Author Topic: Kanaya–Okayama expression for electron penetration  (Read 8408 times)

Probeman

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Kanaya–Okayama expression for electron penetration
« on: April 17, 2018, 02:01:36 PM »
We've all used the Kanaya–Okayama model for electron depth which is very nice for a rough idea of electron depth in a material. But it calculates a radius for 99% of the electrons, when we know that the phi-rho-z emission volume is generally much smaller. For example, in Ti (assuming a density of 4.5) at 20keV we get a K-O depth of about 2.8 um, and with the modified K-O expression including the critical excitation energy, we still see an (99%) electron depth of about 2.5 um.



But using any Monte Carlo model we can see that most of the x-rays are emitted much closer to the surface, e.g., DTSA2 gives an emission volume of about .6 um radius for 50% of the Ti Ka x-rays.

Has anyone played around with modifying the K-O expression (or another expression) to produce a slightly more accurate equation for the estimation of the x-ray emission volume?  I think might be more useful to have a quick and dirty expression for calculating an emission volume radius based on say 90% of the emitted x-rays.
« Last Edit: April 12, 2020, 03:06:34 PM by John Donovan »
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Mike Jercinovic

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Re: Kanaya–Okayama expression for electron penetration
« Reply #1 on: April 17, 2018, 04:00:29 PM »
John, are you basically then just asking for the ionization range for a particular excitation?  In this case, maybe just use the Castaing ionization range - you have it in the Merlet and Llovet IOP paper on low voltage EPMA.  Then the rough analytical resolution for an element will be a quadrature sum of the beam diameter and ionization range.  Jacky Ruste developed an expression for lateral resolution based on this approach.  Then there is the Reed (1966) estimate of resolution d= 0.077(EcExp1.5-EcExp1.5)/rho about which he concludes "...the spatial resolution for quantitative analysis (defined as the particle size required to contain 99% of the X-ray production) is about three times as calculated from equation..." (above).  This seems like the sort of approximation you might be looking for.

Probeman

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Re: Kanaya–Okayama expression for electron penetration
« Reply #2 on: April 17, 2018, 04:23:11 PM »
John, are you basically then just asking for the ionization range for a particular excitation?  In this case, maybe just use the Castaing ionization range - you have it in the Merlet and Llovet IOP paper on low voltage EPMA.  Then the rough analytical resolution for an element will be a quadrature sum of the beam diameter and ionization range.  Jacky Ruste developed an expression for lateral resolution based on this approach.  Then there is the Reed (1966) estimate of resolution d= 0.077(EcExp1.5-EcExp1.5)/rho about which he concludes "...the spatial resolution for quantitative analysis (defined as the particle size required to contain 99% of the X-ray production) is about three times as calculated from equation..." (above).  This seems like the sort of approximation you might be looking for.

Hi Mike,
That might work, but I'm not explaining myself very well.  What I'm after is a sort of "weighted" expression that will provide an estimate of the volume radius that 90-95% x-rays will be emitted/detected from.  In other words possibly something approaching that of the phi-rho-z emission volume radius.  So the x-ray emission radius rather the the electron path radius.

In other words, while the K-O expression (using the critical excitation energy term) gives a result of about 2.5 um radius, the Monte-Carlo x-ray calculation gives something closer to ~1 um radius. So I could just calculate the phi-rho-z radius and normalize to the specified density, but I just wonder if there is a relatively simple analytical expression "out there" that gives one something closer to the Monte-Carlo result? 
john
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Ben Buse

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Re: Kanaya–Okayama expression for electron penetration
« Reply #3 on: April 30, 2018, 05:53:14 AM »
John Fournelle did a good study looking at equations for both x-ray range and lateral resolution and comparing with monte carlo and experimental. See table page 12

PAPER • OPEN ACCESS
Low voltage EPMA: experiments on a new frontier
in microanalysis - analytical lateral resolution
To cite this article: J Fournelle et al 2016 IOP Conf. Ser.: Mater. Sci. Eng. 109 012003

http://iopscience.iop.org/article/10.1088/1757-899X/109/1/012003/pdf

Ben

Probeman

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Re: Kanaya–Okayama expression for electron penetration
« Reply #4 on: April 30, 2018, 08:07:55 AM »
John Fournelle did a good study looking at equations for both x-ray range and lateral resolution and comparing with monte carlo and experimental. See table page 12

PAPER • OPEN ACCESS
Low voltage EPMA: experiments on a new frontier
in microanalysis - analytical lateral resolution
To cite this article: J Fournelle et al 2016 IOP Conf. Ser.: Mater. Sci. Eng. 109 012003

http://iopscience.iop.org/article/10.1088/1757-899X/109/1/012003/pdf

Ben

Hi Ben,
I forgot about this paper.  Excellent, thanks!
john
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Amedeo

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Use of Kanaya-Okayama relation
« Reply #5 on: February 27, 2024, 07:47:04 PM »

Is it a fair use of the modified Kanaya-Okayama range equation:
   
   RK-O (nm) = 27.6 ( A / Z0.89 rho) [E01.67 – EC1.67]

in the format for limiting range of x-ray generation shown here,
as a way to get a “gray” estimate of the detectability of one element beneath an overlayer of another (assuming no absorption)?

For instance, an example of detecting a silicon substrate below a hafnium oxide film. 
For HfO2: A = 210.49 g/mole; rho= 9.68 g/cm3;
Z is replaced by Zbar (using 0.7 exponent) = 52.766;
For Si: Ec = 1.842 keV.

    Si  in  HfO2
E0 (keV)   Range (nm)
1   -31.3
1.5   -14.2
2   7.2
2.5   32.5
3   61.5
3.5   94.0
4   129.6
4.5   168.5
5   210.3
5.5   255.0
6   302.5
6.5   352.8
7   405.7
7.5   461.2
8   519.3

Does this make sense at all, or am I totally misusing this relationship?

Many thanks!


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Probeman

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Re: Use of Kanaya-Okayama relation
« Reply #6 on: February 28, 2024, 09:29:00 AM »

Is it a fair use of the modified Kanaya-Okayama range equation:
...
Does this make sense at all, or am I totally misusing this relationship?

Many thanks!

Your work makes sense.  I am impressed that you utilized the Yukawa potential Z fraction for the average Z calculation, though I cannot say that we've ever tested that relation with the K-O equation.  But since the elements in HfO2 are quite different in A/Z, and the K-O equation was calibrated using pure elements,  I would think that makes sense. 

Though I suspect the accuracy of K-O will be quite limited in general especially at low voltages, if you are attempting to characterize the HfO2 layer thickness.

For those interested here are some calculations in Standard for HfO2 using the Output | Calculate Alternative Zbars menu:

ELEM:       Hf       O
XRAY:      la      ka
ELWT:   84.797  15.203
ATWT:  178.490  16.000
KFAC:    .7705   .0508
ZCOR:   1.1006  2.9944
AT% :   33.333  66.667

ELEMENT:        Hf       O
CONC FRAC:   .8480   .1520

ZFRAC 1.0:   .8182   .1818
C/Z %DIF:  -3.5133 19.5966

ZFRAC 0.7:   .6995   .3005
C/Z %DIF: -17.5094 97.6644

ATOM FRAC:   .3333   .6667
ELAS FRAC:   .9026   .0974
A/Z Ratio:  2.4790  2.0000

Zbar (Mass fraction) =  62.2703
Zbar (Electron (Z^1.0) fraction) =  60.3636
Zbar (Mass/Electron (Z^1.0) fraction Zbar % difference) =  3.06195

Zbar (Electron (Z^0.7) fraction) =  52.7679
Zbar (Mass/Electron (Z^0.7) fraction Zbar % difference) =  15.2600

Zbar (Elastic fraction) =  65.7644
Zbar (Atomic fraction) =  29.3333

Zbar (Saldick and Allen, for backscatter) =  60.3636
Zbar (Joyet et al.) =  42.0793
Zbar (Everhart) =  70.7500

Zbar (Donovan Z^0.5) =  46.4000
Zbar (Donovan Z^0.667, Yukawa Potential, Z^2/3) =  51.7786
Zbar (Donovan Z^0.70) =  52.7679
Zbar (Donovan Z^0.707, 1/SQRT(2)) =  52.9742
Zbar (Donovan Z^0.80) =  55.5888
Zbar (Donovan Z^0.85) =  56.8932
Zbar (Donovan Z^0.90) =  58.1242
Zbar (Bocker and Hehenkamp for continuum) =  38.5584
Zbar (Duncumb Log(Mass) for continuum) =  51.5540

A better effort might be to utilize pr(z) curves such as those calculated using CalcZAF as described here:

https://probesoftware.com/smf/index.php?topic=1070.0

Using this method, at 4 keV we obtain the following curves for HfO2 with a trace of Si:



Note that the curves are all normalized to the surface intensity, but one can see that the emitted Si Ka signal starts to drop around 200 nm.

Of course the most quantitative method would be to model the different thin film thicknesses in PENEPMA or another Monte-Carlo software as described here using a bi-layer geometry model:

https://probesoftware.com/smf/index.php?topic=57.0

But if you are attempting to simply quantify the HfO2 layer thickness (and/or chemistry), I think your best bet would be to utilize multi-voltage measurements with the BadgerFilm (or STRATAGem) software which is specifically designed for this:

https://probesoftware.com/smf/index.php?board=37.0
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Amedeo

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Re: Kanaya–Okayama expression for electron penetration
« Reply #7 on: March 02, 2024, 04:37:46 PM »
Thank you, Probeman.
I appreciate your input!

I will look at your suggestions for use of MC software - which I do not have at this point - and see if I can refine my investigation.
This has been very helpful.

--Amedeo
And all this science, I don't understand.
It's just my job, five days a week.