ES 2502 - Lab Exercise #4
Radiogenic Isotope Geochemistry
Q1.Calculating Sm-Nd Isochron Ages and Information on the Bulk
Earth
(a) Write out the isotope decay scheme for the samarium (Sm) - neodymium (Nd) isotope system.
(b) Shown below is the age equation for the deay of the Rb-Sr isotope system. Using this as an example,
write out the equivalent age equation for the deay of the Sm-Nd isotope system. Remind yourself what the
subscripts "m" and "o" refer to.
(c) For the Sm-Nd age equation in (b), identify which is the "parent" radioactive isotope, the "daughter"
radiogenic isotope, and the stable Nd isotope.
(d) Recall from the course that the Earth accreted from meteorites. The current "best" estimate for the age
of the Earth was determined using Sm-Nd systematics measured for chondritic meteorites. Given the data
measured for several meteorites below, finish the Sm-Nd isochron on the diagram provided and determine
147
-12
-1
the age of the Earth (the decay constant for Sm is λ = 6.54 * 10 yr ). Note that all of the
samples, except "SS dark", are already plotted on the graph - you need only finish it.
chondritic meteorite
SS light
MU_2
GU
JUV
PR
MU_3
MU_1
SS_dark
measured today measured today
143Nd/144Nd
147Sm/144Nd
0.51169
0.1920
0.51174
0.1931
0.51177
0.1946
0.51181
0.1954
0.51183
0.1957
0.51184
0.1964
0.51185
0.1970
0.51194
0.2000
To calculate the age, use the equations outlined in the appendix to the lab. Use the following sheet as a
means of working through the calculations to figure out the slope and Y-intercept.
1
chondritic
meteorite
X2 XY Y X 143Nd/144Nd
147Sm/144Nd
SS light
0.51169
0.192
MU_2
0.51174
0.1931
GU
0.51177
0.1946
JUV
0.51181
0.1954
PR
0.51183
0.1957
MU_3
0.51184
0.1964
MU_1
0.51185
0.197
SS_dark
0.51194
0.2
Σ
n
8
m t t (in Ga) b (e) Does your calculated age for the Earth in (d) appear "reasonable" to you? Explain.
2
0.51200
0.51195
143Nd / 144Nd
0.51190
0.51185
0.51180
0.51175
0.51170
0.51165
0.1910 0.1920 0.1930 0.1940 0.1950 0.1960 0.1970 0.1980 0.1990 0.2000 0.2010
147Sm / 144 Nd
(f) Your Sm-Nd data for meteorites defines, in part, the Sm-Nd characteristics of the "Bulk Earth", also
referred to as CHUR (Chondritic Uniform Reservoir). Extrapolation of your isochron line for the meteorite
data (you do not need to do this) would yield an intercept value (at x = 0) which represents the initial 143Nd
/ 144Nd ratio of 0.50668 for the bulk Earth (or CHUR) at the time the planet accreted.
Given this initial 143Nd / 144Nd value, use the Sm-Nd age equation to calculate what the 143Nd / 144Nd ratio
of CHUR would be as measured today (HINT: assume an age of the Earth of 4.56 Ga (i.e., 4.56 x109
years) and use a parent/daughter ratio 147Sm/144Ndm = 0.1967 for CHUR as measured today).
143
144
143
144
(g) What is the % increase in the Nd/ Nd ratio (today) compared to the initial Nd/ Nd value
147
when the Earth formed (i.e. % increase due to the decay of Sm over 4.56 billion years)? Qualitatively,
what does your % increase indicate about the size of the half-life of 147Sm relative to the age of the Earth
(i.e. has a lot of the parent isotope of Sm decayed in 4.56 Ga)?
3
Q2. Sm-Nd Source Tracing of Ancient Rocks
147
144
Introduction - a sample of gabbro from Labrador has a present-day measured Sm/ Nd = 0.11480
143
144
and present-day measured Nd/ Nd = 0.51084. The slope of a mineral isochron for the gabbro has
yielded a crystallization age for the gabbro of 2.28 Ga (2.28 x 109 yrs) and an intercept (or initial
143
Nd/144Nd) = 0.50920
(a) Using the appropriate present-day value for CHUR from Q1 (f), calculate the present-day εNd value
for the gabbro sample. Recall the formula -
ε
Nd(t)
= { (143Nd/ 144Nd sample(t) - 143Nd/ 144Nd CHUR(t)) } * 104
( 143Nd/ 144Nd CHUR(t))
(b) Use the appropriate present-day values for CHUR from Q1 (f) and the Sm-Nd age equation below to
calculate the 143Nd / 144Nd ratio of CHUR at 2.28 Ga ago.
[143Nd/ 144Nd] m = [143Nd/ 144Nd] o + [147Sm/ 144Nd] m (e t - 1)
λ
[Hint: in this context, where t is given as a value like 2.28 Ga, the [143Nd/ 144Nd] o term in the age equation
refers to the 143Nd/ 144Nd ratio at time t (i.e. 143Nd/ 144Nd] o =[143Nd/ 144Nd] 2.28 Ga)
ε
(c) Use the data given in the introduction to Q2 and your result in (b) above to calculate the initial Nd
value for the gabbro sample at 2.28 Ga.
(d) On the isotope evolution diagram below, plot and label your 143Nd/144Nd ratios for the gabbro sample
ε
today and at 2.28 Ga. Label where your two Nd values lie on the diagram (calculated for today and for
2.28 Ga). Construct and label an evolution vector (i.e. an arrow) on the diagram that portrays the change in
143
Nd/144Nd for the gabbro sample from 2.28 Ga to present-day.
(i) Is the gabbro a product of melting of mantle or crust (i.e. what is the initial source of the melt)?
(ii) Does the data today (and your vector) indicate whether the gabbro is a melt product of upper or
lower crust?
4
147
62
Sm →143
60 Nd
Nd
Sm/Nd CHUR (orSm
PM) = 0.33
Sm/Nd bulk crust = 0.22
Sm/Nd depleted mantle = 0.36
(i.e. Sm prefers to stay in solid)
Lower crust
Bulk crust
Upper crust
or CHUR
5
Sm-­‐Nd Decay Equation and Isochron Age Calculations Sm-­‐Nd system is based on the decay of 147Sm to 143Nd from the following equation: (143Nd/144Nd)m = (143Nd/144Nd)o + (147Sm/143Nd)m(e t-­‐1) (1) where: (143Nd/144Nd)m = measured 143Nd/144Nd ratio of the sample (143Nd/144)o = initial 143Nd/144Nd ratio of the sample (or source) when the rock formed (147Sm/143Nd)m = measured 147Sm/144Nd ratio of the sample = decay constant (6.54 x10-­‐12 yr-­‐1) t = age in Ma. This is an equation of the form y = mx + b, by obtaining the slope of this line we can calculate the age λ
λ
of the sample/rock by using: m = e t –1 (2) and rearranging this we have: λ
t = ln(m+1) / λ (3) Since we are often dealing with multiple datapoints for an isochron we must calculate the “least squares” regression line for the multiple datapoints (i.e., a best-­‐fit line for the data points). The slope of this line can then be substituted into equation (2) to obtain the age of the sample. The slope of the least squares regression (m) and the y-­‐intercept (b) are given by: n ∑ XY − ∑ X ∑Y
(4) m=
2
n ∑ X 2 − (∑ X )
∑Y − m ∑ X
(5) b=
€
n
Where: X = 147Sm/144Nd (measured), Y = 143Nd/144Nd (measured), €
n = number of data points. Set up your calculations as follows
6
Epsilon Notation '
εNd = %
%&
(
(
143
143
)
Nd )
Nd / 144Nd
Nd / 144
rock ,t = 0 Ma
CHUR ,t = 0 Ma
$
− 1" × 10 4 "#
(6) where: 143Nd/144Nd
rock, t=0Ma = measured ratio of rock at present day 143Nd/144Nd
CHUR, t = 0 Ma = 0.512638 = the value of the Chondrite Uniform Reservoir at 0 Ma This is valid for t = 0 Ma but with older rocks we need to recalculate the 143Nd/144Nd ratios of both the rock and CHUR for the time in question by rearranging equation 1, inputting the age of the rock (t) and solving for (143Nd/144Nd)0 such that: (143Nd/144Nd)rock,t = (143Nd/144Nd)rock, t = 0 Ma – (147Sm/144Nd)rock, t = 0 Ma*(e t-­‐1) (7) and (143Nd/144Nd)CHUR, t = (143Nd/144Nd)CHUR,t = 0 Ma – (147Sm/144Nd)CHUR, t = 0 Ma*(e t-­‐1) (8) Since we know: (143Nd/144Nd)rock, t = 0 Ma, (147Sm/144Nd)rock, t = 0 Ma = these are known by measurement and (143Nd/144Nd)CHUR,t = 0 Ma = 0.512638 (147Sm/144Nd)CHUR, t = 0 Ma = 0.1967 We can now “back-­‐calculate” to find the initial 143Nd/144Nd rocks of the rock and CHUR. We then take the results from (7) and (8) and substitute them into equation (6) and we have “εNd at time t” or εNdt; this may also be given as εNd(@ X Ma) or εNd(t). λ
λ
7