All?gre Claude J. Isotope Geology

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Inth is case, wethereforehave:

¼ l

...

¼ l

n1

;

therefore:

¼ l

It is asi fthere wasjust asingle direct decayreaction(1) !(n).

Consid ering the fact that in the course of g eological time natural radioactive chains

rapidly reach equilibrium, the initial products give the end products directly. Dating equa-

tions canthenbewritten:

206

Pb ¼

238

Uðe

 1Þþ

206

207

Pb ¼

235

Uðe

 1Þþ

207

208

Pb ¼

232

Thðe

 1Þþ

208

where constantsl

, l

,andl

arethoseof

238

235

U,and

232

Th.

124 126 128 130 132 134 136 138 140 142 144 146

206

208

3.1 m

207

210

138 d

207

4.8 m

211

2.2 m

211

36 m

215

2x10

–3

219

3.9 s

223

11 d

227

17 d

227

22 yr

224

3.6 d

220

55 s

216

0.16 s

212

3x10

–7

212

61 m

212

11 h

231

26 h

210

50 d

214

20 m

214

27 m

218

3.1 m

222

38 d

214

2x10

-4

210

22 yr

226

1622 yr

235

713 Ma

232

14 Ba

228

6.7 yr

230

75 ka

234

1.2 m

234

24 d

238

4.5 Ba

228

1.9 yr

231

34 ka

234

248 ka

228

6.1 h

Figure 2.2 Radioactive chains represented on the proton–neutron graph. Each slot contains the symbol

of the isotope and its period

. Identify the various types of radioactivity of the three chains for yourself.

37 Radioactive chains

238

U !

206

Pb l

238

¼ l

¼ 1:551 25  10

10

1

235

U !

207

Pb l

235

¼ l

¼ 9:8485  10

10

1

232

Th !

208

Pb l

232

¼ l

¼ 4:9475  10

11

1

This comes downto d i re ct parent^ daug hter dati ng.Warning! Linearapproximation can-

not be used with these chronometers.To be convinced of this, let us compare the values of

1) with (lt)for timeintervalsof 1 and 2 billion years for

235

Uand

238

Letus consider the quantity ½ðe

 1Þlt=½e

 1



100, which is the relative e rror

thatcanbe expressed asapercentage.

With

235

U, for1billion the error is 41.2% and for 2 billion 68.2%! With

238

U, for1billion

the error is 7.5% and for 2 billion14.7%.These are highly non-linear systems, th en, as can

beseen from theshape ofthetypical curvesin Figure2.3.

2.3.3 The special case of lead–lead methods

We have seen that on the geological timescale radioactive chains attain equilibriu m and it

can be considered that

238

U decays directly to

206

Pb,

235

Uto

207

Pb, and

232

Th to

208

Pb.

For uranium-rich minerals li ke zi rcon, uraninite, or even monaz ite, the initial amount of

lead can be considered negligible. Assuming the system has remained closed, we can

write:

206

Pb ¼

238

Uðe

 1Þ

207

Pb ¼

235

Uðe

 1Þ:

Taking the ratio, remembering thatnowadays

the ratio

238

235

U ¼137.8, we get:

207

206

137:8

 1



It can be seen that the

207

Pb/

206

Pb isotope ratio of lead alone gives a direct measurement

of time. This function is impli cit and calculating it requires prior numerical values

(Table 2.1 ).

Exercise

The

206

Pb/

207

Pb ratio of a uranium ore deposit is found to be 13.50. What is the age of the ore

supposing it has remained closed since it crystallized and that common lead can be ignored?

Answer

We shall invert the ratio so Table 2.1 can be used.

207

Pb/

206

Pb ¼0.074. The table shows that

the ratio measured lies between 1 and 1.2  10

years. The result can be improved either by

reﬁning the table by calculating the ratio e

 1=e

 1



in the interval 1 and 1.2, or by

The two uranium isotopes decay each at their own rate from the time that they are ﬁrst formed. Their

ratio is constant as there is no isotopic fractionation between them. However, this ratio has varied over

geological time.

38 The principles of radioactive dating

using a linear approximation between 1 and 1.2  10

years. The value

207

Pb/

206

Pb for 1.2

billion is 0.080 12 and for 1 billion is 0.0725. The variation is therefore:

0:080 12 0:0725

200

¼ 3:8  10

5

per million years:

The difference is between 0.0740 and 0.0725, that is, 1.5  10

3

million years.

0.4

0.2

207

Pb*/

206

Pb*

0.6

0.8

234

Time (Ga)

1.0

0.5

0.0

01 2 3

Age (Ga)

P/P

235

U-

207

K-

238

U-

206

= 0.70

Age of the Earth

4.56 Ga

Rb-

T = 48.8

= 4.47

T = 1.25

235

0.5

2134

207

Pb*

0.5

1234

0.2

206

Pb*

0.4

0.6

1234

0.6

0.4

0.8

238

1234

Figure 2.3 Non-linear decay of

235

U and

238

U and their end products

207

Pb and

206

Pb. (a) The radioactive

constants of these reactions are very different. The curve on the right shows the change in radiogenic

ratio

207

Pb*/

206

Pb* versus time. (b) Comparison in the same ﬁgure of the principal radioactive clocks

used in geology, emphasizing the behavior of U–Pb systems compared to others.

39 Radioactive chains

The approximate age is therefore 1040 Ma. Direct calculation from the

207

Pb/

206

Pb ratio for

1040 Ma gives 0.073 98. The age is therefore 1042 Ma, but such precision is illusory because

error is far greater (see Chapter 5) than the precision displayed.

2.3.4 The helium method

Natural chains feature many instances of  decay, that is, expulsion of

He nuclei. Thus

238

Udecay ultimately produces eight

He ,

235

Ude cay produces seven

He, and

232

Th decay

p roduces six

He.We cantherefore write:

¼ 8l

238

U þ 7l

235

U þ 6l

232

Th:

Remark

This equation underlies the ﬁrst helium dating method thought up by Rutherford (1906).

Integrating the equation, assuming

He(0) ¼0, gives:

He ¼ 8

238

Uðe

 1Þþ7

235

Uðe

 1Þþ6

232

Thðe

 1Þ:

Table 2.1 Numerical values of the radiogenic

207

Pb/

206

Pb isotope ratio

Time (Ga) e

1e

1

207

206



radiogenic

0 0.0000 0.0000

0.2 0.03 15 0.2 177 0.0501 2

0.4 0.0640 0.4828 0.05471

0.6 0. 097 5 0.8056 0.05992

0.8 0.1321 1.1987 0.06581

1.0 0.1678 1.6774 0.07250

1.2 0.2046 2.260 3 0.08012

1.4 0.2426 2.9701 0.08879

1.6 0.2817 3.834 4 0.0 9872

1 .8 0.322 1 4.8869 0. 1 1 000

2. 0 0. 3638 6.1685 0.12298

2.2 0.4 0 67 7.7292 0.13783

2.4 0 . 4511 9.6296 0. 15482

2.6 0.4968 11.9437 0.17436

2.8 0 . 5440 14.76 17 0.1 9680

3.0 0.5926 18.1931 0.22266

3.2 0.6428 22.3716 0.25241

3.4 0.6946 27.4597 0.28672

3.6 0.74 80 33.6556 0.32634

3.8 0.8030 41.20 04 0.37212

4.0 0.8599 50.3878 0.42498

4.2 0.9185 61.57 52 0.48623

4.4 0.9789 75.1984 0.55714

4.6 1 . 04 13 91 .7873 0.63930

40 The principles of radioactive dating

Byadmittinglinearapproximations, which isvalidonly fordurations thatarenotexcessive,

1 lt andobservingthat

235

238

U ¼1/137.8 andthat

232

Th/

238

U 4 weobtain:

He ¼

238

U8l

137:8

þ6 4l



By noting L ¼ 8l

137:8

þ 24l



¼ 14:889  10

10

1

, we obtain

 

238



This formulaisvalidfor you ng ages.

Exercise

Take 1 kg of rock containing 2 ppm of uranium. How many cubic centimeters of

He will it

have produced in 1 billion years?

Answer

At 2 ppm 1 kg of rock represents 10

g 2  10

6

¼2  10

3

g of uranium, and 2  10

3

gof

uranium corresponds to

2  10

3

238

¼ 8:4  10

6

moles. Using the approximate formula:

He ¼ L

U ¼ 14:89  10

10

 10

 8:4  10

6

He  125  10

7

moles:

If 1 mole of

He corresponds to 22.4 liters at standard temperature and pressure, the amount

He produced in 1 billion years is 0.28 cm

A LITTLE HISTORY

The beginnings of radioactive dating

Rutherford performed the ﬁrst radioactive age determination in 1906. He calculated the

amount of helium produced by uranium and radium per year and per gram of uranium and

found 5.2  10

8

1

of uranium.

Ramsay and Soddy had measured the helium content (long confused with nitrogen

because it is inert like the noble gases) in a uranium ore known as fergusonite. The

fergusonite contained 7% uranium and 1.81 cm

of helium per gram. Rutherford calculated

an age of 500 million years. The following year he found uranium ores more than 1 billion

years old! At that time Lord Kelvin was claiming the Earth was less than 100 million years old!

(See previous chapter.)

2.3.5 The ﬁssion track method

Fission reactions emit heavy atoms. The atoms ejected by ¢ssion create £aws in crystals.

Such defects ^ knownastracks ^ canbeshownup on mineralsurfaces throughacid etchi ng

41 Radioactive chains

which preferentiallyattacks damaged areas, those where atoms havebeen displaced (Price

and Walker,1962) (Figure2.4).

Thenumberof¢ssiontracks iswritten:

fission



238

Uðe



 1Þ

fission

¼ 7  10

17

1

1=2

¼ 9:9  10

1

In practice, a thin section is cut and the number of surface tracks (and not in a volume) is

measured.Thevisibleproportion 

is calculated:



¼ q

fission



238

Uðe



 1Þ

where qis ageometric factorforswitching from a surface toavolume.

To measure

238

U, we take advantage of knowledge that

238

235

U ¼137.8 and measure

235

Ubycausinginduced ¢ssion (by placing the ore under study in a reactor), which pro-

duces tracks that are revealed and counted.We then have the number of tracks before (

)

and after (

) irradiation.Thegeometric factordisappears:



fission



137:8

G

ðe



 1Þ:

We calibratethe £uxofneutrons inducing

235

U¢ssionandthe e¡ectivecross-section.

By using a standard sample whose uranium content is kn own an d which is irradiated atthe

same time as the study sample and by counting the ¢ssion tracks p roduced, we can then

calculatet.

Ionization Electrostatic

displacement

Relaxation and

elastic deformation

Figure 2.4 Creation of ﬁssion tracks in an insulating material.

The effective cross-section is the probability of a reaction occurring, that is, here, the probability of

producing a ﬁssion with a given ﬂux of neutrons. The characteristics of this will be seen in Chapter 4.

42 The principles of radioactive dating

2.3.6 Isolation of a part of the chain and dating young

geological periods

The various radioactive isotopes of the radioactive chain belong to di¡erent chemical ele-

mentssothat,undercertaingeologicalconditions,oneortwoisotopesinthechainfraction-

ate chemically and become isolate d, thereby breaking the secular equilibrium. Once

isolated they create a new partial chain in turn. Two straightforward speci¢c cases are of

practicalimportance.

T he radioactive isotope becomes isolated on its own

First, when a radioactive isotope in the chain (but with a long enough period) becomes iso-

lated on its own, it gives rise to a partial chain, but being isolated fro m the parent it dec ays

accordingtothe law:

ðtÞ¼N

ð0Þe

lt

where N

(0) is the number of nuclides ofth e intermediate element at time t ¼0. Ifthis num-

ber can be estimated, the decay scheme can be used as a chronometer. This is equivalent,

then, to datingby theparentisotope.

EXAMPLE

The ionium method and the rate of sedimentation

Thorium is virtually insoluble in sea water. Thus

230

Th (still known as ionium from the

terminology of the pioneers of radioactivity), a decay product of

234

U with

¼ 75 ka,

precipitates on the sea ﬂoor, is incorporated in the sediment and so gradually buried. There,

now isolated, it decays.

At any depth

of sediment from the surface (Figure 2.5) we can write:

230

Thð

Þ¼

230

Thð0Þe

l

230

where

230

Th(0) is the surface content which is assumed constant over time. If the sedimenta-

tion rate is constant, time can be replaced by the ratio

where

is the sedimentation

rate and

the length (depth):

230

Thð

Þ¼

230

Thð0Þexp

½

ðÞ

or in logarithms:

lnð

230

ThðxÞÞ ¼ ln



230

Thð0Þ





The slope of the curve (ln

230

Th,

) gives a direct measure of sedimentation rate and the

ordinate at the origin gives

230

Th(0). (Note its order of magnitude of a millimeter per

thousand years.)

This method only works, of course, if it is assumed that

230

Th(0), that is the thorium content

at the sediment surface, is constant and if the sediment has not been disturbed by chemical,

physical, or biological phenomena.

43 Radioactive chains

Exercise

The lead isotope

210

Pb (as a member of the radioactive chain) is radioactive with a decay

constant

¼3.11  10

2

1

. This natural radioactive lead is incorporated into ice deposited

in Greenland by forming successive layers of ice which can be studied like sedimentary strata.

The activity of

210

Pb is measured at four levels in disintegrations per hour per kilogram of ice

(dph kg

1

). Table 2.2 shows the results.

Calculate the sedimentation rate of the ice. Assuming a constant rate and a compaction

factor of 5, how thick will the glacier be in 5000 years? Calculate the

210

Pb content of fresh ice.

Answer

The dating equation is written noting activity by square brackets:

210



210



l

If the rate of deposition is

and height

, we have

The equation becomes:

210



210



exp l

ðÞ

210



¼ ln

210





If the

210

Pb content has remained constant over time, the data points must be aligned in a (ln

[activity,

]) plot. The slope is therefore l=

. The data points are plotted on the graph and

the slope determined. This gives

¼45 cm yr

1

Depth (cm)

02040

230

(dpm g

–1

)

100

4.3

–1

Figure 2.5 Decreasing

230

Th content in a core from the sea ﬂoor and determination of the

sedimentation rate. Th

is the excess thorium compared with the equilibrium value counted in

disintegrations per minute per gram (dpm g

1

44 The principles of radioactive dating

In 5000 years’ time, allowing for compaction, there will be 5000  45=5 ¼ 450 m of ice.

The

210

Pb content is calculated: surface activity is 75 dph kg

1

ice.

210



210

¼ 75 dph kg

1

hence

210

3:1110

2

 8760ðÞ

1

3:5  10

7

¼ 21:4  10

atoms of

210

Pb nuclides

per kilogram of ice;

8760 being the number of hours in a year. As a mass that gives:

21:4  10

 210

6:023  10

¼ 7:5  10

14

kg of ice;

where Avogadro’s number is in the denominator and the previous atomic mass of

210

Pb in the

numerator.

The

210

Pb content is 7.5  10

17

, which is very little! This shows the incredible sensitivity of

radioactive methods because

210

Pb in Greenland’s glaciers really can be measured and used

for estimating the rate of sedimentation of ice.

Th e parent isotope is isolated an d enge nder s it s daughte r s

Thisiswhathappenswithuranium;forexample, when certain solidphasesl ike calciumcar-

bonate are precipitated uranium is entrained with calcium and then isolated. For the ¢rst

radioactiveproductofanynotablehalf-life, wethen have:

23

ðtÞ¼l

UðtÞ e

l

 e

l

23



where N

2^3

is the number of nuclides in the th ird intermediate product in the chain.Why?

Because the chain includes very short-lived radioactive products such as thorium-234

(

234

Th) or protactinium-234 (

234

Pa) which reach equilibrium very quickly. Thus,

238

decays to

234

Th, an element whose lifespan is 24 days, then

234

Pa, whose lifespan is 1.18

minutes: it canbe considered, then, that

238

Udirectlygives

234

Uwhose half-life is 2.48 10

yearsforal l typesofusualsamples (corals,speleothems, travertines, etc.).

Such a method is applied, for example, to se condary mineralizations of uranium. The

soluble uranium migrates and is deposited further away, leavi ng the insoluble thorium

whereitis. Itthen‘‘resumes’’its decaygiving

230

Th andwe canwrite:

230

Th ¼ l

230

238

l

238

 e

l

230



Thereforetheageofmigration canbe determinedbymeasur ing

230

Thand

238

Table 2.2 Activity of

210

Pb with depth

Depth

0m 1m20 cm 1.50 m 40 cm 2.50m 50 cm

dph k g

1

75 32 5245103

45 Radioactive chains

Exercise

Uranium-238 (

238

U) decays to

234

U which is itself radioactive with

¼2.794  10

6

1

Sea water is not in secular equilibrium relative to uranium isotopes as

234

U weathers better

than

238

U from rocks of the continental crust and is enriched in the rivers ﬂowing into the

ocean. In activity, noted in square brackets [ ],

234

238

U½

seawater

¼ 1:15. When limestone

forms from sea water, it is isotopically balanced with the sea water and so takes the value 1.15.

A fossil mollusk has been found in a Quaternary beach formation and its activity ratio measured

234

238

½

¼ 1:05. Work out the dating equation. What is the age of the mollusk?

Answer

The dating equation is:

234

238



234

238



l

þ 1  e

l

234



because l

234

l

238

and l

238

 0if

> 10

years.

This gives:

234

238

1

234

238



 1

and

¼393 000 years.

2.3.7 General equation and equilibration time

Weshall doan exercisetohelp underst andequilibrationtimeandbythe sametokengivethe

theoretical an swer to theprevious exercise.

Exercise

Establish the general equation for evolution of an isotope in a radioactive chain and where the

parent has a longer half-life and the daughter a markedly shorter half-life. We shall take the

example of

234

U !

230

Th decay to get our ideas straight.

Answer

This is a review exercise for mathematics on integrating a linear differential equation with

constant coefﬁcients:

234

¼ l

238

U  l

234

230

¼ l

234

U  l

230

with

238

U being considered constant.

Integrating the two previous equations in succession for the example in question gives:

234

U ¼ l

234

l

234

þ l

238

U1 e

l

234



230

Th ¼ l

230

l

230

þ l

234

l

234

 e

l

230



These two equations can be written with the activity notation in square brackets:

¼[

which simpliﬁes notation and means the

constants can be dispensed with. This gives:

46 The principles of radioactive dating