All?gre Claude J. Isotope Geology

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thetarget product which, by nuclear reaction, yields the cosmogenic isotope. Sometimes it is

single. This is so in iron meteorites composed of a metallic alloy of iron and nickel.

Sometimes there are more than one, as in ordinary meteorites where krypton isotopes are

producedbyspallationonrubidiu m, strontium, yttrium, and zirconiu m.

Theprimary£uxofcosmicraysiscomposedofprotons(andsomeHe

ions)( a ndofa llthe

isotopesinthe Universeintheionizedstatebutinverysmall abundances). Ithasbothintensity

(numberofparticlesperunitofsurfaceareaandperunittime)andanenergyspectrum,

because in factthere are N

, N

, ..., N

particles, corresponding to energylevels1, 2, ..., n.As

said,thisprimary£uxofprotonsproducesbarelyanythingother than reactionsatthe meteo-

rite’s surface, since as soonas itpenetratesbyafewcentimeters itgeneratesasecondary £uxof

neutrons, which penetrate more deeply.These neutrons also have an energy spectrum which

changes as they penetrate, dimin ishing, of course. The £ux and energy spectrum of cosmic

radiation may vary with time (weknow neither how, nor the magnitude ofsuch £uctuations).

Whatwemeasure istheresultof£uxaggregatedoverseveralmillion orevenbillionyears.

A further, although generally mi nor, complication is that in addition to cosmic

radi ation there is also a £ux of particles from the Sun. Generally, this £ux is weak and

of low energy, but from time to time it may become intense and of high energy. These

bursts in £ux areknown as solar £ares.They too mayengender spallation reactions.

The e¡ective cross-sections for production of new isotopes by nu clear reactions are also

dependent on energy and therefore on penetration inside the meteorite. Figure 4. 15 shows

the production ofvarious isotopesversus depth.

As,inaddition, so meisotopesresultfromspallationonseveraltargetnucleiwhose e¡ects

are cumulative, we can imagine thesheercomplexityofthe phenomenon ifwewishtodeter-

mine all the contributions.That would involve estim ating a mean £ux and its energy spec-

trum and local izing the sample to be measured inside the meteorite. For these reasons, a

simpler wayofmaking the calculation hasbeensought.

Ni,

(dpm kg

–1

)

(dpm kg

–1

)

Distance (cm)

300

100 150

200

100

–10

–20

–30

Figure 4.15 Production of

Cl,

Ni, and

Co by (n,



) reactions in a supposedly spherical chondritic

meteorite. Distance is measured from the meteorite’s surface towards its center. Cl ¼100 ppm,

Ni ¼1.34%, Co ¼700 ppm. 

Cl ¼45 barns, 

Ni ¼4.4 barns, 

Co ¼37 barns. Initial ﬂux

¼0.5

neutron cm

3

1

127 Exposure ages

Radioactive isotopes

In this case, the equation in the preceding subsection must be supplemented by the decay

term l

,whereN

isthe numberofradioactiveatoms:

¼ 

c!r

 l

Let us posit  

c!r

¼T. It is c onsidered to be constant to simplify things, or else it is

assumedweknow itsmeanvalue.The equationbec omes:

¼ T  l

Integrating gives:

T  C e

l

where C is the integration constant.Ifatt ¼0, N

¼0, C ¼T. Fromthis:

1  e

l





c!r

1  e

l



Now supp ose we have two isotopes, one stable and one radioactive, produced byspallation

u ndersimilarconditionsofenergyand £ux.We canestablishthe ratio:

ðtÞ



c!s

c!r

1  e

l



Ifwe waitlong enough for production and destruction ofth e radioactiveproductto achieve

asteadystate, then e



!0.Thisgives:

ðtÞ



c!s

c!r

Notice that we have eliminated the £ux factor. If, in addition, the two isotopes are products

ofthe same element, thenN

c!s

c!r

¼ 1 and the equationbecomes:

ðtÞ



t l

;

henc e :

t ¼



In this way, we determine the time elapsed since the meteorite was subjected to cosmic

radiation without knowing anything more about the cosmic radiation but having just the

ratioofthe e¡ective cross-sections.Theexposureageisthetime sincethe meteoritewasbro-

ken intopieces, leading toits exposureto cosmic rays(Figure 4. 16).

128 Cosmogenic isotopes

One of the conditions for successfully applying this method is that the isotopes used

should have e¡ective cross-sections that are s ensitive to the same ranges of energy and £ux

to justify the simpli¢cation of mathematically eliminating £ux. Accordingly, we are inter-

ested in isotopes either of the same element or of adjacent elements in the periodic table.

The pairs mostusedare

He^

He*,

Ne ^

Na*,

Ar^

Ar*,

Kr^

Kr*,and

(* indicatesthatthe isotope is radioactive).

The only point that remains undetermined if we are to be able to calculate the age

is the ratio of e¡ective cross-sections and their level of constancy depending on

particl e £ux and energy. The determination of e¡ective cross-sections has, of course,

b ene¢ted from the extraordinary research activity in nuclear energy and we have a

good catalogofe¡ective cross-sectionsfromwhich ithasbeenpossibleto deriveprecise laws.

Exercise

The

Kr–

Kr method is used to calculate exposure ages (Marti, 1982). The decay constant of

Kr is

¼0.32  10

5

1

. What is the minimum age we can calculate using the steady state

formula?

Answer

For this, e

lt

must be negligible compared with 1. If we accept that e

lt

must be less than

0.01, then

¼1.4 million years.

Exercise

Can we not use this method for ages of less than 1.4 Ma?

Answer

The evolution equation is



c!r



c!s



1  e

l



Number of nuclei

Radiogenic argon

concentration cm

–3

Exposure age (years) Exposure age (years)

Stable

isotopes

Radioactive

isotopes

Ar*(radioactive)

Ar (stable)

–13

2x10

–13

1000

2000

–1

2λ

Figure 4.16 Pairwise chronometry (stable isotope, radioactive isotope) for cosmogenic isotopes. (a) The

general theoretical case; (b) the example of

Ar/

Ar, which is used extensively in glaciology (see

Oeschger, 1982).

129 Exposure ages

We need to just calculate the curve ð1  e

l

Þ=

and use the value of the ratio

.We

know that

c!r

is identical to

c!s

for krypton. Calculate the curve for the

Kr/

Kr ratio,

given that 

/

 1.6612.

Exercise

The

Kr–

Kr method can be used for calculating exposure ages in stony meteorites by the

formula:





cosmogenic

where 1/l ¼0.307 Ma. To be rid of any complications and given that many Kr isotopes are

produced by spallation, the ratio of effective cross-sections is calculated from the formula:



0:95



Calculate the exposure age of the Juvinas meteorite if the isotopic measurement of Kr in it

is by convention:

Answer

¼10.710

years.

Exercise

We use the

K–

K method to date iron meteorites (Voshage and Hintenberger, 1960). The

only target is therefore iron and we take it that

¼0.5543  10

9

1

. As the effective cross-

sections are 

¼9.4 millibarns and 

¼14.7 millibarns, what is the cosmogenic isotopic

composition of

K for two meteorites whose exposure ages are 100 Ma and 1 Ga?

Answer

For 100 Ma (

cosmogenic

¼1.60; for 1 Ga (

cosmogenic

¼2.038.

Whatarethe main conclusions tobe drawn fromthese measurementsofexposure ages?

Exposureages aregenerallymuchyoungerthantheagesofmeteoriteformation(ormeta-

morphism).Theyrangefromafew millionyearsto afewhundred millionyears.(Ages offor-

mation are closer to 4.5 billion years, as we have seen.) This shows that for most of th eir

lifetime meteorites are inside their parent b ody, where they are shielded from cosmic rays,

and thatthe fracturingof meteoriteshas occur red relativelylate (but longbefore they fall to

Earth !).

1 0.157 0.460 0.0166 0.767 0.968 0.182

130 Cosmogenic isotopes

The distribution of meteorite exposure ages seems to display peaks at 5.7 and 20 million

yearsforordinarystonymeteoritesandat700 millionyearsfor iron meteorites(Figu re 4. 17).

The di¡erenc es in exposure ages for iron and chondritic meteorites is problematic.Why

should iron meteorites have so much older exposure ages? It was ¢rst thought that th ey

came from a di¡erent part of the Solar System. It is now thought that the exposure ages of

iron meteorites are older because iron withstands i mpacts better than the silicate assem-

blages that make up chondritic meteorites, which have been subjected to impacts resulting

in more recent fragmentation. In any event, the results show that meteorites have long life-

times in interplanetaryspace.

4.3.2 Terrestrial rocks

The principle for terrestrial rocks (see reviewby Oeschger,1982) is the same as for meteor-

ites.Galact ic co smic radiatio n causes nuclear reactions in rocks.Th e nuclei produced can

bemeasured, andifweknowthe £u xwe can calculatethe duration ofi rradiation.The di¡er-

ence with meteorites is that the intensity of cosmic radiation at the Earth’s surface is far

lowerbecauseithasbeenattenuatedby theabsorptionofprotonsand neutrons inthe atmo-

sphere. It was not until measurem ent sensitivity had been improved that these methods

could be applied to terrestrial rocks. A second di⁄culty is that until now methods using

isotope ratios employed for meteorites, such as

Kand

Kr/

Kr, have n ot been

applicable for terrestrial rocks. For the ¢rst m ethod, no equivalent has be en found for iron

meteorites, which do not contain large quantities of initial K, which dilutes the e¡ects

produced by irradiation. For the second metho d, sensitivity is still insu⁄cient, but that

may change. Unlike the case of meteor ites, where the particle £ux  is eliminated, here it

mustbe calibrated carefully. Now, the £u x diminishes, ofcourse, as itpenetrates into a rock.

Hence there is a further di⁄culty, which has been solved empirically, that of calibrating the

£ux, or calibratingabsorption.

Iron meteorites

Exposure age (years)

Chondrites

Figure 4.17 Exposure ages of various types of meteorites.

131 Exposure ages

Exploitable chronometers

In practice, we use the four radioactiveisotopes

C*,

Be*,

Al, and

Cl and the two rare

gas es

He and

Ne, which are stable isotopes. For radioactive elements, we use the formula

establishedfor meteorite exposureages:

  N

c!r

1  e

l



Buthere, equilibrium isnotachieved. BynotingT the production rate  N

c!r

, we obtain:

1  e

l



Forexample, forberyllium,

Be ¼

1  e

l



The di⁄cultylies in determining the £ux and assuming ittobe constantover time and then

estimatingthe abundanceoftargetnuclei.

For

He and

Ne, exposure ages are also calculated by the method described for stable

isotopes in meteorites.Thisgives:

¼   N

c!s

The production rate is written as P ¼ N

c!s

.The gases

Hean d

Ne accu mulatelinearly

withtim e,therefore

He ¼Ptor

Ne ¼Pt. Heretoowe needto know the £ux(and to assume

thatitis constant) andtheabsorption laws.

For thetarget,

Heproduction does notdepend much on its compositionwhereas

Ne is

p roduced byspallation of Mg and depends greatlyon the chemical composition of the tar-

get.To make it uniform, olivine, whose Mg content is more or less constant, is separated

and measurements are madeonthis mineral.

Calibration of prod uction rates, erosion rates, etc.

The particle £ux varies with both altitud e and latitude,

as we said when discussing

The £ux has therefore to be calibrated in ea ch place where the measurement is made.

There are twoways to do this. First, gene ral laws h avebeen es tablished giving the value of

£ ux by altitude and latitude. To g et some idea of this, let us say that at an altitude of

5000 m, the £ ux is 20 times greater than at se a level. At the poles, at a constant altitude, it

is 60 times greater than atthe equator.T his iswhy £ux is calibrated locally, where possible,

bymea suring

Be,

Al,

He,or

Ne contentsonsamples whose agehasbeen determined

byother methods.

Let us look at two important examples that will help in understanding the thinking

behind the method but whichwill show its‘‘fruitful complications’’and so sugg est its future

developments.

The ﬁrst measurements were made in the Antarctic where the ﬂux is greatest and, in addition, the erosion

rate is low.

132 Cosmogenic isotopes

Erosionratemeasurements

Our job is to measure the expo sure age of a basalt £ow from the Piton de la Fou rnaise

volcano on the island of Re

union (Staudacher and Alle

gre,1993). This lava £ow is at an

altitude of 2300 m and there is no indicatio n that its altitu de has varied over the course

of time.

We de cide to use the cos mogeni c age method based on

He applied to olivine. We

therefore separate the olivine from a rock sampled from the surface and measure the

He content of the sample. To obtain the

He of cosmogenic origin, we must, of course,

corre ct for

He of internal origin. To do this, we measure the

He/

He ratio of internal

origin on olivine sampl ed at a great distance from the surface. We can calculate the

cosmogenic

He content by measuring the (

He/

He)

total

ratio and the total concentra-

tion i n

He.

Exercise

Given (

He)

total

He/

He)

internal

, and (

He/

He)

total

, establish the formula for calculating

cosmogenic

He.

Answer



cosmo



total



total





internal



internal

When all the calculations havebeen made, we ¢nd1.3 10

12

1

at standard tempera-

ture and pressure of

He of cosmogenic origin.The production rate P

He in the olivine

hasbeen calculated for thelatitude and altitude ofthe sampleas 2.2 10

17

1

per year

atstandardtemperatureand pressure.

Table 4.1 Usable cosmogenic nuclides

Isotope Half-life (yr)

Measurement

method Di⁄culties

Production rate

(atoms/ yr) Sea

level

latitude >558N Time-span

He Stable Massspectrometer Di¡uses easily 160(olivine) 1ka ^ 3 Ma

Be 1.5 10

AMS Atmospheric

contamination

6 (quartz) 3 ka ^ 8 Ma

Al 7.16 10

AMS Atmospheric

contamination

37 (quartz) 5 ka ^ 2 Ma

Cl 3.08 10

AMS Atmospheric

contamination

8 (basalt) 5 ka ^ 1Ma

Ne Stable Massspectrometer Common neon 45(olivine) 7 ka ^ 10 Ma

C 5730 years AMS Contamination 20 (basalt) 1ka ^ 40 ka

Ca 103 10

AMS Verydi⁄ cultto

measure

^300ka

133 Exposure ages

It is easy thento calculatethe exposure age:

cosmo

He/P

¼59 090 years.

In addition, the ‘‘geological’’ age of the basalt £ow has been calculated by the K^Ar

method as 65 200 2000 years.The surface rock has been irradiated for 65 200 years and

yetits exposure ageisjust 6000 yearsattheyoungest. Howcome?

After examining the various source s of error, we accept that the age di ¡erence is due

to erosion.The rock sampled atthe surface today was in fact located at depth for much of

itshistory. Itonlycametothesurface through ablation ofthe m ate rial aboveit, bye rosion.

Now, we k now that the particle £ux decreases exponentially with the thickness of rock it

haspenetrated.Weneed, then, tomodelthe phenomenon.

We canwritethe rateofproduction of

Heintheform:

P ¼ P

exp

XðtÞ



where X(t)isthethickness counting fromthesurfaceand L isthe attenuation factorofradia-

tionwith depth.

Tosimplify matters, we can assum e thatthe rate oferos ion is c onstant.We canwrite:

X ¼ X

2t

where 2is the erosion rateand X

the initialdepthofthesa mple.

We canthereforewrite:



¼ P

exp

 X

2tð



We integrate this equation between 0 and t and then replace X

by its value X

¼X þ2t

(rememberingthat X is counted downwards).This gives:

He ¼

exp

X



1  exp

2t

hi

ThenX is the currentdepth coordinate.

Ourobjectiveis todeterminethe erosion rate, 2.

We k n ow t (65 200 years) and P

(2.2 10

17

1

at standard temperature and

p ressure), butthe re arestill twounknowns: 2and L.

To measure L, we bore a small core and so take samples at various depths.We isolate the

olivine andmeasurethe cosmogenic

Heon eachsample ofit.We can nowdraw theplot:

ln ð

HeÞ

cosmo

¼ fðXÞ:

Ifour assumptions aboutthe £ux an d erosion rate are valid, the relation is a straight line of

the form:

ln ð 3He Þ

cosmo



X

þ A

where A is aconstant.The slopeis (^1/L).

134 Cosmogenic isotopes

Exercise

A core was taken from the lava of the island of Re

union discussed before. Olivine was

extracted at each level and the

He content measured. After correcting for

He of internal

origin, the following results were found as a function of depth

(Table 4.2).

Calculate the attenuation factor

Notice that depth

in this table is expressed in grams per square centimeter, which is an

unusual unit of length! The reason for this is that attenuation is, of course, dependent on the

quantity of matter penetrated and so depends on the density of that matter (attenuation per

unit length penetrated is not the same in rock, soil, or a layer of atmosphere). If attenuation

were expressed per mil length, we would have to multiply it by density, and so the effective

value would be:

length ðcmÞ mass ðgÞ

volume cm

ðÞ

which is equivalent to mass per square centimeter. Naturally enough,

too will be expressed

in g cm

2

Answer

After calculating ln(

He) as a function of

(do it) we ﬁnd for the slope:

L ¼ 165 g cm

2

 5:

How can we now calculate the erosion rate 2?

Let us go back to the expression as a function of

but now start from where

¼0, that is, at

the present-day surface. The expression becomes:

He ¼

1  exp

2



We can develop the exponential in series limiting ourselves to the ﬁrst three terms.

Remembering that

He ¼

cosmo



, we can ﬁnally write:

2¼2







cosm



If we note the relative difference between the geological age and cosmogenic age as 

2¼2





Table 4.2 Relation of

He content to depth in olivine

from lava on Re

union

X (g cm

2

)

He (10

13

3

1

)

10.45 1 3

37 .86 9.12

131.95 6.9

217.42 3.58

3 1 8.4 1 1 . 74

135 Exposure ages

In the present case we ﬁnd 2¼4.7 g cm

2

1

If we accept a density of 2 g cm

3

because the lava sampled is ‘‘bubbly’’ and contains many

cavities, then 2¼2.1 mmyr

1

¼2.1 mm per 1000 yr.

Exercise

Cosmogenic

Ne contents were measured on the same samples of olivine from the island

of Re

union (Table 4.3). Given that the production rate of

Ne is 6.28  10

18

3

1

the latitude and altitude in question, calculate the attenuation factor

and the erosion rate 2.

How do you account for the result for

compared with that found for

He? Which seems to

you to be the more accurate rate of erosion, that determined with

He or that with

Ne? To

what do you attribute the difference?

Answer

¼ 160 gcm

2

It is identical to that for

He because the particle ﬂux at the origin of

He and

Ne is the

same for both. The cosmogenic age of the surface sample is

cosmo

¼52 388 years.

2¼ 9:6g cm

2

1

¼ 4:8 mm per 1000 yr:

The

Ne age is greater than the

He age probably because helium diffuses more readily

than neon and the sample has probably lost a lot of helium. This invites caution when

interpreting helium ages (Figure 4.18).

Table 4.3 Relation of

Ne content to depth in olivine

from lava on Re

union

X (g cm

2

)

Ne (10

13

3

1

)

10.45 3.29

37 .86 1.73

1 31.95 1 .59

21 7.42 0.96

3 18.4 1 0.5 2

–28

–27

–29

–30

–31

100 200 300 400

Depth (g cm

–2

)

In(

)

Figure 4.18 Attenuation curves for the creation of

He and

Ne in rocks of Re

union given by way of

example.

136 Cosmogenic isotopes