Geoscience Reference
In-Depth Information
Since
α
is very close to one, we can use the approximation ln
α
≈ α −
1 and insert it in
the previous equation rewritten as:
1000
18
O
16
O
1
/
18
O
2
−
δ
18
O
1
=
δ
18
O
16
O
ref
(α
−
1
)
≈
1000 ln
α
(3.24)
/
in which we have taken into account that the ratio of two
18
O
16
O ratios is always very
close to one. The difference in delta values between co-existing phases (minerals, liq-
uids) therefore decreases as 1
/
T
2
, which implies that the sensitivity of stable-isotope
thermometers decreases with increasing
T
, hence their important applications to low- and
medium-temperature processes.
For elements with more than two isotopes, at constant temperature, the amplitude of
fractionation
/
increases with the difference in mass of the isotopic ratios. For example,
the difference in the
18
O
α
16
O ratio between two samples is twice the difference in the
/
17
O
16
O ratio. This is because bond energy varies with the mass of the bonding atoms.
Therefore ln
/
α
can be expanded to the first order relative to the difference in mass
m
in
the isotope ratio, giving:
2
m
ln
α (
m
)
=
ln
α (
m
=
0
)
+
f
m
+
O
(3.25)
where
f
, which is the derivative of ln
0, is a coefficient that
is independent of mass, termed mass discrimination. In this equation,
α
relative to
m
for
m
=
means “of the
order of” and we will neglect terms of higher than first order. Moreover, when the mass
difference is zero, there is no fractionation between a mass and itself (
O
1), and the
first term on the right-hand side is therefore zero, giving for fractionation of
18
O and
16
O
between phases 1 and 2 (
α
=
m
=
2):
18
O
16
O
18
O
16
O
O
2
e
2
f
α
1
=
1
=
(3.26)
/
2
When fractionation is small, i.e. when
f
tends toward 0, we utilize the linear approximation
obtained by developing the logarithm of the left-hand side of
(3.25)
to the first order:
O
2
α
1
≈
1
+
2
f
(3.27)
/
We will have, for example, the two linear equations:
17
O
16
O
2
=
17
O
16
O
1
(
1
+
1
f
)
(3.28)
18
O
16
O
2
=
18
O
16
O
1
(
1
+
2
f
)
(3.29)
Alternatively,
(3.27)
can be re-written as
18
O
2
/
1
+
δ
1000
1000
≈
1
+
2
f
(3.30)
18
O
1
/
1
+
δ
)
−
1
We use the approximation
(
1
+
x
≈
1
−
x
valid for small
x
and neglect the products of
deltas to obtain:
18
O
2
−
δ
18
O
1
≈
δ
2000
f
(3.31)
