MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Problem Set 2 Solutions
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1. Diffusion:You are studying Ni diffusion in the insulator MgO fas a potential dilute magnetic semiconductor (i.e.,
spintronic material – Ni has an unpaired electron – potential quantum computing applications). You perform
experiments at various temperatures and determine the diffusivity at these temperatures. The data you have acquired
are tabulated below.
Diffusivity (
cm 2
)
s
Temperature (K)
7.48x10-16
1010
2.78x10-14
1190
1.66x10-13
1305
Using the data above, determine the following parameters for Ni diffusion in MgO:
a. Determine the activation energy in units of eV: Comment on whether or not the value you obtain is physically
valid (hint: look up other values of diffusion activation energies and come to a conclusion to the range of
activation energies for diffusion).
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
To find the activation energy, the slope of the ln of the diffusivity versus the inverse of
temperature should be found. With three data points, there are three ways to determine the slope
assuming the data is linear (which is part c).
∆y ln D2 − ln D1
=
1 1
∆x
−
T2 T1
m
Slope:=
=
Let: T2'
1
1
=
& T1'
T2
T1
So:
−k B ⋅ slope
EA =
=
−kB ⋅
(
ln D2 − ln D1
T2' − T1'
(
)
=
− 8.6173 ×10−5 eV
K
) ⋅ ln(DT −−Tln )D
2
'
2
1
'
1
Implementing this final equation, we actually use D1 and D3 and T1 and T3. Substituting the given
values of diffusivity and temperature, we thus have:
Diffusion activation energies range from 100’s of meV’s to ~3eV, so this activation energy seems
physically valid. A nice place to examine activation energies is in Smithell’s Metals Reference Book –
although for metals, it provides a range of diffusivity data.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
b. Determine the pre-exponential of the diffusivity in units of cm2/s.
We use the activation energy found in part a to find Do
.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
c. Plot the data on two different Arrhenius plots (be sure to provide a figure captions for both):
i. using a log10 scale
Diffusivity (cm2/s)
10-12
10-13
10-14
10-15
10-16
7.5x10-4 8.0x10-4 8.5x10-4 9.0x10-4 9.5x10-4 1.0x10-3
1/Temperature (K-1)
A log10 plot of the diffusivity versus inverse temperature data in the table of
problem 2. The first plot was created using Origin while the second was created
using Mathematica. The figures show a linear relationship between the log10 of
diffusivity and the inverse of temperature indicating Arrhenius behavior.
ii.
using a loge
iii.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
A loge plot of the diffusivity versus inverse temperature data in the table of
problem 2. The figure shows a linear relationship between the loge of diffusivity
and the inverse of temperature indicating Arrhenius behavior.
Although not required (but would be considered extra credit), a fit was performed
with the NonlinearmodelFit[ ] command using Mathematica extracting an activation
energy of EA = 2.09 eV which is very close to the 2.08 eV that was calculated in
part a. The goodness of fit parameters, R2 and adjusted R2, were determined to
be:
R2 = 0.99998
Adjusted R2 = 0.99997
Both goodness of fit parameters show an excellent fit.
2. Na is a monovalent metal (BCC) with a density of 0.9712 g cm-3. Its atomic mass is 22.99
g mol-1. The drift mobility of electrons in Na is 53 cm2 V-1 s-1.
a. Consider the collection of conduction electrons in the solid. If each Na atom donates one electron
to the electron sea, estimate the mean separation (in Angstroms) between the electrons. (Note:
if n is the concentration per volume of particles, then the particles’ mean separation d = 1/n1/3.)
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
b. Estimate (i.e., calculate) the mean separation between an electron (e-) and a metal ion (Na+) in
Angstroms, assuming that most of the time the electron prefers to be between two neighboring
Na+ ions. Calculate the approximate Coulombic interaction energy in eV between an electron
and an Na+ ion.
Answer: Note: In the answer below, FCoulomb should be ECoulomb.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
c. How does this electron/metal-ion interaction energy (in eV) compare with the average thermal
energy per particle (in eV) at room temperature (assume 300K), according to the kinetic
molecular theory of matter?
d.
Calculate the electrical conductivity of Na (in units of Ω-1 m-1) and compare this with the
experimental value of 2.1 × 107 Ω-1 m-1 and comment on the difference. That is, compare values
by examining the accuracy (how is accuracy determined?) of the experimental value compared
to your calculated value.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
To determine the accuracy, use the relative error or % error 1, given by:
experimental value - known value
×100
known value
% Error
( 2.1×10 -2.16 ×10 ) ( Ωcm )
7
7
2.16 ×107 ( Ωcm )
−1
−1
×100
−0.06
×100
2.16
6
=
−
×100
216
1
=
− ×100
36
=
= −2.77%
A negative ~3% error is quite small. Having an experimental value so close to a
calculated value shows that the experimental value is physically substantiated by the
model.
Nota Bene: If one takes the Na+-Na+ separation 2R to be roughly the mean electronelectron separation, then this is 0.37 nm and close to d = 1/(n1/3) = 0.34 nm. In any
event, all calculations are only approximate to highlight the main point. The
interaction PE is substantial compared with the mean thermal energy and we cannot
use (3/2)kT for the mean KE!
1
Found in many Analytical Chemistry textbooks – e.g., Skoog & West & Holler, Fundamentals of Analytical
Chemistry, (Saunders College Publishing, New York; 1988) p. 9-11.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
3. The resistivity of pure aluminum (assume no
Table 1: Resistivity Data for Aluminum
defects are present) as a function of
Temperature (°C)
Resistivity (μΩcm)
temperature will be explored in this problem.
20
2.67
27
2.733
The data in Table 1 lists the resistivity of
100
3.55
aluminum at various temperatures. You will fit
127
3.87
the data (both Mathematica and Matlab both
200
4.78
have nonlinear regression functions and very
227
4.99
300
6.99
good statistical analysis of the fit), examine the
327
6.13
“goodness of the fit”, and assess the “physical
400
7.3
validity of the fit”. Before you do this, you will
427
8.7
thoroughly define both “goodness of fit” and From Smithells Metals Reference Book 8th Edition
“validity of fit”. Hint: Use the two handouts Section 14-3
provided with problem set 2 on course website:
[1] H. Motulsky & A. Christopoulos, “Fitting Models to Biological Data using Linear and
Nonlinear Regression -A practical Guide to Curve Fitting”, Version 4 (GraphPad Prism; 2003) p.
1-351.[2] “Interpreting Regression Results” - from OriginLabs.
a. Mattheissen’s rule gives a linear relationship in temperature for the resistivity of pure metals.
In part b, you will prove that this is the case for the data given in Table 1 by fitting the data to
Mattheissen’s equation. For part a, consider that you are teaching an algebra class and you are
giving a lecture using this problem as an application problem for the subject you are lecturing
on called “Lines and Their Uses in Engineering”. For the algebra students to which you will
lecture, provide a bulleted list of steps (5-6 steps) necessary to demonstrate this application
starting with Mattheissen’s rule. Include mathematical relationships showing the relation of
Mattheissen’s rule to that of a line, defining each of its components, in the steps and connect
Mattheissen’s rule to the line concept so that the students understand the application to “Lines
and Their Uses in Engineering”. In your steps, include the concepts of examining the
“goodness of the fit”, and assessing the “physical validity of the fit”. Hint: remember that this
is a pure metal, so think about what happens at the origin of the plot when you fit the data.
•
•
•
•
Mathiessen’s rule: ρ = ρT + ρI = AT + ρI
Equation of a line: y = m x +b where m = slope and b is the y-intercept
Comparing the two equations, one sees that the slope m = A and the yintercept b = ρI
Mathiessen’s rule for a pure metal: ρI = 0, so ρ = ρT = AT. If ρI is used, its physical
validity would need to be assessed.
•
•
•
Now one can use the equation of a line to fit the data and the y-intercept
should be zero and thus the line should go through the origin.
The fitting parameter is A, which is the slope, and can be extracted from the
fit.
A should be compared to a physical aspect of the resistivity of the data. In
this case, mobility and A are related. Mobility can be
b. Approaches used for fitting data are linear and non-linear regression. A typical regression
method is called “least squares method of fitting”. Define the method of “least squares method
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
fitting” including the assumptions. Use a graph schematic to help you illustrate the “least
squares method of fitting” (i.e., draw a graph). List the assumptions of the “least—squares
method of fitting”. Cite your sources.
[1]
In the figure above, note the following:
yi : the y-value of the data point
yˆi : the y-value of the fit
yi : the average y-value
http://collum.chem.cornell.edu/documents/Intro_Curve_Fitting.pdf
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Assumption of the nonlinear least-squares method: (taken verbatim from [2] page
30).
(1) x is known precisely, all the error is in y.
(2) The variation in y follows a known distribution. Almost always, this is assumed to
be a Gaussian distribution.
(3) Standard nonlinear regression assumes that the amount of scatter (the standard
deviation of the residuals) is the same all the way along the curve. This assumption
of uniform variance is called homoscedasticity. Weighted nonlinear regression
assumes that the scatter is predictably related to the Y value.
(4) Observations are independent. When you collect a y value at a particular value of
x, it might be higher or lower than the average of all y values at that x value.
Regression assumes that this is entirely random. If one point happens to have a
value a bit too high, the next value is equally likely to be too high or too low.
[1] “Interpreting Regression Results” - from OriginLabs.
[2] H. Motulsky & A. Christopoulos, “Fitting Models to Biological Data using Linear and
Nonlinear Regression -A practical Guide to Curve Fitting”, Version 4 (GraphPad Prism;
2003) p. 1-351.
[3] D. Skoog, D. West, & F. Holler, “Fundamentals of Analytical Chemistry, 5th Ed. (Sanders
College Publishing, 1988) Ch. 2.
c. When fitting data, one often examines the “goodness of the fit” and “physical validity of the
fit”. Define “goodness of the fit” and “physical validity of the fit” and describe the difference
between “goodness of the fit” and “physical validity of the fit”. List (i.e., give examples) of
several methods used for “goodness of the fit” and for “physical validity of the fit” and define
two of the examples each for “goodness of the fit” and for “physical validity of the fit”. Be sure
to cite your sources.
•
Goodness of Fit: When data is fit or described by a mathematical relation, a
comparison of (x,y) data to the (x,y) of the fit can be performed (i.e.,
regression method) to see how close the fit data point is to the actual data
point. This is typically done using non-linear regression. The closer the
difference, the better (i.e.,, “gooder”) the fit. From the figure above, we see
that:
yi : the y-value of the data point
yˆi : the y-value of the fit
yi : the average y-value
Using these definitions, we can then define the following:
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
o “Total sum of squares” (TSS) about the mean (variation between data
TSS
points & Mean): =
n
∑ ( yi − yi )
2
i =1
Several methods used to check the “Goodness of Fit” include:
o Residuals or Residual sum of squares or reduced chi-square:

n
∑ ( yˆ − y )
SSreg
Regression sum of squares: =
i
i =1
2
i
is the portion of variation
that is explained by the regression model

RSS
Residual sum of squares: =
n
∑ ( y − yˆ )
i =1
i
i
2
is the portion of variation
that is not explained by the regression model
Residuals should be random. If they are not random, the linear fit should be
replaced with a nonlinear regression fit.
o Coefficient of Determination, R2:
n

∑
SS reg
RSS
R = =
=
1−
1 − i =1
n
TSS
TSS
( yi − yˆi )
2
2
∑ ( yi − yi )
2
i =1
The coefficient of determination, R2, is a measure of the goodness of
a fit. It is determined by taking the difference between the value 1 and
the ratio of two sum-of-squares values: SSreg and SStot. SSreg is the sum
of the square of the difference between the mean y-value and the
actual y-value. SStot is the sum of the square of the difference between
the actual y-value and the best fit curve. Hence, the closer R2 is to 1,
the better the fit.[1]
o Adjusted Coefficient of Determination, Adjusted R2 – When adding fitting
parameters to the model, R2 will increase, however, this does not suggest
a better fit. To avoid this effect, we can examine the adjusted R2:
SS reg
RSS df error
2
= 1−
 R =
TSS
TSS dftotal

 Like R2, the closer AdjR2 is to 1, the better the fit.
o Confidence levels: we need to define the following:
N
 Population standard deviation: σ =
12
∑(x − µ)
i =1
i
N
2
[3]
MSE 410-ECE 340
Electrical Properties of Materials
•
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
where N is the sample number and μ is the mean of the population.
σ is used when N is infinitely large.
N
 Sample standard deviation: s =
∑(x − x )
i =1
2
i
N −1
• where N is the sample number and x is the mean of the sample
population. S is used when N is finite.
When s is a good approximation to σ, then we can define the confidence
limits (in %) by the percent area under a Gaussian curve (figure 2-6). The
x-axis is defined by the difference between the two means and normalized
by the population standard deviation. Hence, each unit on the x-axis is a
standard deviation σ, from the difference between the two means. So if
the deviation is 0.67σ, this equates to a 50% confidence level that the
experimentally determined mean, x , is within that limit.
[3]
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Figure: 95% confidence and prediction limits and intervals of a data set in
which xp is the point of interest. [2]
Physical validity of fit: When data is fit or described by a mathematical
relation, the independent variable (x) is varied and the dependent variable
(y) is determined using a regression method. Other parameters in the
mathematical relation can be used to help fit the data. These relations can
have physical meaning. The physical validity of these variables should be
checked to see if they make physical sense.
o If you have many fits, than one can examine the accuracy of the
fit parameters using the mean, standard deviation, variance, or the
coefficient of variation.
o Precision: used for single values of a parameter and when a known
value for the parameter exists
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MSE 410-ECE 340
Electrical Properties of Materials


School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Relative error =
(fit parameter value – known
value)/(known value) [3]
Absolute error = (fit parameter value – known value) [3]
d. Using the previous parts of this problem, prove that Matthiessen’s rule gives a linear
relationship in temperature for the resistivity of pure metals using the data given in Table 1 by
fitting the data to Matthiessen’s equation. Provide the equation of the fit, define each variable
(i.e., dependent variable, independent variable, fit parameters and names of each) and provide
units of cm·V-1·s-1. Note: Convert the temperature to Kelvin before fitting the data. Show both
the data set and the fit on the same plot. Note: Ensure that the scale of the axis is such that it
allows any divergence of the fit from the data to be observed.
Matthiessen's rule for conduction in metals
ρ(T) = AT + B (Kasap p. 128, equation [2.17]),
ρ(T) is the resistivity and the dependent variable
T is the temperature and the independent variable
A is the slope and a fit parameter
B is the y-intercept and a fit parameter and is zero for pure metals (Kasap p.
128).
Although it is not required, the data was fit with two equations.
First approach: ρ(T) = AT + B where B = 0 (pure metals)
First approach: ρ(T) = AT + B where B not equal to 0 assuming that the data does
not correlate to the origin and thus not constraining that the fit goes through
the origin.
The plot, fit and extracted parameters are below.
15
MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
The plot shows the resistivity for aluminum (points are data) as a function of
temperature. The fits (lines) were generated using Matheissen’s rule for either a
pure metal (B=0, blue line) and by allowing B to vary (green line) which produces a
better fit.
For ρ(T) = AT :
ρ(293K) = 3.17µΩ − cm and A = 0.0108µΩ − cm − K^(−1)
For ρ(T) = AT + B:
Where ρ(293K) = 2.49µΩ − cm − K^(−1) and A = 0.0137 μΩ-cm and B = -1.52561 μΩcm.
A negative B does not make sense, but not constraining the fit to go through the
origin provides a superior R2. See part e.
e. Determine the “goodness of fit” using two of the examples you provided earlier in the problem.
Using the outcome of your chosen “goodness of fits”, argue whether or not the model should
be used to fit the data.
R2 = 0.989 for ρ(T) = AT
Where ρ(293K) = 3.17µΩ − cm and A = 0.0108µΩ − cm − K^(−1)
R2 = 0.995 for ρ(T) = AT + B
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Where ρ(293K) = 2.49µΩ − cm − K^(−1) and A = 0.0137 μΩ-cm and B = -1.52561 μΩcm.
A negative B does not make sense, but not constraining the fit to go through the
origin provides a superior R2. That is, from both the plot and the R2’s , it can be seen
that ρ(T) = AT + B gives the superior fit although it does not assume B = 0. Both R2’s
are fine.
R2 for the fit performed above is 0.989 for B=0 or 0.995 for B not equal to zero,
which indicates that the fit models the data with an accuracy of 98.9% or 99.5%,
respectivel. The above plot shows both the data and the fits, further reinforcing
the validity of the fit. Notice the fits models the behavior of the data well in that
it does not deviate away from any data point(s).
Plot of the Residuals of the fits are below. A residual is the difference between
each original y-data point and its fitted value. The smaller the difference, the better
the fit.[1-2]
In assessing residuals, one looks for patterns. In the first residuals plot, the first
6 residuals show all negative indicating a linear fit may be ill-advised and a
NonLinearFitModel[ ] fit is more appropriate.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
In the second residuals plot, the trend seems to show periodicity from positive to
negative which may indicate that a linear fit may be ill-advised and a
NonLinearFitModel[ ] fit is more appropriate.
f. Discuss the “physical validity of your fit” using the fit model you found earlier in the problem
by calculating the electron drift mobility at 293K and compare the result to table 2.7 in your
textbook by Kasap. Note: Your answer should have units of cm·V-1·s-1.
There are two ways to check the physical validity of your fit:
I. the approach outlined in part f.
II. relating A and B to the TCR, αo, as well as to ρo and To.
A = ρoα o
=
B ρo (1 − α oTo )
So if one knows ρo and To, then one can find αo for Al in Table 2.1 on page 129 of
Kasap and calculate A. That A could be compared to the extracted A.
Here, we shall use approach I.
From equation 2.17 on page 128 in Kasap, it is known that ρ=AT+ρo.
Where:
ρ is the resistivity
ρo is the resistivity at 0°C
A is the constant of proportionality (the slope of the fit found in the Data Analysis
section)
T is the temperature
It is assumed that the aluminum is pure (no impurities). Therefore, Matthiessen's rule
may be used and is given by:
ρtotal(T)= ρT +ρI (Kasap, Page 127, equation 2.15 and 2.16)
which can be written as:
ρtotal(T)= ΑΤ +Β (Kasap, Page 128), equation 2.17)
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
In pure metals, we know that ρI=0 (thus, B=0) (Kasap, Page 127 and 128), equation
2.15), so we can write:
ρtotal(T) = ΑT (ρI=0 or B=0 in pure metals).
We know that σ=1/ρ, and that σ=enµd.
∴ µd=1/(e n ρtotal(T)) = 1/(e n ΑT).
n is the number of free carriers per unit volume (page 119 Kasap).
From example 2.2 on page 119, it is known that the number of carriers per volume is:
dN A
n=v
M atomic
where:
d is the density (g/cm3)
NA is Avogadro's Number (mol-1)
Mat is the atomic mass (g/mol)
v is the number of valence electrons ( electrons/cm3).
One can also use dimensional analysis to determine n if given d, NA, Mat and v.
Using Mathematica, n = 1.81 x 1023 electrons/cm3.
The drift mobility, µmob, is given by: µdrift (T ) =
1
cm 2
in
units
of
e ⋅ n ⋅ ρ (T ) ⋅10−6
Vs
cm 2
Using Mathematica: µdrift (T ) = 13.9
Vs
So the electron drift mobility in aluminum as calculated using the fit is 13.9
cm2/(V·s). According to table 2.7 in Kasap, the measured drift mobility of electrons
in aluminum at 293K is ~ 12 cm2/(V s). The % error, calculated using Mathematica,
is given by:
% error =
µactual − µ fit
×100 = −15.6%
µactual
A -15.6% percent error as determined using the value listed in Kasap, is less than
20%, which indicates that the mobility extracted by the fit is physically valid; in
other words, it is within an order of magnitude of the measured value and not an
unrealistic value. The ~16% error is somewhat high but the extracted mobility is
certainly fine to use when assessing whether or not the material should be used
for electronic applications.
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
4. In Kasap, extract the data from the figure 2.11 and perform the following using a mathematical
program. Provide a figure caption for all figures. Perform the following:
a. Plot the data as seen in figure 2.11.
Data (green solid circles) of resistivity versus fractional percent of Ni in a Cu matrix.
Data extracted from figure 2.11 of Kasap.
b. Fit the data using Nordheim`s rule and determine the Nordheim coefficient and plot both the
data and the fit on the same plot. Evaluate the “goodness of your fit” and comment. your fit and
you need to evaluate your fit using statistical means.
Quadratic fit:
Data (green solid circles) of resistivity versus fractional percent of Ni in a Cu matrix
and a quadratic fit (using Mathematica) to the data (green solid line). Data extracted
from figure 2.11 of Kasap.
Equation of Quadratic fit: -0.329 + 1860.51 XNi – 1794.96 (X Ni)2
Note that the Nordheim coefficient cannot be found in this manner because
Nordheim’s relation has the functionality of:
CXNi-C(XNi)2
which is not the same as the quadratic formula above. For the quadratic formula,
note that: 1860.51 ≠ 1794.96 . So, we must use another approach.
Fitting using the Nordheim’s relation:
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MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
CXNi-C(XNi)2
So we are using C as a fitting variable. We obtain:
Data (green solid circles) of resistivity versus fractional percent of Ni in a Cu matrix
and a Nordheim’s relation fit (using Mathematica) to the data (blue solid line). Data
extracted from figure 2.11 of Kasap.
Equation of C XNi (1-XNi) fit: 1916.5 (1-XNi)XNi
Nordheim’s Coefficient is the slope: 1916.5 nΩ m
Note that the quadratic fit is better than the Nordheim’s relation fit.
21
MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Using Origin:
Data
Quadratic Fit
Nordheim Fit
500
Resistivity (nΩ.m)
400
Model: Nordheim
Weighting: y No weighting
Chi^2/DoF
= 1068.51446
R^2
= 0.96475
C = 1916.50119 +/-61.51772
300
Quadratic Fit
Y = A + B1*X + B2*X^2
Parameter
Value Error
-----------------------------------------------------------A
-0.32902
14.78471
B1
1860.50832 88.82537
B2
-1794.95655 90.83098
------------------------------------------------------------
200
100
0
0.0
R-Square(COD)
SD
N
P
-----------------------------------------------------------0.97992
27.27085
12
<0.0001
------------------------------------------------------------
0.2
0.4
XNi
0.6
0.8
1.0
Data (green solid circles) of resistivity versus fractional percent of Ni in a Cu matrix
and fit (using Origin Graphical Analysis software) to a quadratic equation (green line)
and the Nordheim’s relation (blue solid line). Data extracted from figure 2.11 of
Kasap.
c. Now plot the data linearly. You will need to determine how to linearize the equation (hint: its
not taking the log of the equation).
Data (red solid circles) of resistivity versus XNi (1-XNi) where XNi is the fractional
percent of Ni in a Cu matrix (using Mathematica). Data extracted from figure 2.11
of Kasap.
22
MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
d. Fit the data using a linear fit and determine Nordheim`s coefficient. Yes, you need to show the
plot of the data and the fit and you need to evaluate your fit.
Equation of linear fit: 0.361 + 1890.1𝑥𝑥
Nordheim′ s Constant is the slope: 1890.1nΩ m
Data (red solid circles) of resistivity versus XNi (1-XNi) where XNi is the fractional
percent of Ni in a Cu matrix and a Nordheim’s relation fit (using Mathematica) to
the data (red solid line). Data extracted from figure 2.11 of Kasap.
Using Origin:
500
400
Resistivity (nΩ.m)
Data
Linear Fit
300
Linear Regression for Data5_B:
Y=A+C*X
200
Parameter
Value Error
-----------------------------------------------------------A
0.36054
17.70849
C
1890.05957 113.05993
------------------------------------------------------------
100
0
R
SD
N
P
-----------------------------------------------------------0.98599
30.82867
10
<0.0001
------------------------------------------------------------
0
0.05
0.10
0.15
XNi(1-XNi)
23
0.20
0.25
MSE 410-ECE 340
Electrical Properties of Materials
School of Materials Science & Engineering
Fall 2016/Bill Knowlton
Data (red solid circles) of resistivity versus XNi (1-XNi) where XNi is the fractional
percent of Ni in a Cu matrix and a Nordheim’s relation fit (using Origin) to the data
(red solid line). Data extracted from figure 2.11 of Kasap.
e. Assess the “physical validity of your fits” by comparing (then commenting on) the differences
or similarities of the Nordheim coefficients you determined in b and d. Compare your values of
the Nordheim coefficient to those in the literature. Comment on the comparison.
Fitting the data using the Nordheim’s rule, we obtained 1916 nΩ m (both
Mathematica & Origin) while the linear fit gave 1890 nΩ m (both Mathematica &
Origin). On page 136, table 2.3 of Kasap, for dilute alloys, Nordheim’s coefficient
for Ni in a Cu matrix is listed as 1200 nΩ m. Examined below is the physical validity
of both the linear fit and the Nordheim Rule fit.
Linear Fit:
experimental value - known value
×100
known value
(1890-1200 )( Ωnm ) ×100
=
1200 ( Ωnm )
% Error
= 57.5%
Nordheim Rule fit:
experimental value - known value
×100
known value
(1960-1200 )( Ωnm ) ×100
=
1200 ( Ωnm )
% Error
= 59.71%
Obviously, the difference between the Nordheim coefficient obtained from fitting
and the one from literature are significantly large. The most likely reason is that for
dilute systems, the resistivity follows the functionality:
ρ = CoX
since X is very small and so 1- X ~1. If we compare the above equation to:
ρ = CX(1-X)
Co is actually C(1-X), thus C = Co/(1-X).
The R2 value of 0.985 for the linear fit is superior to the R2 value of 0.97 for the
fit using the Nordheim relation. So C obtained via the linear fit is probably a
better estimate.
24