# Sm-Nd Decay#

High-Temperature Geochemistry

# import relevant modules

%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from IPython.display import display
from math import log10, floor

# create our own functions

# function to round a value to a certain number of significant figures
def round_to_n_sf(value, no_of_significant_figures):
value_rounded = round(value, no_of_significant_figures-1-int(floor(log10(abs(value)))))
if value_rounded == int(value_rounded):
value_rounded = int(value_rounded)
return value_rounded



## Sm-Nd Decay System#

${^{147}Sm \longrightarrow {^{143}Nd} + \alpha} \qquad t_{\frac{1}{2}} = 106\,Gyr$

Both $$Sm$$ (samarium) and $$Nd$$ (neodymium) are Rare Earth Elements ($$REEs$$). In nature, both elements generally occur in dispersed form, with typical concentrations in mantle and crustal rocks of less than $$\sim20\,ppm$$.

Most $$REEs$$, including $$Sm$$ and $$Nd$$, occur as trivalent ($$3+$$) ions with ionic radii that decrease systematically with increasing atomic number, so $$Sm$$ has a smaller ionic radius than $$Nd$$.

Both $$Nd$$ and $$Sm$$ are moderately incompatible elements, but $$Nd$$ is slightly more incompatible than $$Sm$$ during mantle melting because it has a slightly larger ionic radius.

The $$REEs$$ are generally considered to be relatively resistant toward mobilization by fluids – they are fluid-immobile elements.

## Dating of Terrestrial Rocks#

The continental crust in general and siliceous rocks in particular have low and relatively uniform $$Sm/Nd$$ ratios (= parent/daughter ratio), so the $$Sm$$-$$Nd$$ system is not particularly suitable for dating such rocks. On the other hand, mafic and ultramafic rocks have variable and high $$Sm/Nd$$ ratios, so they are good candidates for $$Sm$$-$$Nd$$ dating.

Due to the fluid immobility of the $$REEs$$, $$Sm$$-$$Nd$$ ages and initial $$Nd$$ isotope compositions are not very sensitive to weathering and metamorphism.

Notice how the $$Sm$$-$$Nd$$ system is opposite to the $$Rb$$-$$Sr$$ decay system in many ways!

# Sm-Nd decay equation - very similar to the Rb-Sr one
# each return depends on what we want to find from the equation
def Sm_Nd_decay_equation(Nd143_Nd144_ratio, initial_Nd143_Nd144_ratio, Sm147_Nd144_ratio, t):
decay_const_Sm = 6.54 * 10**-12  # yr^-1  # decay constant of Sm-147
if Nd143_Nd144_ratio == '?':
return initial_Nd143_Nd144_ratio + Sm147_Nd144_ratio*(np.exp(decay_const_Sm*t)-1)
elif initial_Nd143_Nd144_ratio == '?':
return Nd143_Nd144_ratio - Sm147_Nd144_ratio*(np.exp(decay_const_Sm*t)-1)
elif Sm147_Nd144_ratio == '?':
return (Nd143_Nd144_ratio - initial_Nd143_Nd144_ratio)/(np.exp(decay_const_Sm*t)-1)


## Problem Set 5#

### Question 1#

A pigeonite basalt ($$12039$$, $$19$$) from the Moon yielded the following results:

# create a dataframe to show the obtained results
samples = ["Whole rock", "Plagioclase", "Pyroxene"]
Sm147_Nd144_ratio = [0.2090, 0.1727, 0.2434]
Nd143_Nd144_ratio = [0.513142, 0.512365, 0.513861]

dict1 = {'Samples' : samples,
'$${^{147}Sm}/{^{144}Nd}$$' : Sm147_Nd144_ratio,
'$${^{143}Nd}/{^{144}Nd}$$' : Nd143_Nd144_ratio}
df1 = pd.DataFrame(dict1)
df1.loc[:, '$${^{147}Sm}/{^{144}Nd}$$'] = df1['$${^{147}Sm}/{^{144}Nd}$$'].map('{:g}'.format)
df1.loc[:, '$${^{143}Nd}/{^{144}Nd}$$'] = df1['$${^{143}Nd}/{^{144}Nd}$$'].map('{:g}'.format)
display(df1.style.hide_index())

Samples $${^{147}Sm}/{^{144}Nd}$$ $${^{143}Nd}/{^{144}Nd}$$
Whole rock 0.209 0.513142
Plagioclase 0.1727 0.512365
Pyroxene 0.2434 0.513861

a) Plot the data in an isochron diagram. Scale the y-axis from about $${^{143}Nd}/{^{144}Nd}$$ = $$0.5080$$ to $$0.5150$$ and the x-axis from $${^{147}Sm}/{^{144}Nd}$$ = $$0$$ to about $$0.3$$.

b) Determine the age and initial $${^{143}Nd}/{^{144}Nd}$$ ratio of this rock from the slope and y-intercept of the isochron.

c) Calculate the $$Nd$$ isotope ratio of CHUR for the age of rock.

d) Express the initial $$Nd$$ isotope composition of the rock as an $$\epsilon_{Nd}$$ value relative to $$CHUR$$. What does the initial $$\epsilon_{Nd}$$ value tell you, if you assume that the Moon as a whole has chondritic $$Sm/Nd$$ and $$Nd$$ isotope ratios?

Solution:

a) See below

# Question 1a

# set figure size
plt.figure(figsize=(8,6))
# Plot data points
plt.plot(Sm147_Nd144_ratio, Nd143_Nd144_ratio, 'ro', label="Data points")
# plot isochron by fitting a polynomial degree 1 - ie a straight line.
poly_coeffs=np.polyfit(Sm147_Nd144_ratio, Nd143_Nd144_ratio, 1)
p1 = np.poly1d(poly_coeffs)
slope = poly_coeffs           # e^(\lambda t) - 1
y_intercept = poly_coeffs   # initial Nd143/Nd144 ratio
x = np.linspace(0, 0.3, 10)  # Sm147/Nd144 ratio
plt.plot(x, p1(x), 'b', label="linear fit ($y = %gx + %g$)" % (round_to_n_sf(slope, 3), round_to_n_sf(y_intercept, 3)))
# label and title the plot
plt.xlabel('${^{147}Sm}/{^{144}Nd}$')
plt.ylabel('${^{143}Nd}/{^{144}Nd}$')
plt.title('Isochron plot', fontsize=14)
plt.legend(loc='best', fontsize=10)

<matplotlib.legend.Legend at 0x1f58e092ca0> b) Now we have already got the slope ($$m$$) and y-intercept ($$b$$) of the isochron, so we can estimate the age ($$t$$) and initial $$Nd$$-$$143/Nd$$-$$144$$ ratio ($$({^{143}Nd}/{^{144}Nd})_0$$) since

$m = e^{\lambda t} - 1 \quad \longrightarrow \quad t = \frac{1}{\lambda}\ln(m+1)$
$b = \left(\frac{^{143}Nd}{^{144}Nd}\right)_0$
# Question 1b

decay_const_Sm = 6.54 * 10**-12  # yr^-1  # decay constant of Sm-147
t = (1/decay_const_Sm) * np.log(slope + 1)  # age
b = y_intercept  # initial Nd143/Nd144 ratio
print("The age implied by the isochron is %.2e yr." % t)
print("The initial Nd-143/Nd-144 ratio is %.6f." % b)

The age implied by the isochron is 3.20e+09 yr.
The initial Nd-143/Nd-144 ratio is 0.508713.


As the distribution and evolution of $$Sm$$ and $$Nd$$ in the Earth are not explained in this page where we focus on quantitative parts, we will recall the concepts of CHUR and $$\epsilon_{Nd}$$ from the lecture slide before attempting questions c and d.

CHUR (Chondritic Uniform Reservoir) is defined by the average present-day $$Sm/Nd$$ ratio and $$Nd$$ isotope composition of chondritic meteorites. It is representative for the $$Nd$$ isotope composition and evolution of the bulk Earth and bulk silicate Earth. The present-day $$Nd$$ isotope composition of chondrites is $$({^{143}Nd}/{^{144}Nd})_{CHUR} = 0.512638$$. The average $$({^{147}Sm}/{^{144}Nd})_{CHUR} = 0.1967$$.

It is convenient to consider past and present variations in $$Nd$$ isotope compositions relative to the isotopic evolution of CHUR. This is done using the $$\epsilon$$ notation:

$\epsilon_{Nd} = \frac{({^{143}Nd}/{^{144}Nd}) - ({^{143}Nd}/{^{144}Nd})_{CHUR}}{({^{143}Nd}/{^{144}Nd})_{CHUR}} \times 10^4$

$$\epsilon_{Nd}$$ values denote relative differences in $$Nd$$ isotope compositions (relative to CHUR) in parts per $$10,000$$.

c) From b), the age of the basalt is $$3.20\,Gyr$$. So, in this question, we are going to calculate $$({^{143}Nd}/{^{144}Nd})_{CHUR}$$ at $$3.20\,Gya$$ using the $$Sm$$-$$Nd$$ decay equation (very similar to $$Rb$$-$$Sr$$ decay equation):

$\frac{^{143}Nd}{^{144}Nd} = \left(\frac{^{143}Nd}{^{144}Nd}\right)_0 + \frac{^{147}Sm}{^{144}Nd}(e^{\lambda t} - 1)$

In this case:

$$\quad {^{143}Nd}/{^{144}Nd} = ({^{143}Nd}/{^{144}Nd})_{CHUR,\,present} = 0.512638$$

$$\quad ({^{143}Nd}/{^{144}Nd})_0 = ({^{143}Nd}/{^{144}Nd})_{CHUR,\,3.20\,Gyr} =$$ haven’t know yet

$$\quad {^{147}Sm}/{^{144}Nd} = ({^{147}Sm}/{^{144}Nd})_{CHUR,\,present} = 0.1967$$

$$\quad \lambda = 6.54 \times 10^{-12}\,yr^{-1}$$

$$\quad t = 3.20 \times 10^9\,yr$$

# Question 1c

# Given values
Nd143_Nd144_ratio = 0.512638
Sm147_Nd144_ratio = 0.1967
t = 3.2 * 10**9  # yr
initial_Nd_ratio_CHUR = Sm_Nd_decay_equation(Nd143_Nd144_ratio, '?', Sm147_Nd144_ratio, t)  # Nd isotope ratio of CHUR at 3.20 Gya
print("The Nd isotope ratio of CHUR at 3.20 Gya is %g." % round_to_n_sf(initial_Nd_ratio_CHUR, 6))

The Nd isotope ratio of CHUR at 3.20 Gya is 0.508478.


d)

From b), the initial $$Nd$$ isotope ratio of the rock is $$0.508713$$.

From c), the initial $$Nd$$ isotope ratio of CHUR is $$0.508478$$.

Calculate the initial value of $$\epsilon_{Nd}$$ (at $$t = 3.20\,Gyr$$)

# function to calculate epsilon Nd
def epsilon_Nd(Nd_ratio_rock, Nd_ratio_CHUR):
return (Nd_ratio_rock - Nd_ratio_CHUR)/Nd_ratio_CHUR * 10000

# Question 1d

# Given values
initial_Nd_ratio_rock = 0.508713
initial_Nd_ratio_CHUR = 0.508478
print("The initial differences in Nd isotope compositions (relative to CHUR) of the rock in parts per 10,000 is %g." \
% round_to_n_sf(epsilon_Nd(initial_Nd_ratio_rock, initial_Nd_ratio_CHUR), 3))

The initial differences in Nd isotope compositions (relative to CHUR) of the rock in parts per 10,000 is 4.62.


The positive $$\epsilon_{Nd}$$ initial value means that the rock comes from a depleted source. The depleted source is probably the lunar mantle, which was depleted by the formation of the lunar crust.

### Question 2#

Neodymium model ages or crustal residence ages are obtained by calculating the intersection of the $$Nd$$ isotope evolution of a rock sample with the $$Nd$$ isotope evolution of a reservoir (RES) that is assumed to have a composition akin to CHUR (for $$t_{CHUR}$$) or a depleted mantle (DM) composition (for $$t_{DM}$$).

The respective Nd isotope evolutions evolution curves are given by:

Sample:

$\left(\frac{^{143}Nd}{^{144}Nd}\right)_{Sam} = \left(\frac{^{143}Nd}{^{144}Nd}\right)_{Sam,0} + \left(\frac{^{147}Sm}{^{144}Nd}\right)_{Sam}(e^{\lambda t} - 1)$

CHUR or DM reservoir:

$\left(\frac{^{143}Nd}{^{144}Nd}\right)_{Res} = \left(\frac{^{143}Nd}{^{144}Nd}\right)_{Res,0} + \left(\frac{^{147}Sm}{^{144}Nd}\right)_{Res}(e^{\lambda t} - 1)$

The intersections is obtained by equating $$\left(\frac{^{143}Nd}{^{144}Nd}\right)_{Sam,0} = \left(\frac{^{143}Nd}{^{144}Nd}\right)_{Res,0}$$

After rearranging this yields for $$t$$:

$t = \frac{1}{\lambda} \ln\left[\frac{\left(\frac{^{143}Nd}{^{144}Nd}\right)_{Sam} - \left(\frac{^{143}Nd}{^{144}Nd}\right)_{Res}}{\left(\frac{^{147}Sm}{^{144}Nd}\right)_{Sam} - \left(\frac{^{147}Sm}{^{144}Nd}\right)_{Res}} + 1\right]$

Calculate the $$t_{DM}$$ for a sample characterized by $${^{147}Sm}/{^{144}Nd} = 0.102$$ and $${^{143}Nd}/{^{144}Nd} = 0.511552$$. The depleted mantle is assumed be characterized by $${^{147}Sm}/{^{144}Nd} = 0.222$$ and $${^{143}Nd}/{^{144}Nd} = 0.513114$$.

# Question 2

# decay constant of Sm-147
decay_const = 6.54 * 10**-12  # yr^-1
# isotopic ratios of the sample
Sm147_Nd144_ratio_sample = 0.102
Nd143_Nd144_ratio_sample = 0.511552
# isotopic ratios of the depleted mantle
Sm147_Nd144_ratio_DM = 0.222
Nd143_Nd144_ratio_DM = 0.513114
# calculate t
t = (1/decay_const) * np.log(((Nd143_Nd144_ratio_sample - Nd143_Nd144_ratio_DM)/(Sm147_Nd144_ratio_sample - Sm147_Nd144_ratio_DM)) + 1)

The Nd model age is 1.98e+09 yrs.