2/1
(b) How long will it take to reduce the activity of 241Am from 1.0 to 0.50 µCi?
(b) From 1.0 µCi to 0.5 µCi is a reduction of one-half in the activity, so 1 half-life is required. For 241Am this is 432.7 y.
137Cs decays via β- emission to 137mBa. An experiment is begun with 5.00 x 106 Bq of pure 137Cs. Calculate the activity due to 137mBa after a decay period of 50 min.
? Bq 105.00 A
s 104.53 s 107.28
m 2.552 y 30.1 t
Ba Cs
6
1-3-1-10-
1/2
m137137
Problem
dps 1071.3)atoms 1021.8)(s 1053.4(NA
atoms 1021.8)257.99999)(.1087.6)(1061.1(N
)e()e)(atoms 1087.6()107.281053.4(
s 107.28N
Cs atoms 1087.6s 107.28
Bq 105AN
.zeroequals term last the ,start the at present is Ba no Since
eN)ee(N)(
N
6813
8157Ba
)s 300)(s 1053.4()s 300)(s 107.28(15-103
-1-10
Ba
15-1-10
6oCs
toBa
ttoCs
CsBa
CsBa
131-10-
BaBaCs
A sample contains a mixture of 239Pu and 240Pu in unknown proportions. The activity of the mixed sample was found to be 4.35 x 107 dpm for a sample of 0.125 mg of Pu. Calculate the weight % of each Pu isotope present.
Problem
-1102
240
-1111
239
240239
4-total
2402397total
m 1001.2 y 6,563 Pulife -half
m 1047.5 y 24,119 Pulife -half
Puto refer "2" and Puto refer "1"let
g 101.25mg 125.0m
Pu Pudpm 1035.4A
240Pu 57.4%42.6-100 239Pu %6.42100g 1025.1
g 1033.5Pu%
239Pu g 1033.5x
]x 1004.51030.6[]x 10[1.38dpm 1035.4
240
)1002.6)(x1025.1)(m 1001.2(
239
)1002.6)(x)(m 10(5.47dpm 1035.4
mol/atoms 1002.6mol/g 240
g )x1025.1(N mol/atoms 1002.6
mol/g 239
g xN
g )x1025.1(mass g xmass let g 1025.1massmass
NNdpm 1035.4A
4
5239
5
117117
234110231-117
234
223
1
421
421
22117
total
In the decay chain A B Cstable the half-lives of A and B are 17.75 and 43.9 min, respectively. If we start with pure A, how long a decay period would be required for the activity of B to become equal to the activity of A?
Problem
m 7.38)039.0016.0(
039.0
016.0ln
t
)(
ln
t
ee :t at
0dt/dN ,time that At
).t( aximumm a reaaches B of activity the that
time same the at A of activity the equal will B of Activity
Cm 016.0
m 9.43t B
m 039.0
m 75.17t A
m
AB
A
B
m
tB
tAm
B
m
1B
2/11
A
2/1
mBmA
218Po decays with a half-life of 3.10 min to 214Pb, which in turn decays with a half-life of 26.8 min to 214Bi. Assuming we have a source of pure 218Po at the start of our experiment, what decay time will be required for the activity to 214Pb to reach its maximum value?
Problem
m 9.10)224.00259.0(
224.0
0259.0ln
)(
ln
t
0dt/dN
,occurs Pbwhen activity maximum the when
AB
A
B
m
B
214
A piece of wood from the ruins of an ancient dwelling was found to have a 14C activity of 13 disintegrations per minute of carbon content. The 14C activity of living wood is 16 disintegrations per minute per gram. How long ago did the tree die from which the wood sample came?
Problem
y10x7.113
16ln
693.0
y5760
R
Rln
1 t
R
Rlnt
R
R e eAA
30
00tt0