The u series disequilibrium method of dating

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 0


− ln N = λt − ln N 0
ln N − ln N 0 = −λt
N
ln
= − λt
N0

N
− λt
=e
N0

N = N 0e

− λt

D* = N 0 − N

D* = N 0 − N 0e
D* = N 0(1 − e
− λT1
1
N 0 = N 0e 2
2

ln( 2) = λT1

2

− λt

− λt

)

1
ln( ) = −λT1
2
2
T1

2

ln 2 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 1
= λ1 N 1
If it is a decay series: −
dt

N1 =

N 10 e −λ1t

dN 2
= λ1 N 1 − λ2 N 2
dt

dN 2
+ λ2 N 2 − λ1 N 10 e − λ1t = 0
dt

λ1
N2 =
N 10 ( e − λ1t − e − λ2t ) + N 20 e − λ2 t
λ2 − λ1

If t = 0

λ1
N2 =
N 10 ( e − λ1t − e − λ2 t )
λ2 − λ1


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 << λ2
λ2 − λ1 ≅ λ2
N1
N2

=

λ2
λ1

λ1 N 1 = λ2 N 2

λ1 N 1 = λ2 N 2 = λ3 N 3 = ....... = λn N n
Nn =

λ1 N 1
λn

λ1 N 1 λ2 N 2
=
= ...... = 1.000
λ2 N 2 λ3 N 3



α

β−

U
→ Th →

238

234

234

β−

Pa → 234U

U A = 234U AS + 234U AX

234

UA=activity of 234U per unit weight of sample at the present time
234
UAS=activity of 234U in secular equilibrium with 238U
234
UAX=activity of excess 234U per unit weight of sample
234

0
U AX = 234U AX
e − λ234 t

234

0
U AX
= 234U A0 − 234U AS

234


Since secular equilibrium

U AS = 238U A

234

U A = 238U A + ( 234U A0 − 238U A )e − λ234 t

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 a given ocean basin has
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 = 230ThAx + 230ThAs

For Excess:
0
ThAx = 230ThAx
e − λ230 t

230

230

230
Th
Th 0 − λ230 t
( 232 ) Ax = ( 232 ) Ax e
Th

Th

R = R0 e − λt
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 = 238U A

234

λ230 − λ234 ≈ λ230

ThAs = 238U A (1 − e − λ230 t )

230

e

− λ234 t

≈1