The u series disequilibrium method of dating

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The U-series Disequilibrium Method of Dating • Uranium series, or Ionium dating, is based on the radioactive decay of uranium in calcium carbonate and other minerals that precipitate from solution • Natural 238U decays into 234Th, while the other isotope of U, 234U, decays into 230Th Because the U is soluble in water and the Th is not, minerals that precipitate from solution often contain U but very little Th Through time Th is formed in the mineral as the U decays • The Th is itself a radioactive element and it decays into daughter products The 234Th decays into 234U and the 230Th decays into 226Ra • Ra is also radioactive, and it decays into Rn in a very short time If the mineral is of sufficient age the entire suit of U decay products, from 238U through to 206Pb (238U - 234Th - 234U 230 Th - 226Ra - 222Rn - 218Po - 214Pb - 214 Bi - 214Po - 210Pb - 210Po - 206 Pb = the U series), will be present • The critical elements in this series are 238U, 234U and 230Th because of their half lives For any mineral that initially contained only U, the time since its formation can be calculated from the 238U/ 234U and 234U/ 230 Th ratios dN − ∝N dt dN − = λN dt dN −∫ = λ ∫ dt N − ln N = λt + C C = − ln N − ln N = λt − ln N ln N − ln N = −λt N ln = − λt N0 N − λt =e N0 N = N 0e − λt D* = N − N D* = N − N 0e D* = N 0(1 − e − λT1 N = N 0e 2 ln( 2) = λT1 − λt − λt ) ln( ) = −λT1 2 T1 ln 0.693 = = λ λ Dating Assumptions •The rock or mineral system has neither gained nor lost either parent or daughter atoms so that the ratio of D*/N has changed only as a result of radioactive decay This condition is often expressed by the statement that the rock or mineral sample must be a “closed system” with respect to the parent and daughter •It must be possible to assign a realistic value to D0 This can usually be done reliably, especially when D* is much greater than D0 •The value of the decay constant (λ) must be known accurately •The measurements of D and N must be accurate and representative of the rock or mineral to be dated dN = λ1 N If it is a decay series: − dt N1 = N 10 e −λ1t dN = λ1 N − λ2 N dt dN + λ2 N − λ1 N 10 e − λ1t = dt λ1 N2 = N 10 ( e − λ1t − e − λ2t ) + N 20 e − λ2 t λ2 − λ1 If t = λ1 N2 = N 10 ( e − λ1t − e − λ2 t ) λ2 − λ1 Assumptions: • The 230Th/232Th ratio in the water mass adjacent to the sediment in a given ocean basin has remained constant during the last several hundred thousand years • 230 • 230 • Th isotopes not migrate in the sediment Th and 232Th have the same chemical speciation in sea water and there is no isotopic fractionation between sea water and the mineral phases with which the Th is associated in the sediment Th and 232Th occurring in detrital mineral particles are excluded from the analysis 230 ThA = 230ThAx + 230ThAs For Excess: ThAx = 230ThAx e − λ230 t 230 230 230 Th Th − λ230 t ( 232 ) Ax = ( 232 ) Ax e Th Th R = R0 e − λt R0 t = ln( ) λ R For Supported: λ1 N2 = N 10 ( e − λ1t − e − λ2 t ) λ2 − λ1 λ234 234 − λ234 t Ths = U (e − e − λ230 t ) λ230 − λ234 230 If secular equilibrium reaches U λ234 = 238U A 234 λ230 − λ234 ≈ λ230 ThAs = 238U A (1 − e − λ230 t ) 230 e − λ234 t ≈1 Sum up excess and supported ThA = 230ThAx e − λ230 t + 238U A (1 − e − λ230 t ) 230 230 230 238 Th Th − λ230 t U ( 232 ) A = ( 232 ) Ax e + ( 232 ) A (1 − e − λ230 t ) Th Th Th Consider supported is negligible − λh λh R= R e a ln R = ln R0 − m=− λ a λ a=− m a U/Th method *Calcium Carbonate is free of 230Th *If secular equilibrium reaches: ThAs = 238U (1 − e − λ230 t ) 230 If excess 234U presents Thx = 230 λ234 234 − λ234 t U x (e − e − λ230 t ) λ230 − λ234 ThAx = 230 230 Thx λ230 λ230 234 − λ234 t = U Ax ( e − e − λ230 t ) λ230 − λ234 U AX = 234U A0 − 234U AS 234 230 ThAx λ230 = ( 234U A0 − 234U As ) × ( e − λ234 t − e − λ230 t ) λ230 − λ234 234 234 Th λ230 U A − U As − λ234 t − λ230 t ( 238 ) Ax = ( ) × ( e − e ) 238 λ230 − λ234 U UA 230 Th λ230 − λ230 t ( 238 ) A = (1 − e )+ × (γ − 1)( e − λ234 t − e − λ230 t ) λ230 − λ234 U 230 234 U A0 Where γ = 238 UA U As = 238U A 234 If there is no excess 230 ( Th − λ230 t ) = − e A 234 U Fortunately it turns out that the presence of excess 234 U makes a very small difference [...]... α β− U → Th → 238 234 234 β− Pa → 23 4U U A = 23 4U AS + 23 4U AX 234 UA=activity of 23 4U per unit weight of sample at the present time 234 UAS=activity of 23 4U in secular equilibrium with 23 8U 234 UAX=activity of excess 23 4U per unit weight of sample 234 0 U AX = 23 4U AX e − λ234 t 234 0 U AX = 23 4U A0 − 23 4U AS 234 Since secular equilibrium U AS = 23 8U A 234 U A = 23 8U A + ( 23 4U A0 − 23 8U A )e... 234 234 234 0 238 U U A − U A − λ234 t ( 238 ) A = 1 + ( )e 238 U UA 234 U 0 ( 238 ) A = γ 0 = 1.15 U 234 U 0 ( 238 ) A = 1 + (γ 0 − 1)e − λ234 t U Th/232Th Method 230 *On the earth surface U becomes uranyl ions UO22+ such as (UO2)(CO3)3-4 *Based on the assumption 230Th and 232Th simultaneously removed from sea water Assumptions: • The 230Th/232Th ratio in the water mass adjacent to the sediment in... ( 23 4U A0 − 23 4U As ) × ( e − λ234 t − e − λ230 t ) λ230 − λ234 234 0 234 Th λ230 U A − U As − λ234 t − λ230 t ( 238 ) Ax = ( ) × ( e − e ) 238 λ230 − λ234 U UA 230 Th λ230 − λ230 t ( 238 ) A = (1 − e )+ × (γ 0 − 1)( e − λ234 t − e − λ230 t ) λ230 − λ234 U 230 234 U A0 Where γ 0 = 238 UA U As = 23 8U A 234 If there is no excess 230 ( Th − λ230 t ) = 1 − e A 234 U Fortunately it turns out that the presence... Consider supported is negligible − λh λh R= R e a ln R = ln R0 − 0 m=− λ a λ a=− m a U/ Th method *Calcium Carbonate is free of 230Th *If secular equilibrium reaches: ThAs = 23 8U (1 − e − λ230 t ) 230 If excess 23 4U presents Thx = 230 λ234 234 0 − λ234 t U x (e − e − λ230 t ) λ230 − λ234 ThAx = 230 230 Thx λ230 λ230 234 0 − λ234 t = U Ax ( e − e − λ230 t ) λ230 − λ234 0 U AX = 23 4U A0 − 23 4U AS 234... remained constant during the last several hundred thousand years • 230 • 230 • Th isotopes do not migrate in the sediment Th and 232Th have the same chemical speciation in sea water and there is no isotopic fractionation between sea water and the mineral phases with which the Th is associated in the sediment Th and 232Th occurring in detrital mineral particles are excluded from the analysis 230 ThA... 1 R0 t = ln( ) λ R For Supported: λ1 N2 = N 10 ( e − λ1t − e − λ2 t ) λ2 − λ1 λ234 234 0 − λ234 t Ths = U (e − e − λ230 t ) λ230 − λ234 230 If secular equilibrium reaches U 0 λ234 = 23 8U A 234 λ230 − λ234 ≈ λ230 ThAs = 23 8U A (1 − e − λ230 t ) 230 e − λ234 t ≈1 Sum up excess and supported 0 ThA = 230ThAx e − λ230 t + 23 8U A (1 − e − λ230 t ) 230 230 230 238 Th Th 0 − λ230 t U ( 232 ) A = ( 232 ) Ax...A series of n: N n = C1e − λ1t + C 2 e − λ2t + C ne − λn t −λ t −λ t −λ t If n=3 N 3 = C1e 1 + C 2e 2 + C 3e 3 λ1λ2 N 10 C1 = ( λ2 − λ1 )( λ3 − λ1 ) λ1λ2 N 10 C2 = ( λ1 − λ2 )( λ3 − λ2 ) λ1λ2 N 10 C3 = ( λ1 − λ3 )( λ2 − λ3 ) λ1 − λ1t − λ2 t 0 N2 = N1 (e −e ) λ2 − λ1 λ1 < λ 2 N =0 0 2 λ1 N2 ≈ N 10 e − λ1t λ2 − λ1 λ1 N2 ≈ N1 λ2 − λ1 N 1 λ2 − λ1 = = constant N2 λ1 238 92 U → λ1
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