The Rb-Sr Method of Dating in Geochemistry
Geo 564 Dr. Bassam Abuamarah
1
•
We will learn how the
Radioactive decay Of Rb is
confined in the
geological materials (i.e rocks, minerals.. Etc)
, that can
be used to
date the geological processes, via the following topics
:
I.
We will have a brief notes on the geochemistry of Rb
and Sr .
II.
Nature of occurring the Rb and Sr Isotopes
III.
Radioactive decay of Rb and the growth of
87
Sr.
IV.
Graphical presentation of isotope data – is called
Isochron diagram
V.
Application on the use of the method to date
magnetism and metamorphism and sedimentation
2
These elements
commonly form divalent
ions
.
Proton (P
+
)
Neutron (N)
87
Rb
Increasing in Ionic Radius (sizebigger
1
st
column
2
nd
column
K
Rb
K
1+
1.33
Å
Rb
1+
1.48
Å
Ca
Sr
Ca
2+
smallest
Å
Sr
2+
bigger ionic radii
Å 1.32
3
Neutral atoms of the alkali
metals are characterized by 1
electron in outer shell,
therefore, they are monovalent
ions.
أكبر
Ca radii
Å 1.32
Atomic
Mass
37
38
Atomic
Number
4
The Periodic table arranges the chemical element upon
their Atomic Number, and electron configuration, and
then elements are presented according to the increase in
atomic number horizontally.
Thus,
Thus,
The Atomic Radius is the distance
from the atomic nucleus to the
outermost Electron Orbital in the
Atom.
In general,
the atomic
radius decreases
as we move
from left to right in a period, and
it increases when we go down a
group.
5
This is because, in
periods, the valence
electrons are in the same
outermost shell.
The atomic number
increases within the same
period while moving from
left to right, which in turn
increases the
effective
nuclear charge
.
Thus,
Rubidium (Rb)
belong to alkali metals (
column 1
above)
which also include
Sodium (Na) and Potassium (K).
Strontium (Sr
)
belong to the
earth alkali metals
Column
2
, which include Magnesium (Mg) and Calcium (Ca)
Thus,
the ionic radii
of the
alkali elements
increase
towards higher atomic number.
The
ionic radii of Rb
(1.48
Å) is
sufficiently
similar to that
of K (1.33 Å)
to allow Rb to substitute for K
in all K-bearing
minerals
.
6
and
and
Thus, minerals suitable or used for the
Rb-Sr method
dating rocks
which involve the following minerals:
The
ionic radius
of
Sr
is
slightly larger
than of that Calcium
(Ca)
, which
can replace it in many common minerals
, such
as
plagioclase, apatite and calcite
7
# of protons
39
38
36
37
# of Neutrons
46
47
48
49
50
51
Rb
has
2 natural
occurring isotopes
(
87
Rb)
and
(
85
Rb)
(
87
Rb)
is
radioactive
, whereas the
(
85
Rb)
is stable
8
whereas the
(
85
Rb)
is stable
The
87
Rb
nucleus
consists of :
37 protons ,
and
50 neutron
.
Thus,
87
Rb is
unstable
unstable
and it radioactive decays to
87
Sr
., by emitting
particles:
87
Rb
87
Sr
87
Rb
9
Thus, from the above
Atomic
Mass
37
38
Atomic
Number
i.e.
one neutron
of the nucleus
transform to PROTON by
emitting a (
) particles
,
and releasing antineutron
(
).
The
results the
87
Sr
nucleus
consist of
38 PROTONS
and
49 NEUTRONS
, which corresponds
to
87
Sr.
87
Sr
87
Sr
87
Rb
The decay of
87
Rb to stable
87
Sr is described as follows the formula:
87
Rb =
87
Sr +
+
+ Q
Where
= represents the Beta Particles or neutrons,
is the antineutron
, and
Q = is the decay energy released during
the decay reactions
10
87
Rb
has a
half-life (T
½
) = 4.88 X 10
10
years
,
87
Rb
87
Rb
87
SR
# of Atoms
Time
11
And Half-life = (
T
½
= 0.693/
),
For instance;
the above right box filled with 32 atoms of
87
Rb .
which corresponds to the
decay constant
= 1.42 X 10
-11
y
-1
Thus, what will happen to
87
Rb atoms as years passed by?
87
Rb &
87
Sr
87
Sr
87
Rb
# of Atoms
Time Ga
4.88
9.76
14.64
19.52
24.40
After 4.88 billion years half-life of the
87
Rb atoms
of that
initially was found in the box have decayed via radioactive
processes to
87
Sr
And then
12
This range in time corresponds to one half (½) of the
initial
87
Rb found in the box
88
Sr = 82.53%
87
Sr= 7.04%
86
Sr=9.87&
84
Sr==0.56%
In nature:
•
87
Rb=27.83 %
•
85
Rb= 72.17
13
87
Sr by
emitting B
Half-life =
48.8 Billion
years
Decayed to
Thus, for
every 4.88 billion years
the number of
87
Rb
will be halved
, and
the number of
87
Sr increases
(growth) accordingly.
As a result,
after a period of 5 half-lives
,
only 1 atom of
87
Rb will
remains
in the box.
So, the number of
87
Sr atoms found in the box is 31
atoms
14
Thus,
87
Sr and
87
Rb
represent
the number of the
respective isotope atoms in the box TODAY
.,
87
Sr
i
represents the initial content of the
87
Sr in the
box
( Zero in the example above)
So the equation can be arranged as follows:
t = (1/
) ln
(
87
Sr –
87
Sr
i
) /
87
Rb) + 1
15
We fill the numbers that are known:
t
= (1/
1.42 X 10
-11
y
-1
) ln
(31 –
0.0
i
/ 1) + 1
The age is t= 24.4 years
16
So
How do we calculate the age ?
The tool we are going to use for Rb-Sr method of dating is
applying the following equation
87
Sr =
87
Sr
i
+
87
Rb ( e
t
– 1)
87
Sr
i
represents the initial content of the
87
Sr in the box
( Zero
in the example above)
87
Sr and
87
Rb
represent
the number of the respective isotope
atoms in the box TODAY
Where :
17
87
Sr =
87
Sr
i
+
87
Rb ( e
t
– 1)
if we know
how many
87
Sr
atoms
were
incorporated
in the
mineral at its formation
or
initiation time
(
87
Sr
i
),
So as to
calculate the
age
of the mineral (i.e. the rock)
the equation above will be applied:
therefore
we have to analyses the percentage (Ratio) (%) of the
87
Rb
and
87
Sr
content
in the mineral (Concentration).
then
18
Practically
, the
concentration of the isotopes as
input
is
required to apply the above equation
.
87
87
Sr/
Sr/
86
86
Sr
Sr
=
=
(87
(87
Sr
Sr
/
/
86
86
Sr
Sr
)
)
i
i
+
+
87
87
Rb/
Rb/
86
86
Sr
Sr
( e
( e
t
t
– 1)
– 1)
Equation 1
Mass-spectrometry
, is
the instrument that
performs an
isotopes analyses
by giving
isotopes
ratios
as an output,
So, in order to have the ratio; we have to
DIVIDE THE
BOTH SIDE OF THE EQUATION (1) BY
86
Sr as
a stable isotope
, So, the equ. Becomes:
19
and
86
Sr
is
not produced by decay, it is a naturally
occurring isotope
of other elements.
Thus, the
86
Sr content
in the mineral does not
change with time.
The
equation 1
is considered as
source for age
calculation
by using
Rb – Sr method
I believe
before we apply this
equation, we have to
have an idea about
using
ISOCHRON DIAGRAM
,
and have some of its
basic mathematics , in order
basic mathematics , in order
to be
to be
used for
20
Equation 2:
87
Sr/
86
Sr =
(87
Sr
/
86
Sr)
i
+
87
Rb/
86
Sr ( e
t
– 1)
Y
=
A
+ X
K
Y
= 1 +
0.5
X
Isochron is a line plotted in X-Y
diagram,
This line intersect
with Y–axis
at
y
= 1= A
,
the
slope of the line
is
0.5 = k
The equation used for this line
is Y = 1 + 0.5 X
Thus, the general expression
for a
line in X-Y diagram
is
Y =A + XK (linear equation)
(A)
point corresponds
to the
point of intersection between line
point of intersection between line
and Y-axis
and Y-axis
,
while K
corresponds to
the slope of the line
.
Measured
Measured
When you crystallize a rock,
you will always have some Sr
Presented
X axis
Y axis
slope
Y = A + X K
21
87
Sr/
86
Sr =
(
87
Sr/
86
Sr)
i
+
87
Rb/
86
Sr ( e
t
– 1)
Now, let us compare the Equ. 2 With the general equation for a line
of a X-Y diagram.
87
Rb/
86
Sr
87
Sr/
86
Sr
On the basis of equation 2
, we can plot a diagram with
87
Rb/
86
Sr
along the
horizontal axis (x-axis
)
, and
87
Sr/
86
Sr along the vertical
axis (Y-axis)
In this diagram, equation 1 represent a line :
22
87
Sr/
86
Sr
87
Rb/
86
Sr
(
87
Sr/
86
Sr)
i
( e
t
– 1)
87
Sr/
86
Sr
=
(
87
Sr/
86
Sr
)
i
+ (
87
Rb/
86
Sr
) ( e
t
– 1)
Isochron diagram
shows
value of (
87
Rb/
86
Sr)
plotted
along on the horizontal axis
(X-axis),
while
the
values of
87
Sr/
86
Sr
plotted along the
vertical axis (Y-axis).
And
•
The
values of (
87
Sr/
86
Sr
i
)
as initial content
corresponds
to the points of intersection
between
the line and the vertical axis (Y-axis),
•
while
( e
t – 1
) value
corresponds to the slope of
the line
.
23
Now we will take a look at the results
of Rb-Sr isotopes.
24
of the a geological map of the granitic
area showed
87
Sr/
86
Sr
87
Rb/
86
Sr
(
87
Sr/
86
Sr)
I
= 0.70404
( e
t
– 1) = slope= 0.02523
87
Sr/
86
Sr =
(
87
Sr/
86
Sr)
i
+ (
87
Rb/
86
Sr) (e
t
– 1)
In the diagram 7 samples
of the granite were
prepared for the analyses
by a Mass-Spectrometry:
070
0.71
0.72
0.73
0.74
0.75
0.76
0.77
Samples 1 & 2
Samples 1 & 2
were biotite poor granite
,
3 to 5
3 to 5
were characterized by
intermediate biotite
content,
while samples
6 & 7
6 & 7
were rich in biotite mineral.
25
When the analyses had been carried out, the data
were plotted in the diagram, as shown above.
Note:
that
that
the
the
Rb
Rb
content of the samples correlates with
content of the samples correlates with
the biotite contents of the rocks
the biotite contents of the rocks
, why?
, why?
26
87
Sr/
86
Sr
87
Rb/
86
Sr
(
87
Sr/
86
Sr)
I
= 0.70404
( e
t
– 1) = slope= 0.02523
070
0.71
0.72
0.73
0.74
0.75
0.76
0.77
A line can be drawn through the data points.
The
line intersects with the
vertical axis (Y-axis) at the
value of
87
Sr/
86
Sr = 0.70404
.
This is corresponding to the
87
Sr/
86
Sr
ratio of granite (and
all minerals it is composed of)
immediately after formation.
The
slope of the line gives us the equation ( e
t
– 1) =
0.02523
in this equation there is only an unknown, namely (t)
(Time
), which can be found by solving the equation.
1
2
7
The
line in this diagram is called a ISOCHRON
, which
mean
“ similar age”
27
In the ISOCHRON Diagram, rocks and minerals of the
similar age plot along a single-an Isochron
1.
When
the granite
crystallized all minerals in
the rocks
get
the same
value of
87
Sr/
86
Sr ratio =
(0.70404),
Because
they all formed
from the same isotopically
homogenized magma
.
The slope of the isochron
was Zero which corresponds
to the age = Zero
28
1.0
2.0
3.0
4.0
0.69
0.70
0.71
0.73
0.74
1
2
1
2
3
4
5
6
7
Sample
with higher
Rb
Sample
with lower
Rb
The bottom line;
High Rb/Sr
contains
more
87
Sr
Low Rb/Sr
contains
less
87
Sr
2.
(Line 2 )
(Line 2 )
In the diagram
above:
Since the time of the
crystallization
,
atoms of
87
Rb have continuously
undergone radioactive
decay
to
87
Sr.
In rocks with
initial low
initial low
content
of
87
Rb
, They
relatively have a small
amount of
87
SR formed.
29
0.7
1
0.7
5
0.7
2
0.7
4
0.7
3
0.7
6
0.7
8
0.77
0.7
4
0.7
9
1.5
2.0
2.5
3.0
3.5
1.0
0.5
4.0
87
Rb/
86
Sr
87
Sr/
86
Sr
87
Sr/
86
Sri
T = 1706
15 Ma
87
Sr/
86
Sr
= (
87
Sr
/
86
Sr
)
i
+
87
Rb/
86
Sr
(e
t
– 1)
Equation 1
30
In
the rocks which initially contained more
87
Rb,
it
proportionally showed a higher amount of
87
Sr
formed.
Note:
Seven samples were
plotted on a line at any stage
,
during
the
evolution
of the rocks.
This
line is hinged to the vertical Y-
axis, at the value of (
87
Sr/
86
Sr).
The
slope of the line
increases
as
time goes by present slope
indicate that the Granite is 1.706 Ma old
31
Rocks
87
Sr/
86
Sr =
0.702
87
Rb/
86
Sr
ratios for Various
Rocks are:
Ultra basic =
0.2
Basaltic =
0.06
Granite =
0.25 -1.7
Shale =
0.46
Sandstone =
3
32
87
Sr/
86
Sr ratios for igneous Rocks:
MORB =
0.7025
Continents =
0.7119
Oceanic Island =
0.704
Meteorites
=
0.699
Mantle
87
Sr/
86
Sr =
0.702
Rb/Sr
=0.8
Rb/Sr
=1.2
Rb/Sr
=0.6
33
0.70
0.72
0.71
0.75
0.76
0.74
0.73
1.0
2.0
3.0
0.77
M 1
M 3
M 2
R 2
R 3
R 1
87
Rb/
87
Sr
87
Sr/
86
Sr
R 1
R 2
R 3
M 1
M 2
M 3
1
2
3
we will
focus on how the Rb-Sr
method
can be used to date
metamorphic rocks
,
The mountain
on the lift Is composed of metamorphic
igneous rocks (R), while M1, M2, and M3
the red
circle represent three
minerals contained in R2
1)
line 1:
Now
The diagram to the right
shows the isotopic
composition of the rock
volumes and the minerals
formed shortly after
crystallization
34
Both
the rocks volume and
minerals show
the same
87
Sr/
86
Sr ratio
,
since they all formed in
the same ISOTOPICALLY
homogenous melt.
35
The
87
Rb/
86
Sr ratios
, however,
are different
because
because
various minerals
fractionate Rb and Sr differently
.
2) Line 2 formed
2) Line 2 formed
due to
, after, 1000 Ma,
the line will be hinged as shown
in the above diagram.
i
.e.
during the period of 1000 Ma
,
the slope of
the
Isochron
increases
increases
in response to the
RADIOACTIVE DECAY of
87
Rb
and the growth of
87
Sr.
36
3) Line 3 :
3) Line 3 :
Then the rock was heated (in the
Then the rock was heated (in the
diagram) under metamorphic
diagram) under metamorphic
events of short duration,
events of short duration,
The
increased temperature
(T
o
)
led to a
mobilization
mobilization
of
Sr
on a
small scale
(between the
minerals of a piece of the rock),
which resulted in an
ISOTOPIC
HOMOGENIZATION of
minerals in the sample of R2
R 2
R 3
M 1
M 2
M 3
R 1
T= 1000 Ma
37
However,
the mobility of Sr was not
large enough to
homogenize larger volume of the rock.
0.70
0.71
0.75
0.74
0.73
0.72
0.78
0.77
0.76
3
2
1
4
87
Sr/
86
Sr
87
Rb/
86
Sr
R 1
R 3
R 2
M1
M3
M2
t
1
= 1500 Ma
t
m
= 500 Ma
After
cooling,
the Sr
was
immobilized,
and
87
87
Sr/
Sr/
86
86
Sr ratios
Sr ratios
continued to evolve
continued to evolve
in the minerals and
in the minerals and
rocks volumes
rocks volumes
.
The
age 1500 Ma
,
after
the crystallization
of
the
rocks specimen define
an ISOCHRON
that gives
the of the crystallization of
the rocks (t).
The
ISOCHRON
defined by the
sample R2 and the
3 minerals from
R2,
However it gives
the age of the
metamorphic
event, which
occurred 500 Ma
ago as (t
m
)
38
0.70
0.71
0.75
0.74
0.73
0.72
0.78
0.77
0.76
3
2
1
87
Sr/
86
Sr
R 1
R 3
R 2
M1
M2
4
M3
t
1
= 1500 Ma
t
m
= 500 Ma
87
Rbr/
86
Sr
Many Organisms living in the Ocean water
have
shells or Hard parts
that are
made of CaCO
3
. As we
know the
Sr
2+
is a divalent element and the Ca
2+
has
similar charges and Ionic Radii,
39
Consequently,
the
Sr
2+
is readily substituted for the Ca
2+
cation in
the carbonate structure
.
Rb
, however, is not incorporated in Carbonates
The
87
Sr/
86
Sr
ratio in the carbonaceous shell of a
living organism is similar to the
87
Sr/
86
Sr ratio of the
water
living in.
Consequently,
when the organism
died
and is
incorporated
into the bottom’s sediments of the sea
,
the
87
Sr/
86
Sr
signature belongs
to the environment it was living in and
is
preserved
in the sediments that formed.
40
i.e. since the shell does not contains Rb.
So,
the
87
Sr/
86
Sr
signature is not altered with time
Conclusion:
The marine Carbonate can be used only to
monitor the
variation in
variation in
78
78
Sr/
Sr/
86
86
Sr
Sr
of
the Ocean water throughout
the time
Now, let us have a quick
look at the
87
Sr/
86
Sr
ratios
of the Earth’s
crust and Mantle.
41
0.702
0.704
0.710
0.708
0.706
87
Sr/
86
Sr
87
Rbr/
86
Sr
Present
1
4
3
2
0.689
Crust
Mantle
Since
the formation
of the earth 4.6 Ga
ago;
The
mantle has
been homogenous
with respect to the
87
Rb/
86
Sr and
87
Sr/
86
Sr ratios.
Shortly,
after the formation of the earth
, the
87
Sr/
86
Sr
ratios of the Mantle was about 0.699
.
42
The reason
The reason
is that of the
RADIOACTIVE DECAY
of
87
Rb to
87
Sr
, and
due to
87
Sr/
86
Sr ratios of the
mantle
.
The mantle has evolved
to
present an
87
Sr/
86
Sr
value of about 0.704,
0.702
0.704
0.710
0.708
0.706
87
Sr/
86
Sr
87
Rbr/
86
Sr
Present
1
4
3
2
0.699
Mantle
During
the partial melting
of the mantle.
Materials extracted
from
the mantle
and
added to
the Earth’s crust;
43
and
Immediately
after the formation of the new crust
,
its Sr
isotopic
composition
is similar to the composition of the
mantle
it was extracted from
Formation of
new crust
Partial
melting
During
the
partial melting
of the mantle,
Rb is
partitioned, and
segregated into a melt
relative Sr.
44
This
happened
immediately
after
its
formation;
the newly
formed crust
shows a
high Rb/Sr
ratio
than
the
mantle it was
extracted from
.
0.702
0.704
0.710
0.708
0.706
87
Sr/
86
Sr
87
Rbr/
86
Sr
Present
1
0.689
Mantle
Depleted Mantle
Crustal Rocks
Formation of
new crust
Partial melting
2
3
4
Similarly,
the part of the
mantle
that has
undergone partial melting
( it is depleted; i.e.
quantity not able to fulfil)
shows a lower Rb/Sr ratio
than the un-depleted
mantle,
0.702
0.704
0.710
0.708
0.706
87
Sr/
86
Sr
87
Rbr/
86
Sr
Present
1
0.689
Mantle
Depleted Mantle
Crustal Rocks
3
2
4
Formation of
new crust
Partial melting
Because
and due
to the differences in their
Rb/Sr ratio, the
87
Sr/
86
Sr
ratios of the crust’
The depleted
(no longer sufficient)
and un-depleted
mantle evolved differently.
46
So,
crustal rocks show
, in general, considerably
higher
87
Sr/
86
Sr ratios
than
mantle rocks
The Mantle is; as pointed out on a previous slide
homogenous concerning its Sr isotope composition.
The reason for this
is
that
the various
components of the
crust
have been
extracted from the
mantle at a
different time
as
shown in the diagram
47
0.702
0.704
0.710
0.708
0.706
87
Sr/
86
Sr
87
Rbr/
86
Sr
Present
1
0.689
Crustal Rocks
3
2
4
mantle
Partial melting
A
B
C
Different components of the
earth’s crust , and
; however; show large
vartion in the ratios of
87
Sr/
86
Sr
A
B
C
The
Sr of Marine water
is
derived
from 3 principles
sources:
48
1.
Water on the crust
87
Sr/
86
Sr signature that
enters the sea via rivers,
and runoff water.
2.
Water
from mantle
87
Sr/
86
Sr signature
that enters the sea
through the hydrothermal system along the Mid-oceanic
ridge (MORB), and
Sr from
continent
crust
Sr from marine
Carbonate
Sr from mantle
Sea level
3.
Recirculation of
Sr from Pre-existing Carbonate rich rocks
Although
the
ocean of the world continuously receives
Sr of high variable isotopic compositions
from the
source mentioned above.
49
The
isotopic composition of
the Oceans are everywhere
the same of:
87
Sr/
86
Sr = 0.70924
The
strong current of the Ocean’s waters is very
important for the isotopic homogenization of Sr on
a Global Scale
.
Equally important is
, however,
the high Sr solubility
in marine waters
Sr
that is
introduced to the SEA from external sources
exist in
for sufficiently long time in the sea to
homogenize isotopically
on a
global scale.
50
Less
soluble elements
such as (Nd) participate
long
before the
global
isotopic
homogenization is
attained
0.7070
0.7075
0.7090
0.7085
0.7080
87
Sr/
86
Sr
Age Ma
0.7065
25
75
50
100
150
200
0.7095
Modern ocean water
87
Sr/
86
Sr = 0.70924
87
Sr/
86
Sr = 0.7082=25Ma
87
Sr/
86
Sr = 0.7076=75 &200 Ma
Sr Isotope source of the Oceans
That
the ocean
receive high amount
of Sr
with
Mantle
signature (
87
Sr/
86
Sr =
0.704) in that period
51
The
diagram shows
the
87
Sr/
86
Sr
variation
ratios of the ocean
water
back to
about
210 Ma
.
The
low
87
Sr/
86
Sr
ratio
at
about 155 –
165 Ma
I
ndicating
This contrast is due to modern oceanic water
87
Sr/
86
Sr) =
0.709
52
Which indicating
High input of Sr with CRUSTAL signature (High
87
Sr/
86
Sr)
Thus, the ocean water
87
Sr/
86
Sr ratio diagram can be used to date
marine carbonate although the method does not always give
unique results.
A
87
Sr/
86
Sr ratio of 0.7082 corresponds to an age of about 25 Ma.
And an
87
Sr/
86
Sr of 0.7076 corresponds to an age 75 and 200 Ma.
Rb substitute for K (Potassium) in minerals
. Thus, K-
bearing minerals such as Muscovite, Biotite, amphibole,
and K-feldspar are suitable for Rb-Sr method dating.
53
Rb/Sr dating method is based on the radioactive decay of
87
Rb/
86
Sr
, thus the decay of
87
Rb to stable
86
Sr
occurs by the
β
-
decay mechanism, and is described by the Equation:
87
Rb =
87
Sr +
+
+ Q
The decay of
87
Rb and production of
87
Sr with time is
normalized by dividing the equation by
86
Sr then can be
expressed as follows:
54
87
87
Sr/
Sr/
86
86
Sr
Sr
=
=
(87
(87
Sr
Sr
/
/
86
86
Sr
Sr
)
)
i
i
+
+
87
87
Rb/
Rb/
86
86
Sr
Sr
( e
( e
t
t
– 1)
– 1)
Equation 1
Where t = time in only unknown
When
87
87
Rb/
Rb/
86
86
Sr
Sr
and
and
87
87
Sr/
Sr/
86
86
Sr ratios of a geological system
Sr ratios of a geological system
(Minerals, Rocks samples) is known, then the time elapsed since
(Minerals, Rocks samples) is known, then the time elapsed since
the system closed to Rb and Sr can be calculated from the equation
the system closed to Rb and Sr can be calculated from the equation
(1) above.
(1) above.
The
Isochron diagram of
87
87
Rb/
Rb/
86
86
Sr
Sr
is drawn on a horizontal axis
is drawn on a horizontal axis
and
and
87
87
Sr/
Sr/
86
86
Sr is drawn in the graph as a vertical axis is the graphical
Sr is drawn in the graph as a vertical axis is the graphical
presentation of the equation (1).
presentation of the equation (1).
55
The calculated age may represent the time elapsed since
crystallization of the rocks, or the time elapsed since the rock
underwent metamorphose
.
.
The
87
87
Sr/
Sr/
86
86
Sr of
Sr of
marine carbonates can be used to date the
time of sediments depositional
.
.
•
Rubidium
is the twenty-third
most abundant element
in the Earth's crust
, roughly as abundant as
zinc
and
rather more common than
copper
.
•
It occurs naturally in the minerals
leucite
,
pollucite
,
carnallite
, and
zinnwaldite
, which contain up to 1% of its
oxide
.
Lepidolite
contains between 0.3% and 3.5% rubidium, and is the commercial
source of the element.
•
Some
potassium
minerals and
potassium chlorides
also contain the
element in commercially significant amounts.
•
Seawater
contains an average of 125 µg/L of rubidium compared
to the much higher value for potassium of 408 mg/L and the much
lower value of 0.3 µg/L for caesium
.
56
•
Because of its large
ionic radius
, rubidium is one of the
"
incompatible elements
.”
•
During
magma crystallization
, rubidium is concentrated
together with its heavier analogue caesium in the liquid
phase and crystallizes last. Therefore the largest deposits of
rubidium and caesium are zone
pegmatite
ore bodies
formed by this enrichment process.
•
Because rubidium substitutes for
potassium
in the
crystallization of magma, the enrichment is far less effective
than in the case of caesium.
•
Zone pegmatite ore bodies containing mineable quantities of
caesium as
pollucite
or the lithium minerals
lepidolite
are
also a source for rubidium as a by-product.
57
58
Strontium commonly occurs in nature, being the 15th
most
abundant element
on Earth (its heavier congener barium
being the 14th), estimated to average approximately 360
parts
per million
in the
Earth's crust
and is found chiefly as
the
sulfate
mineral
celestine
(SrSO
4
) and
the
carbonate
strontianite
(SrCO
3
).
Of the two, celestine occurs much more frequently in
deposits of sufficient size for mining. Because strontium is
used most often in the carbonate form
59
strontianite
would be the more useful of the two common
minerals, but few deposits have been discovered that are
suitable for development.
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strontium
only exists in nature when combined to form minerals.
Naturally occurring strontium is stable, but its synthetic
isotope Sr-90 is only produced by nuclear fallout.
In groundwater,
strontium
behaves chemically much like calcium.
At intermediate to acidic
pH
Sr
2+
is the dominant
strontium
species.
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forms
coprecipitates
with calcium minerals such as
calcite
and
anhydrite at an increased pH. At intermediate to acidic pH,
dissolved strontium is bound to soil particles by
cation exchange
.
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Explore the principles of the Rb-Sr dating method in geochemistry, focusing on radioactive decay of Rubidium and Strontium isotopes in geological materials. Learn about isotope data presentation, application in dating geological processes, and the significance of Rb and Sr in the periodic table.
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Presentation Transcript
The Rb Sr Method of Dating Geo 564 Geochronology Course Course Director Dr. Bassam A. A. Abuamarah Geo 564 Dr. Bassam Abuamarah 1
The Rb-Sr method of Dating - Introduction We will learn how the Radioactive decay Of Rb is confined in the geological materials (i.e rocks, minerals.. Etc), that can be used to date the geological processes, via the following topics: I. We will have a brief notes on the geochemistry of Rb and Sr . II. Nature of occurring the Rb and Sr Isotopes III. Radioactive decay of Rb and the growth of 87Sr. IV. Graphical presentation of isotope data is called Isochron diagram V. Application on the use of the method to date magnetism and metamorphism and sedimentation 2
Rb and Sr in the Periodic Table 2nd 1st Increasing in Ionic Radius (sizebigger Atomic Mass column column 87 Rb 1 H 38 3 Li 4 Be 11 Na 12 Mg Proton (P+) Neutron (N) Atomic Number 37 19 K 20 Ca Neutral atoms of the alkali metals are characterized by 1 electron in outer shell, therefore, they are monovalent ions. 37 Rb 38 Sr 55 Cs 56 Ba K1+ K 87 Fr 88 Ra 1.33 These elements commonly form divalent ions. Rb Rb1+ Ca Ca2+ Ca radii 1.32 1.48 Sr2+ smallest Sr 3 bigger ionic radii 1.32
Thus, The Periodic table arranges the chemical element upon their Atomic Number, and electron configuration, and then elements are presented according to the increase in atomic number horizontally. The Atomic Radius is the distance from the atomic nucleus to the outermost Electron Orbital in the Atom. In general, the atomic radius decreases as we move from left to right in a period, and it increases when we go down a group. 4
This is because, in periods, the valence electrons are in the same outermost shell. The atomic number increases within the same period while moving from left to right, which in turn increases the effective nuclear charge. 5
Contd: Rb and Sr in the Periodic Table Thus, Rubidium (Rb) belong to alkali metals (column 1 above) which also include Sodium (Na) and Potassium (K). and Strontium (Sr) belong to the earth alkali metals Column 2 , which include Magnesium (Mg) and Calcium (Ca) Thus, the ionic radii of the alkali elements increase towards higher atomic number. The ionic radii of Rb (1.48 ) is of K ( of K (1.33 1.33 ) ) to allow Rb to substitute for K to allow Rb to substitute for K in all K minerals minerals. . ) is sufficiently sufficiently similar to that in all K- -bearing similar to that bearing 6
Contd: Rb and Sr in the Periodic Table Thus, minerals suitable or used for the Rb-Sr method dating rocks which involve the following minerals: K-feldspar Micas Clay minerals and Evaporate minerals such as sylvite and Carnallite The ionic radius of Sr is slightly larger than of that Calcium (Ca), which can replace it in many common minerals, such as plagioclase, apatite and calcite 7
Rb) is stable Cont d: Rb and Sr in the Periodic Table Y 85 Sr 84 Rb 83 K 82 46 Y 86 Sr 85 Rb 84 K 83 Y 87 Sr 86 Rb 85 K 84 48 Y 88 Sr 87 Rb 86 K 85 Y 89 Sr 88 Rb 87 K 86 Y 90 Sr 89 Rb 88 K 87 39 # of protons 38 37 36 49 47 50 51 # of Neutrons Rb has 2 natural occurring isotopes (87Rb) and (85Rb) (87 Rb) is radioactive, whereas the (85 Rb) is stable Sr has four natural occurring isotopes, which all are STABLE Sr 84, Sr 86,Sr 87,Sr 88 8
Contd: Rb and Sr in the Periodic Table Atomic Mass 87Sr 38 87Rb Atomic Number 37 87 Rb Thus, from the above The 87Rb nucleusconsists of : 37 protons ,and 50 neutron. Thus, 87Rb is unstable and it radioactive decays to 87Sr., by emitting particles: 9
The decay Mechanism 87Sr 87Rb 87 Sr i.e. one neutron of the nucleus transform to PROTON by emitting a ( ) particles , and releasing antineutron ( ). The results the87Sr nucleus consist of 38 PROTONS and 49 NEUTRONS, which corresponds to 87Sr. The decay of 87Rb to stable 87 Sr is described as follows the formula: 87Rb = 87 Sr + + + Q Where = represents the Beta Particles or neutrons, is the antineutron, and Q = is the decay energy released during the decay reactions 10
Radioactive Decay of 87Rb and the growth of 87 Sr 87Rb # of Atoms 87SR 87Rb Time 87Rb has a half-life (T ) = 4.88 X 1010 years, which corresponds to the decay constant = 1.42 X 10-11 y-1 And Half-life = (T = 0.693/ ), For instance; the above right box filled with 32 atoms of 87Rb . Thus, what will happen to 87Rb atoms as years passed by? 11
Radioactive Decay of 87Rb and the growth of 87 Sr 87 Rb 9.76 4.88 24.40 # of Atoms 19.52 14.64 87Rb & 87Sr 87 Sr Time Ga After 4.88 billion years half-life of the 87 Rb atoms of that initially was found in the box have decayed via radioactive processes to 87 Sr This range in time corresponds to one half ( ) of the initial 87 Rb found in the box And then 12
87Rb/87Sr Decay Scheme In nature: 87Rb=27.83 % 85Rb= 72.17 87Sr by emitting B Half-life = 48.8 Billion years 88 Sr = 82.53% 87 Sr= 7.04% 86Sr=9.87& 84Sr==0.56% 13
Contd: Radioactive Decay of 87Rb and the growth of 87 Sr Thus, for every 4.88 billion years the number of 87Rb will be halved, and the number of 87 Sr increases (growth) accordingly. As a result, after a period of 5 half-lives, only 1 atom of 87Rb will remains in the box. So, the number of 87 Sr atoms found in the box is 31 atoms 14
Contd: Radioactive Decay of 87Rb and the growth of 87 Sr Thus, 87 Sr and 87Rb represent the number of the respective isotope atoms in the box TODAY., 87Sri represents the initial content of the 87Sr in the box( Zero in the example above) So the equation can be arranged as follows: t = (1/ ) ln (87Sr 87Sri) / 87 Rb) + 1 We fill the numbers that are known: t= (1/1.42 X 10-11 y-1) ln (31 0.0i / 1) + 1 The age is t= 24.4 years 15
Contd: Radioactive Decay of 87Rb and the growth of 87Sr So How do we calculate the age ? The tool we are going to use for Rb-Sr method of dating is applying the following equation 87Sr = 87Sri + 87Rb ( e t 1) Where : 87Sri represents the initial content of the 87Sr in the box( Zero in the example above) 87 Sr and 87Rb represent the number of the respective isotope atoms in the box TODAY 16
The Fundamental Equation 87Sr = 87Sri + 87Rb ( e t 1) therefore if we know how many87Sr atoms were incorporated in the mineral at its formation or initiation time (87Sri), then we have to analyses the percentage (Ratio) (%) of the 87Rb and 87Sr content in the mineral (Concentration). So as to calculate the age of the mineral (i.e. the rock) the equation above will be applied: 17
Contd: The Fundamental Equation Practically, the concentration of the isotopes as input is required to apply the above equation. Mass-spectrometry, is the instrument that performs an isotopes analyses by giving isotopes ratios as an output, So, in order to have the ratio; we have to DIVIDE THE BOTH SIDE OF THE EQUATION (1) BY 86Sr as a stable isotope, So, the equ. Becomes: 87Sr/86Sr = (87Sr/86Sr)i + 87Rb/86Sr( e t 1)Equation 1 18
Contd: The Fundamental Equation and 86Sris not produced by decay, it is a naturally occurring isotope of other elements. Thus, the 86Sr content in the mineral does not change with time. The equation 1 is considered as source for age calculation by using Rb Sr method I believe before we apply this equation, we have to have an idea about using ISOCHRON DIAGRAM, and have some of its basic mathematics , in order to be used for 19
Mathematics used-The Isochrom Diagram Equation 2: 87Sr/86Sr = (87Sr/86Sr)i + 87 Rb/86Sr ( e t 1) When you crystallize a rock, you will always have some Sr Presented Measured Measured Isochron is a line plotted in X-Y diagram, The equation used for this line is Y = 1 + 0.5 X Y = A + X K Y axis slope This line intersect with Y axis at y = 1= A, the slope of the line is 0.5 = k Thus, the general expression for a line in X-Y diagram is Y =A + XK (linear equation) (A) point corresponds to the point of intersection between line and Y-axis, while K corresponds to the slope of the line. Y = 1 + 0.5 X X axis 20
The Isochrom Diagram Y = A + X K 87 Sr/86Sr =(87Sr/86Sr)i + 87 Rb/86Sr ( e t 1) 87 Sr/86Sr 87 Rb/86Sr Now, let us compare the Equ. 2 With the general equation for a line of a X-Y diagram. On the basis of equation 2, we can plot a diagram with 87Rb/86Sr along the horizontal axis (x-axis), and 87Sr/86Sr along the vertical axis (Y-axis) 21
The Isochrom Diagram In this diagram, equation 1 represent a line : 87Sr/86Sr =(87 Sr/ 86Sr)i + ( 87Rb/86Sr) ( e t 1) Isochron diagram shows value of (87Rb/86Sr) plotted along on the horizontal axis (X-axis), while the values of 87Sr/86Sr plotted along the vertical axis (Y-axis). (87 Sr/86Sr)i 87 Sr/86Sr And 87 Rb/86Sr 22
Contd: The Isochron Diagram The values of (87Sr/86Sri)as initial content corresponds to the points of intersection between the line and the vertical axis (Y-axis), while ( e t 1) value corresponds to the slope of the line. 23
Dating of igneous Rocks Now we will take a look at the results of Rb-Sr isotopes. of the a geological map of the granitic area showed 87Sr/86Sr =(87Sr/86Sr)i + (87 Rb/86Sr) (e t 1) 0.77 87 Sr/86Sr In the diagram 7 samples of the granite were prepared for the analyses by a Mass-Spectrometry: 0.76 0.75 0.74 0.73 0.72 (87 Sr/86Sr)I = 0.70404 0.71 070 87 Rb/86Sr 24
Contd: Dating of igneous Rocks Samples 1 & 2 were biotite poor granite, 3 to 5 were characterized by intermediate biotite content, while samples 6 & 7 were rich in biotite mineral. When the analyses had been carried out, the data were plotted in the diagram, as shown above. Note: that the Rb content of the samples correlates with the biotite contents of the rocks, why? 25
Contd: Dating of igneous Rocks A line can be drawn through the data points. The line intersects with the vertical axis (Y-axis) at the value of 87 Sr/86Sr = 0.70404. 87 Sr/86Sr 0.77 0.76 7 0.75 0.74 This is corresponding to the 87Sr/86Sr ratio of granite (and all minerals it is composed of) immediately after formation. 2 0.73 1 0.72 (87 Sr/86Sr)I = 0.70404 0.71 070 87 Rb/86Sr The slope of the line gives us the equation ( e t 1) = 0.02523 in this equation there is only an unknown, namely (t) (Time), which can be found by solving the equation. 26
Contd: Dating of igneous Rocks The line in this diagram is called a ISOCHRON, which mean similar age In the ISOCHRON Diagram, rocks and minerals of the similar age plot along a single-an Isochron 27
Contd: Dating of Igneous Rocks 1. When the granite crystallized all minerals in the rocks get the same value of 87Sr/86Sr ratio = (0.70404), Becausethey all formed from the same isotopically homogenized magma. The slope of the isochron was Zero which corresponds to the age = Zero The bottom line; High Rb/Sr contains more 87Sr Low Rb/Sr contains less 87Sr 7 2 6 0.74 5 0.73 4 23 1 0.71 1 0.70 0.69 1.0 2.0 3.0 4.0 Sample with higher Rb Sample with lower Rb 28
Contd: Dating of Igneous Rocks 2. (Line 2 )In the diagram above: Since the time of the crystallization, atoms of 87Rb have continuously undergone radioactive decay to 87Sr. In rocks with initial low content of 87Rb, They relatively have a small amount of 87SR formed. 0.79 0.78 0.77 0.74 0.76 87Sr/86Sr 0.75 0.74 0.73 0.72 0.71 3.5 0.5 1.52.02.53.0 87Rb/86Sr 4.0 1.0 87Sr/86Sr = (87Sr/86Sr)i + 87Rb/86Sr(e t 1)Equation 1 29
Contd: Dating of Igneous Rocks In the rocks which initially contained more 87Rb, it proportionally showed a higher amount of 87Sr formed. Note: Seven samples were plotted on a line at any stage, during the evolution of the rocks. This line is hinged to the vertical Y- axis, at the value of (87Sr/86Sr). The slope of the line increases as time goes by present slope indicate that the Granite is 1.706 Ma old 30
Contd: Dating of Igneous Rocks 87Sr/86Sr ratios for igneous Rocks: MORB = 0.7025 Continents = Oceanic Island = Meteorites = Rb/Sr =0.8 87Rb/86Sr ratios for Various Rocks are: Ultra basic = 0.2 Basaltic = Granite = Shale = Sandstone = 3 0.7119 0.704 0.699 0.06 0.25 -1.7 0.46 Rb/Sr =1.2 Rocks 87Sr/86Sr = 0.702 Rb/Sr =0.6 Mantle 87Sr/86Sr = 0.702 32
Dating of Metamorphic Rocks R 2 0.77 2 0.76 0.75 0.74 87Sr/86Sr 0.73 3 M 1 M 3 R 1 1)line 1: Now M 2 R 3 0.72 0.71 1 we will focus on how the Rb-Sr methodcan be used to date metamorphic rocks, The mountain on the lift Is composed of metamorphic igneous rocks (R), while M1, M2, and M3 the red circle represent three minerals contained in R2 0.70 1.0 R 1 3.0 2.0 M 1 M 3 R 2 R 3 M 2 87Rb/87Sr 33
Contd: Dating of Metamorphic Rocks The diagram to the right shows the isotopic composition of the rock volumes and the minerals formed shortly after crystallization Both the rocks volume and minerals show the same 87Sr/86Sr ratio, since they all formed in the same ISOTOPICALLY homogenous melt. 34
Contd: Dating of Metamorphic Rocks The 87Rb/86Sr ratios, however, are different because various minerals fractionate Rb and Sr differently. 2) Line 2 formed due to, after, 1000 Ma, the line will be hinged as shown in the above diagram. i.e. during the period of 1000 Ma, the slope of the Isochronincreases in response to the RADIOACTIVE DECAY of 87Rb and the growth of 87Sr. 35
R 2 Cont d: Dating of Metamorphic Rocks 3) Line 3 : Then the rock was heated (in the diagram) under metamorphic events of short duration, R 1 M 1 M 2M 3 R 3 The increased temperature (T o) led to a mobilization of Sron a small scale (between the minerals of a piece of the rock), which resulted in an ISOTOPIC HOMOGENIZATION of minerals in the sample of R2 36
Contd: Dating of Metamorphic Rocks However, the mobility of Sr was not large enough to homogenize larger volume of the rock. After cooling, the Sr was immobilized, and 87Sr/86Sr ratios continued to evolve in the minerals and rocks volumes. 0.78 0.77 0.76 R 3 0.75 87Sr/86Sr M1 M3 M2 0.74 0.73 tm= 500 Ma R 2 The age 1500 Ma, after the crystallization of the rocks specimen define an ISOCHRON that gives the of the crystallization of the rocks (t). 0.72 0.71 R 1 0.70 1 3 2 4 87Rb/86Sr 37
The ISOCHRON defined by the sample R2 and the 3 minerals from R2, However it gives the age of the metamorphic event, which occurred 500 Ma ago as (tm) 0.78 0.77 0.76 R 3 0.75 87Sr/86Sr M1 M3 M2 0.74 0.73 tm= 500 Ma R 2 0.72 0.71 R 1 0.70 1 3 2 4 87Rbr/86Sr 38
Dating Of Sedimentary Processing using Sr isotope of Marine Carbonate Many Organisms living in the Ocean water have shells or Hard parts that are made of CaCO3. As we know the Sr2+ is a divalent element and the Ca 2+ has similar charges and Ionic Radii, Consequently, the Sr2+ is readily substituted for the Ca2+ cation in the carbonate structure. Rb, however, is not incorporated in Carbonates The 87Sr/86Sr ratio in the carbonaceous shell of a living organism is similar to the 87Sr/86Sr ratio of the water living in. 39
Contd: Dating Of Sedimentary Processing using Sr isotope of Marine Carbonate Consequently, when the organism died and is incorporated into the bottom s sediments of the sea, the 87Sr/86Sr signature belongs to the environment it was living in and is preserved in the sediments that formed. i.e. since the shell does not contains Rb. So, the 87Sr/86Sr signature is not altered with time Conclusion: The marine Carbonate can be used only to monitor the variation in 78Sr/86Sr of the Ocean water throughout the time 40
Sr isotopes of the mantle and the crust Mantle Crust Now, let us have a quick look at the 87Sr/86Sr ratios of the Earth s crust and Mantle. Since the formation of the earth 4.6 Ga ago; The mantle has been homogenous with respect to the 87Rb/86Sr and 87Sr/86Sr ratios. 0.710 0.708 87Sr/86Sr 0.706 0.704 0.702 0.689 Present 1 3 2 4 87Rbr/86Sr 41
Contd : Sr isotopes of the mantle and the crust Shortly, after the formation of the earth, the 87Sr/86Sr ratios of the Mantle was about 0.699. 0.710 The reason is that of the RADIOACTIVE DECAY of 87Rb to 87Sr, and due to 87Sr/86Sr ratios of the mantle. The mantle has evolved to present an 87Sr/86Sr value of about 0.704, 0.708 87Sr/86Sr 0.706 0.704 0.702 0.699 Present 1 3 2 4 87Rbr/86Sr 42
Contd : Sr isotopes of the mantle and the crust Formation of new crust Partial melting and During the partial melting of the mantle. Materials extracted from the mantle and added to the Earth s crust; Immediately after the formation of the new crust, its Sr isotopic compositionis similar to the composition of the mantle it was extracted from 43
Contd : Sr isotopes of the mantle and the crust Formation of new crust During the partial melting of the mantle, Rb is partitioned, and segregated into a melt relative Sr. This happened immediately after its formation; the newly formed crust shows a high Rb/Sr ratio than the mantle it was extracted from. Partial melting 0.710 0.708 87Sr/86Sr 0.706 0.704 0.702 0.689 Present 1 4 3 2 87Rbr/86Sr 44
Contd : Sr isotopes of the mantle and the crust Similarly, the part of the mantle that has undergone partial melting ( it is depleted; i.e. quantity not able to fulfil) shows a lower Rb/Sr ratio than the un-depleted mantle, Formation of new crust Partial melting 0.710 0.708 87Sr/86Sr Because and due 0.706 0.704 Present 0.702 to the differences in their Rb/Sr ratio, the 87Sr/86Sr ratios of the crust 0.689 1 4 3 2 87Rbr/86Sr
Contd : Sr isotopes of the mantle and the crust The depleted (no longer sufficient) and un-depleted mantle evolved differently. So, crustal rocks show , in general, considerably higher 87Sr/86Sr ratios than mantle rocks The Mantle is; as pointed out on a previous slide homogenous concerning its Sr isotope composition. 46
Contd : Sr isotopes of the mantle and the crust Different components of the earth s crust , and ; however; show large vartion in the ratios of 87Sr/86Sr 0.710 C A B 0.708 87Sr/86Sr 0.706 C The reason for this is that the various components of the crust have been extracted from the mantle at a different time as shown in the diagram 0.704 B A 0.702 0.689 1 4 3 2 Present 87Rbr/86Sr A C 47 mantle Partial melting
Sr Isotope source of the Oceans The Sr of Marine water is derived from 3 principles sources: 1. Water on the crust 87Sr/86Sr signature that enters the sea via rivers, and runoff water. Sea level Sr from continent crust Carbonate Sr from marine Sr from mantle 2. Water from mantle 87Sr/86Sr signature that enters the sea through the hydrothermal system along the Mid-oceanic ridge (MORB), and 3. Recirculation of Sr from Pre-existing Carbonate rich rocks 48
Sr Isotope of the Oceans Although the ocean of the world continuously receives Sr of high variable isotopic compositions from the source mentioned above. The isotopic composition of the Oceans are everywhere the same of: 87Sr/86Sr = 0.70924 The strong current of the Ocean s waters is very important for the isotopic homogenization of Sr on a Global Scale. Equally important is, however, the high Sr solubility in marine waters 49
Sr Isotope source of the Oceans Sr that is introduced to the SEA from external sources exist in for sufficiently long time in the sea to homogenize isotopically on a global scale. Less soluble elements such as (Nd) participate long before the global isotopic homogenization is attained Modern ocean water 87Sr/86Sr = 0.70924 0.7095 0.7090 0.7085 87Sr/86Sr = 0.7082=25Ma 87Sr/86Sr = 0.7076=75 &200 Ma 87Sr/86Sr 0.7080 0.7075 0.7070 0.7065 25 50 75 100 150 50 200 Age Ma
