Applications of Stable & Radioactive Isotopes in Geochemistry: O, N, S, Study notes of Geochemistry

Download Applications of Stable & Radioactive Isotopes in Geochemistry: O, N, S and more Study notes Geochemistry in PDF only on Docsity! Chapter 16 Stable and Radioactive Isotopes Each atomic element (defined by its number of protons) comes in different flavors, depending on the number of neutrons. Most elements in the periodic table exist in more than one isotope. Some are stable and some are radioactive. Scientists have tallied more than 3600 isotopes, the majority are radioactive. The Isotopes Project at Lawrence Berkeley National Lab in California that gives detailed information about all the isotopes. Stable and radioactive isotopes are the most useful class of tracers available to geochemists. In almost all cases the distributions of these isotopes have been used to study oceanographic processes controlling the distributions of the elements. Radioactive isotopes are especially useful because they provide a way to put time into geochemical models. The chemical characteristic of an element is determined by the number of protons in its nucleus. Different elements can have different numbers of neutrons and thus atomic weights (the sum of protons plus neutrons). The atomic weight is equal to the sum of protons plus neutrons. The chart of the nuclides (protons versus neutrons) for elements 1 (Hydrogen) through 12 (Magnesium) is shown in Fig. 16-1. The Valley of Stability represents nuclides stable relative to decay. Examples: Atomic Protons Neutrons % Abundance Weight (Atomic Number) Carbon 12C 6P 6N 98.89 13C 6P 7N 1.11 14C 6P 8N 10-10 Oxygen 16O 8P 8N 99.76 17O 8P 9N 0.024 18O 8P 10N 0.20 Several light elements such as H, C, N, O, and S have more than one stable isotope form, which show variable abundances in natural samples. This variability is caused by isotopic fractionation during chemical reactions. Heavier elements like Pb also have several stable isotopic forms but their distributions are controlled more by their different sources than by fractionation. Docsity.com Fig 16-1 VALLEY oF STHaBICITY 8 g e 5 Zz g ogeny & oa Isotones trodes eck Open ees guest die beste ee ere men) 10 n fee, Neulron Number (N) FIGURE Partial chart of the nuclides. Each square represents a particular nuclide which is defined in terms of the number of protons (Z) and neutrons (N) that make up its nucleus. The shaded squares represent stable atoms, while the white squares are the unstable or radioactive nuclides. Isotopes are atoms having the same Z but different values of N. Isotones have the same N but different values of Z. _lsobars have the same A but different values of Z and N. Only isotopes are atoms of the same element and therefore have nearly identical chemical properties. izes 14 Docsity.com This process, in which the isotopic composition of the product varies as a result of the extent of reaction is called a Raleigh Distillation (see Appendix 16-A1 for the derivation). Its effect on the isotope composition of the remaining reactant is given by: Rf / Ro = f (α-1) Where f is the fraction of remaining reactant, Rf is the δ value of the reactant at some f and Ro is the δ value when f = 1 (i.e. the initial value)(Fig 16-3). In additional to fractional removal, temperature has a large effect. As temperature decreases the fractionation α increases (Fig 16-4). This effect has been taken advantage of to calculate past variations in temperature in Greenland and Antarctic ice core records. Docsity.com B-1. Ice Volume The Raleigh Distillation of water vapor also causes polar ice to be depleted in 18O relative to seawater. Thus an increase in ice volume causes the ocean to become enriched in 18O. During periods of maximum glaciation the δ18O of seawater increased to +0.90%o (SMOW). If all present-day continental glaciers melted the δ18O of seawater would decrease to -0.6%o (SMOW). The calibration relationship for glacial-interglacial sea level change is 0.011%o per meter change in sea level (Fairbanks and Matthews, 1978). The δ18O signature of sedimentary carbonates is perhaps the most widely used record of paleoclimate change. The basis for this application is that biogenic calcite forms in isotopic equilibrium with ambient seawater. The δ18O composition of planktonic foraminifera is a function of temperature as well. The value is usually reported on the PDB scale as the standard is also a fossil marine carbonate. The δ18O record for carefully dated coral reefs from Barbados have provided the first continuous and detailed record of sea level change during the last deglaciation (Fig. 16- 5)(Fairbanks, 1989). Sea level was 121+5 meters below present during the last glacial maximum. The deglacial sea level rise was not monotonic. It was marked by two intervals of rapid rise that occurred approximately 13,070 yr BP and 10,445 yr BP. Docsity.com 16-I.C. Carbon Carbon has only two stable isotopes with the following natural abundances: 12C 98.89% 13C 1.11% The values of δ13C for geologically important materials are shown in Fig 16-6 (Hoefs, 1980). The standard used for carbon isotopes is the same PeeDee Belemnite shell used for oxygen isotopes. Below are some typical δ13C values on the PDB scale in %o. Standard (CaCO3; PDB) 0 Atmospheric CO2 -8 Ocean ΣCO2 +2 (surface) 0 (deep) Plankton CaCO3 +0 Plankton organic carbon -20 Trees -26 Atmospheric CH4 -47 Coal and Oil -26 Photosynthesis produces organic matter which is "depleted" in 13C relative to the substrate CO2. Compare the values for trees (-26) versus atmospheric CO2 (-8) and marine plankton (-20) versus ΣCO2 (+2)(Fig 16-7). Representative δ13C values are shown for living organisms (Fig 16-8) and biochemical constituents (Fig 16-9) (from Degens, 1968). C-1. Primary Production (to be added) C-2. Marine versus Terrestrial (use examples by Sackett, Hedges etc) C-3. Methane Production (see Reeburgh, Martens, Quay for examples) Docsity.com 16-I.E. Sulfur Sulfur has four stable isotopes with the following natural abundances: 32S 95.02% 33S 0.75% 34S 4.21% 36S 0.02% Sulfur isotope compositions are expressed as per mil differences relative to sulfur in an iron meteorite called Canyon Diablo Troilite (CDT) according to: δ34S = [(34S/32S)sample - (34S/32S)CDT] / (34S/32S)CDT x 1000 Thode et al (1949) was one of the first to document wide variations in the abundances of sulfur isotopes. The range of δ34S in geological materials is shown in Fig 16-14 (from Hoefs, 1980). It is interesting to note that δ34S of seawater sulfate is +20%o. This implies that some process is preferentially removing 32S from the ocean. This process is probably sulfate reduction which produces 32S enriched sulfide which can precipitate be removed from the ocean as iron sulfide minerals (pyrite, gregite etc). The δ34S of sedimentary sulfides over geological time are highly variable but trends exist which led Canfield (1998) to suggest that there was a sulfide-rich deep ocean in the middle- to late- Proterozoic. E-1. Sulfate reduction Sulfate reducing bacteria produce sulfide depleted in 34S (enriched in 32S) by 4 to 46 per mil compared with the starting sulfate. The depletions are variable because this is a kinetic fractionation that occurs during bacterial sulfate reduction. Based on this fractionation we expect sulfide to be depleted in 34S by large but variable amounts. Field data show much larger fractionation suggesting that the real system even is more complicated. For example, the Black Sea is the classic marine anoxic basin where sulfate reduction is occurring. Enrichment cultures of sulfate-reducing bacteria produce sulfide depleted in 34S by 26 to 29 per mil, yet water column sulfide is depleted by 62 per mil (Fry et al, 1991). Canfield and Thamdrup (1994) showed that these large fractionations in sulfide could be produced by a scheme where elemental sulfur (S°) is disproportionated to H2S and SO42- by sulfur- disproportionating bacteria followed by oxidation of the sulfide to S° (Fig 16-16). E-2 Paleo variations in δ34S By comparison with sulfides, sulfur is removed to gypsum (CaSO4.2H2O) with little fractionation of sulfur isotopes. Thus the δ34S of gypsum in the geological record should reflect the isotopic composition of ambient seawater sulfate. Variations of δ34S in marine sulfate minerals from the Precambrain to the present are shown in Fig 16-17. The distribution of sulfur isotopes in seawater has clearly changed due to variations in the relative proportions of the oxidized and reduced reservoirs of sulfur (Holland, 1973). Variations in burial and weathering of pyrite are probably the key processes. The present Docsity.com value of δ34S-SO4 is about +20 %o but during the Permian the value was about +10 and at the Cambrian/Precambrian boundary it was greater than +30%o. Garrels and Lerman (1984) and Berner (1987) have modeled these variations (together with variations in δ13C) in terms of burial and weathering of 1) organic matter (as CH2O), 2) sedimentary carbonates (Ca,MgCO3), 3) sedimentary pyrite (FeS2) and 4) evaporitic gypsum (CaSO4.H2O). The global redox reaction can be written as (Garrels and Perry, 1974): 4FeS2 + 8 CaCO3 + 7 MgCO3 + 7 SiO2 + 31 H2O = 8 CaSO4.2H2O + 2 Fe2O3 + 15 CH2O + 7 MgSiO3 The coupled records of δ13C and δ34S (Fig 16-17) have been used to reconstruct the concentration of atmospheric oxygen over geological history. During the Paleozoic, the ocean became progressively enriched in 13C and depleted in 34S because of preferential removal of ocean carbon to organic matter rather than as CaCO3 and of ocean sulfur to CaSO4 rather than as FeS2. There is little doubt that atmospheric O2 has varied over Phanerozoic time and was higher during the mid-to-late Paleozoic due to increased organic carbon burial however accurate calculations of the atmospheric O2 concentrations remain difficult (Berner, 1987; 1989). Docsity.com 16-II. Radioisotopes 16-II.A. Fundamentals of Radioactive Decay Radioactive decay is a fundamental natural property of many elements in the periodic table. The isotopes of these elements enter the sea from the atmosphere, from rivers, by diffusion from the underlying sediments, and by radioactive decay of other elements within the ocean itself. Radioactive isotopes useful in marine chemistry can be of natural origin (e.g., 14C, 10Be, 238U, 232Th) or they can be the products of bombs (e.g., 14C, 90Sr, 240Pu) or nuclear reactors (e.g., 239Pu). Because of their decay, radioactive isotopes are instrumental in the study of many oceanographic and geochemical processes. Decay provides a means of incorporating time and rates into geochemical models. In many instances, radioactive isotopes can also serve as analogues for interpreting the chemistry of non-radioactive elements (e.g., 234Th, 210Pb for marine scavenging of reactive elements). In marine chemistry, radioactive isotopes have been used to study the rates of mixing of water and sediments, to calculate the age of water and sediments, and to estimate rates of scavenging and production. 1. The Process of Decay Radioactive decay is the result of the adjustment of the nuclei of atoms from unstable to more stable states. The original radioactive atom is called the parent. The product of the decay is called the daughter. Frequently, a series of successive decay transformation will constitute a decay chain. The use of isotopes to study chemical reactions is facilitated by the fact that the decay process is a function of the stability of the nucleus and is independent of the chemical state. As a result of radioactive decay energy is released as radiation in the form of alpha particles, positive and negative beta particles (positrons and electrons), gamma rays and x-rays. Alpha particles result primarily from the decay of isotopes of high atomic number. These particles consist of two protons and two neutrons and thus are identical to helium nuclei. They are commonly emitted from natural radioactive materials with an energy of three to eight million electron volts (mev). An electron volt is the kinetic energy acquired by an electron falling through a potential difference of 1 volt. It is equivalent to 1.602 x 10-12 erg. The resulting daughter product is an element two protons lighter in the periodic table (see Fig 16-1). Negations or electrons (e–) are emitted during decay of isotopes with excess neutrons relative to the stable state. This change involves the transition of one of the neutrons into an electron, proton, and neutrino. The proton stays in the nucleus, increasing the atomic number by one, and the electron and the neutrino are ejected from the nucleus. The ejected electron is referred to as a beta particle. If the nucleus has a neutron-proton ratio lower than the stable forms of the element, two processes compete to return the nucleus to stability. One pathway is for a proton in the nucleus to combine with or capture an electron from one of the extranuclear orbitals (usually from the K level). This process is commonly referred to as electron captive (EC). As a result of this process, a characteristic x-ray of the daughter isotopes is emitted when the vacant energy level produced by the electron capture is filled by an Docsity.com ln (2) = λ t1/2 (16.12) 0.693 = λ t1/2 (16.13) so t1/ 2 = 0.693 λ (16.14) 4. Mean Life The mean life is equivalent to the average life of a radioactive atom. Numerically it is the sum of the times of existence of all the atoms initially present divided by the initial number or: τ = 1 No tdN No 0 ∫ (16.15) = 1 No tλNoe −λtdt 0 ∞ ∫ (16.16) = λ te−λt dt 0 ∞ ∫ (16.17) = – λt +1 λ e−λt    0 ∞ (16.18) = 1 λ The mean life, then, corresponds to the time required for the activity to decrease by the factor of e-1. Thus the mean life is greater than the half life by 1/0.693. The difference arises because of the weight given in the averaging calculation to the fraction of atoms that by chance survive for a long time. Example 16-1 One gram of 238U contains 2.53 x 1021 nuclei, all of which were formed about 4.5 x 109 years ago with the rest of our planet. By coincidence the half life of 238U is also 4.5 x 109 years, which means we presently have on earth only about half of the 238U initially present. The mystery of radioactive decay is that for billions of years a 238U nucleus behaves like any other non-radioactive nucleus until suddenly it emits an alpha particle and transforms in 234Th. Docsity.com 5. Parent-Daughter Relationships Example 16-2 An important oceanographic example of secular equilibrium is the decay of 226Ra with a half life of 1620 years to 222Rn with a half life of 3.824 days. The growth of 222Rn from 226Ra is shown graphically in Figure 16.18. The initial growth of 222Rn is a function of the half life of 222Rn as can be seen from the general equation for in growth of the daughter (as derived from Equation 16.5): A B = AA,0( ) 1− e −0.693 t / t1/ 2,B( ) (16.19) If necessary, the half life of the daughter can be determined from the rate of in growth of the daughter. No information about the half life of the long-lived parent can be obtained from the growth of the short lived daughter. However, there are frequently situations where the activity of a parent is conveniently measured by measuring the equilibrium activity of its daughter. An important example is 228Ra which emits a weak β– and is more conveniently analyzed by measuring the alpha decay of its daughter 228Th. At secular equilibrium the activities of 228Ra and 228Th are equal. Using the equation given above, we can see that after five daughter half lives the activity of the daughter is within 3% of that of the parent. t / t1/ 2 A B / AA,0 1 .50 2 .75 3 .87 4 .937 5 .969 10 .999 Docsity.com Example 16-3 For decay chains involving several nuclides, all isotopes in the chain will be in secular equilibrium (AA = AB = AC = AD = etc.) if the parent has a long half life relative to all of the daughters. The total disintegration rate will be either λAA or λAA0 depending on whether the decay of the parent is negligible. The 238U decay chain is a prime example (Figure 16-19). The half life of the parent 238U is more than 104 greater than the second largest half life in the decay chain (234U: 2.5x105 yr). Within a few million years of the formation of the earth, the total activity from this decay chain was equal to 12 λ(238U). Now that about one half life of 238U has elapsed, the total radiation from this decay chain is about half what it was during the early days of the earth's history. Example 16-4: 228Ra-228Th 228Ra with a half life of 6.7 years decays to 228Th which has a half life of 1.9 years. 228Ac is an intermediate with a half life of 6.1 hours which we will ignore in this example. The in growth of 228Th and decay of both isotopes is shown in Figure 16-20 (not shown here). The equilibrium activity ratio of 228Th to 228Ra is 1.39. Docsity.com B/A - B°/A° = f α-1 - 1 16.28 B°/A° The expression for the per mil change in isotopic composition, δ , is given by (1000) x B/A - B°/A° 16.29 B°/A° Thus δ = 1000 (fα-1 - 1) 16.30 The classic example of the Raleigh distillation is the fractionation of oxygen isotopes between water vapor in clouds and rain released from the cloud (Dansgaard, 1964) (Fig 16-3). 18O is enriched by 10 per mil in the rain drops. Hence the 18O will be depleted at a rate 1.010 times faster than 16O ( α = k18O/k16O = 1.010). δ = - 1000 ( 1 - f0.10) 16.31 The resulting depletion of the 18O/16O ratio in the residual cloud water is given as a function of the fraction of the original vapor remaining in the cloud ( f ) (Fig A16-1). As f depends on cloud temperature, the relationship can also be given between isotopic composition and cloud temperature. You can see that the initial drops will be rich in 18O. As rain is removed the cloud is progressively depleted in 18O. The 18O in the rain drops decrease also but they are always enriched relative to the cloud vapor. So the fractionating of oxygen isotopes in rain will depend on cloud temperature but it could also be a function of the history of the air mass. In the studies of Greenland and Antarctic ice cores δ18O is usually used to indicate temperature. But the interpretation can be complicated when a site receives rain from air masses with different histories (e.g. Charles et al 199x). The δ18O in average annual rainfall is a function of mean annual air temperature as shown in Figure 16-4. Docsity.com Appendix 16-A2 - Parent-Daughter Relationships In many problems of interest in oceanography it is useful to compare the activity of one isotope with its parent or daughter. For example, mixing near the ocean floor is studied using the excess of 222Rn over its parent 226Ra. In situ scavenging is studied using the depletion of 210Pb relative to its parent 226Ra. For the simple case where a radioactive parent (A) decays to a stable daughter (B) (e.g., 210Po → 206Pb), the rate of formation of the daughter is equal to the rate of decay of the parent at steady state or: dNB dt = λA NA = λ ANA ,0 e −λ At (16.32) If some of the stable daughter is present initially, then in integrated form: N B = (N A,0 ) (1– e −λAt) (16.33) For the more complicated case where the daughter is also radioactive, the rate of formation of the daughter is equal to the rate of production from the parent minus the rate of decay of the daughter. dNB dt = λA NA − λ BNB (16.34) Integration of Equation 16.34 for the case where there are no daughter atoms present at t=0 leads to: N B = λ A(NA,0 ) λB − λ A e−λ At − e−λ Bt( ) (16.35) In terms of activity: Docsity.com A B = λB(AA,0 ) λB − λA e−λA t – e−λ Bt( ) (16.36) If some daughter atoms are present initially, we simply add the term AB,0 e-λBt to Equation 16.36. In terms of these equations we can visualize three limiting cases for the relationship between parent and daughter activity. These three conditions are: 1) t1/2(A) > t1/2(B) or λA < λB 2) t1/2(A) = t1/2(B) or λA = λB 3) t1/2(A) < t1/2(B) or λA > λB Regardless of whether the daughter has a longer or shorter half life, an expression can be derived for the time at which the daughter activity reaches a maximum by setting dAB/dt equal to zero and solving for t. Thus: t(AB,max) = 3.323(t1/ 2,A )(t1 /2,B) t1 /2,A − t1/ 2,B log t1 /2,A t1/ 2,B (16.37) Case I: Half life of parent greater than that of daughter. When the half life of the parent is greater than that of the daughter the term e−λ At with increasing time and becomes negligible in Equations 6.35 and 6.36. Thus: A B = λB(AA,0 ) λB − λA d−λ A t( )= λBλ B • λA AA (16.38) (1 ) 1B B A A B A B A A λ λ λ λ λ = = − − − (16.39) Docsity.com