4. Radioactive Decay radioactive transmutation and decay are synonymous expressions 4 main series 4n 232 Thorium 4n + 2 238 Uranium-Radium 4n + 3 235 Actinium

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


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


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


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


4.2c 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


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


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


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


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.


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


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


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


4.2d 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


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


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


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


4.4 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


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


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


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


specific activity of 90Y is

therefore mass of 90Y is

g 264.0

g106.2Ci/g 105.50

Ci 145.0 75


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


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


4.6 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


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


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


with the coefficients given as:

01

1n1312

1n211 N

)())((c

01

2n2321

1n212 N

)())((c

01

n1nn2n1

1n21n N

)())((c


))((

e

))((

e

))((

e

NN

3132

t

2321

t

1312

t

01213

3

21

for example if n = 3 we get


21013 NNNN

if nuclide 3 is stable then

t

21

1t

12

2013

21 ee1NN


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


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


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 ….)


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


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


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)


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)


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)


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.


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


4. measurement of isotope ratios of stable radiogenic isotopes

5. measurement of radioactive disequilibria

6. measurement of fission tracks


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.



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.


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


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


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.


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.


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


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.


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.


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


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:



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.


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


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)


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)


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


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


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)


applications of this technique are summarized in the following table


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


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)


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


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)


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


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.


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.


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.


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


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


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


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


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

Documents

4. Radioactive Decay radioactive transmutation and decay are synonymous expressions 4 main series 4n 232 Thorium 4n + 2 238 Uranium-Radium 4n + 3 235 Actinium