Course 12.141: Electron Microprobe
Analysis – Hints and solution for Problem
Set 4
(1) Note that in the equations, if “i” is the element being considered, “j” stands for all the other elements in the
sample. Our samples have only two elements, Fe and Ni. So, when calculating parameters for Fe, “i” is Fe
and “j” is Ni. For example, Si for Fe will be,
SFe = CNi * SFe-Ni
where, SFe-Ni = (const) [(2ZNi/ANi )/(E0+Ec(Fe K-shell))]ln[583(E0+Ec(Fe K-shell))/JNi ] See p. 14, Eq 3.12 of Course Notes. (2) When calculating a parameter for the specimen, you need to find a concentration-weighted average. For
example,
(  )
FeK
specimen
= CFe * ( 
FeK
+ CNi * ( 
FeK
 ) Ni
Note when the concentrations of Fe and Ni are equal, ( 
(  )
FeK
specimen
= [ (
 ) Fe
FeK
 ) Fe
+ (
FeK
 ) Ni
FeK

) specimen
becomes a simple average,
]/2
Also note that the standards are pure metals. So,
(  )
FeK
Fes tan dard
= (
 ) Fe
(  )
FeK
Ni s tan dard
= (
 ) Ni
FeK
= 71.4 cm2/g
FeK
= 90 cm2/g (3) Note that fluorescence occurs only when the energy of the X-ray being absorbed is higher than the critical
excitation energy of the atomic shell of the element it is fluorescing. Since,
EFeK (6.404 keV) < Ec (Ni K-shell) (8.332 keV), FeK cannot fluoresce Ni. So the fluorescence correction factor of Ni is equal 1. (4) Your graphs should show the following:
(A) Atomic number correction deviates more and more from 1 as the composition of the sample deviates
more and more from the pure element standards.
(B) There is a high positive absorption correction for NiK at high Fe contents because NiK is absorbed
in Fe.
N. Chatterjee: January 2012, MIT, Cambridge, MA, USA
1
(C) There is a high negative fluorescence correction for FeK at high Ni contents because NiK
fluoresces FeK.
Following are the graphs you should obtain:
1.015
1.070
A
FeKa A
NiKa A
1.060
FeKa Z
1.010
B
NiKa Z
1.050
1.005
1.040
1.000
1.030
0.995
1.020
0.990
1.010
0.985
1.000
0.980
0.990
0.0
1.050
0.2
0.4
0.6
N i ( wt - f r ac)
0.8
1.0
0.0
FeKa F
C
1.100
0.2
0.4
0.6
N i ( wt - f r ac)
0.8
D
NiKa F
1.0
FeKa ZAF
NiKa ZAF
1.000
1.050
0.950
1.000
0.900
0.950
0.850
0.900
0.800
0.850
0.750
0.800
0.0
0.2
0.4
0.6
N i ( wt - f r ac)
0.8
N. Chatterjee: January 2012, MIT, Cambridge, MA, USA
1.0
0.0
2
0.2
0.4
0.6
N i ( wt - f r ac)
0.8
1.0
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12.141 Electron Microprobe Analysis
January (IAP) 2012
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