### Author Topic: monte carlo bse coefficient  (Read 2259 times)

#### Ben Buse

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##### monte carlo bse coefficient
« on: July 24, 2018, 06:47:01 am »
Hi All, a brain teaser

Here's a plot from penepma of metals, alloys and oxides. Errors approximate symbol size

It suggests oxides plot lower than metals or alloys, any ideas?

Win x-ray gives good agreement to penepma

Materials simulated

I should point out that density is not the problem see screenshot results for diamond and amorphous carbon are the same

Interestingly in the paper 'Mean Atomic Number Quantitative Assessment in Backscattered Electron Imaging'
https://doi.org/10.1017/S1431927612013566
They measure different BSE intensities for minerals and metals but attribute this to variations in BSE from day to day

With subsurface charging in insulators you might expect a higher BSE in insulators (e.g. Ghorbel et al. 2005) - but this is not included in monte carlo simulations, and is the reverse of what is simulated
« Last Edit: July 24, 2018, 07:09:12 am by Ben Buse »

#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #1 on: July 24, 2018, 08:41:50 am »
I should point out that density is not the problem see screenshot results for diamond and amorphous carbon are the same

Ha!   This brings back memories...

You are correct Ben, the problem is *not* different densities. Also see some Penepma calculations I did many years ago assuming different densities for gold. See attached plot below.

It is also true that measuring reliable BSE values is problematic.  But that is not the problem we are seeing with the Monte Carlo calculations.  The problem lies in another area of calculation.

Let me give another brain teaser that answers the first brain teaser: how is average atomic number calculated?  For pure elements the answer is obvious, it's simply the atomic number of the element, but what about for compounds?  How should average Z be calculated for compounds?   It's not obvious until you think through the physics.

Think about the physics of elastic scattering. What exactly is involved?  It's true that there is a tiny mass effect which is 0.05% energy loss in a 180 degrees momentum exchange between an electron and a hydrogen nucleus  (1/2000 mass ratio).  And how often does one get a high angle BSE in hydrogen?  Not very often. And for heavier elements it's a *smaller* mass effect.

See attached pdf below.  Remember to login to see attachments...  Here's another clue:  the neutron has almost no effect on elastic scattering (~10-7 or so).  It is almost purely an electrostatic effect.

So how should average Z be calculated for compounds?
« Last Edit: July 24, 2018, 03:47:19 pm by Probeman »
The only stupid question is the one not asked!

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##### Re: monte carlo bse coefficient
« Reply #2 on: July 24, 2018, 02:28:32 pm »
err....whats on the X and y axis?  Is that part of the brain teaser?

#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #3 on: July 24, 2018, 02:48:08 pm »
err....whats on the X and y axis?  Is that part of the brain teaser?

Hi Jon,
Eta (BSE coefficient) on the Y axis, and Z (atomic number) on X.
john
« Last Edit: July 24, 2018, 03:45:37 pm by Probeman »
The only stupid question is the one not asked!

#### JonF

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##### Re: monte carlo bse coefficient
« Reply #4 on: July 25, 2018, 01:07:54 am »
Oooh a brain teaser. Perfect for my train commute!

I'll have a crack at this. My initial thought is that its something to do with the fact that eta for mixed phases is calculated as n[mixed] = sum(n(i)*C(i)) where n= eta, i is an index of the elements and C is concentration in weight percent.*

In my mind, the weight percent weighting doesn't make sense for mixed phases - the likelihood that a primary beam electron encounters a particular element depends on the atomic percent, rather than the weight percent. For example, if an electron interacts with an atom in Si metal, it has a 100% chance of that atom being Si. If it interacts with an atom in SiO2, it has a 1 in 3 chance of interacting with Si and a 2 in 3 of being weakly scattered by O.

*From Scanning Electron Microscopy and X-ray Microanalysis, 4th Edition, Goldstein et al

#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #5 on: July 25, 2018, 12:33:29 pm »
Oooh a brain teaser. Perfect for my train commute!

I'll have a crack at this. My initial thought is that its something to do with the fact that eta for mixed phases is calculated as n[mixed] = sum(n(i)*C(i)) where n= eta, i is an index of the elements and C is concentration in weight percent.*

In my mind, the weight percent weighting doesn't make sense for mixed phases - the likelihood that a primary beam electron encounters a particular element depends on the atomic percent, rather than the weight percent. For example, if an electron interacts with an atom in Si metal, it has a 100% chance of that atom being Si. If it interacts with an atom in SiO2, it has a 1 in 3 chance of interacting with Si and a 2 in 3 of being weakly scattered by O.

*From Scanning Electron Microscopy and X-ray Microanalysis, 4th Edition, Goldstein et al

Hi Jon,
You're on the right track!

You are correct that weight fraction averaging for BSE is really only a proxy for what we should actually be doing.  As you surmise, elastic scattering is not affected by neutrons so atomic weight isn't a very physically accurate model for calculating average Z for BSE.

But should we be using atomic fraction for average Z of the BSE signal?  That assumes that each atom contributes equally to the BSE signal and we already know that simply isn't the case.

Think about this: a sample of uranium carbide or UC.  Is the BSE signal of UC more like carbon (Z=6) or more like uranium (Z = 92) or halfway between carbon and uranium (Z=49)?  The answer of course is very much like uranium (Z=92).

So that is why people started using weight fraction instead of atomic fraction. Because weight fraction averaging was a much more accurate model for average Z of BSE than atomic fraction averaging.

But again, atomic weight is not a physically accurate model, so what would be more physically accurate?  Think about it from the perspective of the incident electron... what does the BSE (elastically scattered electron) interact with, inside the atom?
« Last Edit: July 25, 2018, 07:37:37 pm by Probeman »
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#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #6 on: July 26, 2018, 01:51:00 pm »
Here's another clue for Ben's brain teaser:

He's trying to plot BSE coefficient vs. average Z (atomic number) for pure elements and compounds. For pure elements average Z is just the atomic number (because 1 atom times the atomic number (Z) equals Z). But for compounds the average Z needs to include the atom fraction of each element in the compound, times what fractional quantity?

E.g., in uranium carbide, we have 1 atom of uranium and 1 atom of carbon, so we have a .5 atoms of each element for this UC compound. Now we want to multiply the atom fraction of each element times what physical parameter?  Remember, that the BSE coefficient of UC is *much* more like U than C, so what other physical parameter (other than weight fraction) scales like that? Yes, there is no commonly accepted name for this physical quantity, but it pops up in all sorts of physics calculations...

Think of the analogy to weight fraction. We calculate the weight fraction of a compound how?
« Last Edit: July 26, 2018, 04:33:20 pm by Probeman »
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#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #7 on: July 31, 2018, 09:52:01 am »
OK, so here is the answer to Ben's "brain teaser" for plotting BSE coefficients of pure elements and compounds as a function of Z, as described in the first post above.

We know from physics that mass has almost no effect for BSE production. At most the momentum exchange of an incident electron interacting with a hydrogen nucleus (with a 180 degrees elastic scatter) produces an energy loss (besides a change in angle) of around 1/2000 or 0.05%. For higher atomic number elements the energy loss effect is even smaller.

So since the number of protons in the nucleus dominates the BSE effect we should not be using weight fraction (as it's based on atomic weight) to calculate the average Z for comparing BSE coefficients even though it is traditional, and consider a different scaling parameter based on atomic number.

But first let's consider how we calculate weight fraction for a element i in a compound:

wtfrac(i) = n(i) * A(i)/ SUM(n() * A())

where n(i) is the number of atoms of element (i) in the compound, and A(i) is the atomic weight of element (i) in the compound.

But since we know that atomic weight does not scale exactly with atomic number, by utilizing atomic weight for our average Z calculation for plotting our BSE coefficients, we are introducing an error equal to the degree by which the ratio A/Z varies across the periodic table. And remember, for some increases in atomic number, the atomic weight actually *decreases*!  Do you know where they are in the periodic table? It's a good trivia question.

But what if we simply substitute atomic number for atomic weight in the above calculation? In that case we have:

zfrac(i) = n(i) * Z(i)/ SUM(n() * Z())

where n(i) is the number of atoms of element (i) in the compound, and Z(i) is the atomic number of element (i) in the compound.

So let's refer to a figure from a paper we published in 2003 where we plotted Monte Carlo calculations (using NIST"s MQ software) of BSE coefficients versus average Z for a number of pure elements and compounds as Ben did in his original post:

Note that here we are using weight fractional averaging to calculate average Z for the compounds just as Ben did and it appears very similar to the two other Monte Carlo softwares. Now let's do the same plot but use the Z based fractional averaging (called here, electron fraction averaging, but protons/electrons- it's still Z based):

It's a better fit, but still not perfect.  What could be the issue?

Well as we have all observed, as the atomic number increases, the amount of backscattered electrons does not increase linearly.  The BSE coefficient curves starts to bend down beginning around zinc (Z = 30 or so). Why is this?

It's because of partial "screening" of the nuclear (proton) charge by the inner electron orbitals of atoms with higher Z (as pointed out to us by Dale Newbury from NIST at the time). Basically it means that incident electrons do not always get to "see" the full electrostatic charge of the nucleus in the higher atomic number elements.  Therefore they are elastically scattered a little less than one would think, given their atomic number.

So we need to correct for this nuclear screening by the inner orbital electrons effect, and so we tried a couple of exponents and found the following fit of Z^0.8 works beautifully as seen here:

Now it's true, that we don't commonly utilize this Z fractional weighting everyday, but it's essentially what the Monte Carlo models do when they calculate the elastic scattering cross sections for compounds.

If you are interested in the details I've attached the PDFs of our original paper, Reed's comments to our paper and our response to his comments. The latter is fun reading.  The sharp eye will notice that we stated the maximum mass effect for an electron interacting with a hydrogen nucleus is 0.2% but that was a typo, since 1/2000 is actually 0.05%.  Doh!

Finally, if anyone is interested in comparing these different average Z calculations, simply open the Standard app in the CalcZAF distribution and check the menu Output | Calculate Alternative Zbars. Here is the calculation of alternative Zbars for the uranium carbide compound I mentioned in a previous post:

ELEM:        U       C
CONC:    .9520   .0480
ELEC:    .9388   .0612
%DIF:  -1.3854 27.4551
ATOM:    .5000   .5000
ELAS:    .9741   .0259
A/Z :   2.5873  2.0018

Zbar (Mass/Electron fraction Zbar % difference) =  1.29078
Zbar (Mass fraction) =  87.8689
Zbar (Electron fraction) =  86.7347
Zbar (Elastic fraction) =  89.7720
Zbar (Atomic fraction) =  49.0000

Zbar (Saldick and Allen) =  86.7347
Zbar (Joyet et al.) =  65.1920
Zbar (Everhart) =  91.7179

Zbar (Donovan Z^0.5 for continuum) =  74.5053
Zbar (Donovan Z^0.80 for backscatter) =  83.2973
Zbar (Donovan Z^0.85 for backscatter) =  84.3084
Zbar (Donovan Z^0.90 for backscatter) =  85.2124
Zbar (Bocker and Hehenkamp for continuum) =  64.3464
Zbar (Duncumb Log(Mass) for continuum) =  80.6927
« Last Edit: August 01, 2018, 08:59:42 am by Probeman »
The only stupid question is the one not asked!

#### JonF

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##### Re: monte carlo bse coefficient
« Reply #8 on: August 06, 2018, 04:30:06 am »
It's interesting that nuclear screening has come up with regards to BSE coefficient.

I wonder whether it would be possible to see atomic screening 'artefacts', such as the poor screening by 4f electrons (cause of the 'lanthanide contraction') or the 'd-block' contraction, and if this difference would be discernible with our BSE detectors. Could be an interesting experiment!

#### Ben Buse

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• Posts: 415
##### Re: monte carlo bse coefficient
« Reply #9 on: August 06, 2018, 06:39:54 am »
Hi John,

Thanks for your responses - I didn't know the answer. To summarise, Basically the offset to compounds is explained by nuclear screening - which you account for using a 0.8 exponent and electron fraction equation.  The 0.8 is determined by fitting to monte carlo data.

As you point out atomic fraction is rubish, UO2 MAN = 36 compared to 82 using wt fraction, and 79.5 using electron fraction and 73.448 using electron fraction and 0.8 exponent.

Nice paper I'd never come across it before, I'm just recalculating the results using eq. 20. And will post the new results

I've attached a spreadsheet which calculates the MAN using the different equations

« Last Edit: August 06, 2018, 06:42:04 am by Ben Buse »

#### JonF

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##### Re: monte carlo bse coefficient
« Reply #10 on: August 06, 2018, 07:25:12 am »
Agreed, using atomic fraction alone is daft.

Regarding screening, I was wondering whether an alternative approach would be to see if the atomic radii (itself a function of nuclear screening) would be an alternative?
« Last Edit: August 06, 2018, 10:25:29 am by JonF »

#### Ben Buse

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• Posts: 415
##### Re: monte carlo bse coefficient
« Reply #11 on: August 07, 2018, 05:49:51 am »
Here's the data for the metals and the binary compounds. You can see using electron fraction and 0.8 exponent, gives good agreement to metals' line.

Ben

And following Jon's good advice, I've added some axes labels, now we've solved the brain teaser

#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #12 on: August 07, 2018, 08:21:27 am »
Hi Ben,
So the Z fraction to the 0.8 exponent are the light symbols?   And the dark blue M symbols are the pure elements?

Now just do a bit more work on your legend and you'll have an actual plot!
john
« Last Edit: August 07, 2018, 08:43:57 am by Probeman »
The only stupid question is the one not asked!

#### Ben Buse

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##### Re: monte carlo bse coefficient
« Reply #13 on: August 07, 2018, 08:39:50 am »
Yes I think your right! It certainly wouldn't get published!

light blue: exponent 0.8,
pinky purple - mass % calc.
black 'M': metals

Ben

#### Probeman

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##### Re: monte carlo bse coefficient
« Reply #14 on: August 07, 2018, 11:06:44 am »
Agreed, using atomic fraction alone is daft.

Regarding screening, I was wondering whether an alternative approach would be to see if the atomic radii (itself a function of nuclear screening) would be an alternative?

Hi Jon,
It would be worth a try.

The electron fraction Z to the 0.8 power is just a model, as weight fraction Z is just a model. The advantage of the electron fraction (or proton fraction or Z fraction whatever you want to call it) model, is that it makes one less faulty assumption than the weight fraction Z model: which assumes that mass affects elastic scattering of electrons.

By the way, the discerning eye will notice something else from the output of alternative Z-bars in the post above:

Zbar (Mass/Electron fraction Zbar % difference) =  1.29078
Zbar (Mass fraction) =  87.8689
Zbar (Electron fraction) =  86.7347
Zbar (Elastic fraction) =  89.7720
Zbar (Atomic fraction) =  49.0000

Zbar (Saldick and Allen) =  86.7347
Zbar (Joyet et al.) =  65.1920
Zbar (Everhart) =  91.7179

Zbar (Donovan Z^0.5 for continuum) =  74.5053
Zbar (Donovan Z^0.80 for backscatter) =  83.2973
Zbar (Donovan Z^0.85 for backscatter) =  84.3084
Zbar (Donovan Z^0.90 for backscatter) =  85.2124
Zbar (Bocker and Hehenkamp for continuum) =  64.3464
Zbar (Duncumb Log(Mass) for continuum) =  80.6927

Does anyone see it?

When we did our original literature search for different ways to calculate average Z, we found that none of them worked that well for calculating BSE trends. The one that was closest was Saldick and Allen, 1954.  That was when we started thinking about what would be a better model and then we came up with the electron fraction model.

Then we noticed that our model and the Saldick and Allen model gave the same results, even though the equations looked quite different. Then we realized that their expression and our expression were algebraically equivalent!  See our paper here for that discussion:

http://epmalab.uoregon.edu/publ/Compositional%20Averaging%20of%20Backscatter%20Intensities%20in%20Compounds%20(M&M,%202003(.pdf

Of course as we pointed out above, the electron fraction Z to the power of 1.0 model was only marginally better than weight fraction Z model, and that was when we tried some different exponents to account for nuclear screening, and found that electron fraction Z to the 0.8 worked really well.

So why didn't the Saldick and Allen get the attention of people in the EPMA community?  Well, here's the title of their paper from 1954:

"The Yield of Oxidation of Ferrous Sulfate in Acid Solution by High-Energy Cathode Rays"

That's why!  Maybe next time they would choose a title with broader appeal...
« Last Edit: February 04, 2019, 04:11:49 pm by Probeman »
The only stupid question is the one not asked!