Author Topic: Explain This If You Can!  (Read 28252 times)

Probeman

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Re: Explain This If You Can!
« Reply #30 on: August 27, 2016, 09:20:41 AM »
This is a beam damage issue related I think to the beam damage issues in SiO2 as described here:

http://probesoftware.com/smf/index.php?topic=418.msg2269#msg2269

But this is zircon, a much more robust material I think everyone will agree. And the data was acquired at 100nA but with a 5 um defocussed beam. To me it looks like a 5 um carbon ring, with a smaller central area that appears more damaged. So what I'm wondering is whether the damage pattern seen here:



represents an uneven distribution of electrons in the defocussed beam or if the central area that appears damaged is simply an effect of the heat being concentrated in the center, since the outer regions are also being heated by the defocussed beam so the center area should get the hottest...

I guess my question is: has anyone attempted to profile the electron distribution in a defocussed beam spot?  In other words, is the electron density distribution profile across the defocussed beam like A or B:



I assume we all hope it is like A, but I suspect it is a lot more like B...

I recently ran across this issue again on a very beam sensitive sample. This was a thin film of ZnSn oxide on a polyester material that showed significant damage in the center of the beam spot, even when the beam was defocused to over 20 microns. Even when the beam current was reduced to 15 nA!

I ended up having to use a scanning mode at 10,000x to avoid damaging the sample. The difference was that the beam distribution seems much more uniform with the scanning beam than the defocused beam.

I wonder if anyone has tried to measure the beam flux profile on our EPMA instruments...  perhaps using an imaging CCD array as a sample?  Or would that just cook the CCD?
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Probeman

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Re: Explain This If You Can!
« Reply #31 on: August 27, 2016, 12:34:36 PM »
I recently ran across this issue again on a very beam sensitive sample. This was a thin film of ZnSn oxide on a polyester material that showed significant damage in the center of the beam spot, even when the beam was defocused to over 20 microns. Even when the beam current was reduced to 15 nA!

I ended up having to use a scanning mode at 10,000x to avoid damaging the sample. The difference was that the beam distribution seems much more uniform with the scanning beam than the defocused beam.

I wonder if anyone has tried to measure the beam flux profile on our EPMA instruments...  perhaps using an imaging CCD array as a sample?  Or would that just cook the CCD?

I just chatted with Ed Vicenzi and he pointed out that by looking carefully at a defocused electron beam on a fluorescent sample, we should be able to see if the electron flux is uniform over the defocused area... and now that I think of it, this is what I think we in fact see.

So that might mean that the damage in the center of the defocused beam area must be due to differential heating of the inner area by the outer area.  That is consistent with seeing less sample damage when we scan the beam over the same size area since we are distributing the heating over the scan area...

He also mentioned that any finite element modeler should be able to calculate the temperature distribution in a material given the electron flux, area and the thermal conductivity of the material.  Anyone out there interested?
« Last Edit: August 27, 2016, 12:40:42 PM by Probeman »
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Nick Bulloss

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Re: Explain This If You Can!
« Reply #32 on: August 28, 2016, 12:25:55 PM »
Is the image an otolith?

Probeman

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Re: Explain This If You Can!
« Reply #33 on: August 28, 2016, 12:50:31 PM »
Is the image an otolith?

Hi Nick,
How's it going?

Which image are you asking about?  This one?

http://probesoftware.com/smf/index.php?topic=144.msg871#msg871

john
« Last Edit: August 28, 2016, 12:55:03 PM by Probeman »
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Nick Bulloss

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Re: Explain This If You Can!
« Reply #34 on: August 29, 2016, 04:16:55 PM »
Is the image an otolith?

Hi Nick,
How's it going?

Which image are you asking about?  This one?

http://probesoftware.com/smf/index.php?topic=144.msg871#msg871

john

Hi John,
I was looking at the image in the third post on page one, now I look more at the attachment I don't think it is an otolith.
Cheers,
Nick

Probeman

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Re: Explain This If You Can!
« Reply #35 on: August 29, 2016, 04:30:29 PM »
Hi John,
I was looking at the image in the third post on page one, now I look more at the attachment I don't think it is an otolith.
Cheers,
Nick

Ah, OK. 

That is a SIMS etch pit in a zircon grain.  The area around the hole is coated with gold, but the etch pit is uncoated and somehow creates a charging cavity of some type.

The thing that is really weird is that this charging artifact is quite stable and persistent.
john
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Les Moore

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Re: Explain This If You Can!
« Reply #36 on: December 29, 2017, 06:22:07 AM »
Regarding the odd spots.

The C rich contamination spot deposited at high currents on a steel sample is 'as expected' nicely Gaussian.

The C rich contamination spot on a refractory mineral under the same conditions is a donut.
Well, a Gaussian donut made by spinning the distribution about one tail on the sample.
I have some maps showing this behaviour somewhere (a long time ago).

The only idea that makes sense is that the sample is becoming so hot that it stops contaminating right under the beam but deposits on the cooler regions immediately adjacent. The high conductivity of metals does not allow sufficient elevation in temp for this to occur.

Cool.

Quite what this would do to the C yield is a worry as the C is deposited right where you don't want it re secondary fluorescence.
 

Probeman

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Re: Explain This If You Can!
« Reply #37 on: August 25, 2019, 02:54:46 PM »
I just finished re-reading Victor's Weisskopf's most excellent book "The Privilege of Being a Physicist" (1990):

https://www.amazon.com/Privilege-Being-Physicist-Victor-Weisskopf/dp/0716721066

And this isn't really an "Explain This if You Can" but it does make one wonder what explanation Victor Weisskopf could have for leaving out energies of 10^4 eV in his "Quantum Ladder":



This is after all, the realm of microanalysis!  He even says in the text that atomic physics deals with energies up to several thousand eV, but then jumps to 10^5 eV, calling that the beginning of the nuclear realm.  He must have heard of inner shell ionizations!
« Last Edit: August 25, 2019, 02:57:52 PM by Probeman »
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Probeman

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Re: Explain This If You Can!
« Reply #38 on: December 19, 2019, 10:45:48 AM »
Here's a fun science trivia quiz question: there are three places in the periodic table where the atomic number increases, yet the atomic weight decreases.  Can you name them without looking at the table?

But the more interesting question is: why does this occur in those three places?   Explain this if you can!   :)
« Last Edit: December 19, 2019, 12:12:15 PM by Probeman »
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Probeman

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Re: Explain This If You Can!
« Reply #39 on: December 23, 2019, 12:59:00 PM »
Here's a fun science trivia quiz question: there are three places in the periodic table where the atomic number increases, yet the atomic weight decreases.  Can you name them without looking at the table?

But the more interesting question is: why does this occur in those three places?   Explain this if you can!   :)

OK, I guess you all better check the periodic table if you don't have it memorized!   ;)

The element pairs with an increasing atomic number and a decreasing atomic weight are:

Ar-K
Co-Ni
Te-I

Did anyone know this bit of science trivia?  In fact the Te-I pair is historically interesting because Mendeleyev, who arranged his periodic table by atomic *weight* (because the concept of atomic number had not yet been invented!), made a not often remembered (and wrong) prediction as seen here in a slide I prepared some years ago:



But now the question is why does this decrease in atomic weight occur in these three places only in the periodic table?  Seriously.  I'm asking for help here!
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Mike Matthews

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Re: Explain This If You Can!
« Reply #40 on: December 25, 2019, 05:21:55 AM »
The atomic weight is the sum of the nucleons minus the amount of ‘mass’ that’s been taken up as binding energy in the nucleus (it’s that e=mc2 equation) so I presume there must be a larger relative increase in binding energy at these points.

Probeman

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Re: Explain This If You Can!
« Reply #41 on: December 26, 2019, 09:05:50 AM »
The atomic weight is the sum of the nucleons minus the amount of ‘mass’ that’s been taken up as binding energy in the nucleus (it’s that e=mc2 equation) so I presume there must be a larger relative increase in binding energy at these points.

Hi Mike,
That's what I thought also. But I wrote to my two "card carrying" physicist friends, Andrew Westphal and Zack Gainsforth at Berkeley, and asked about these binding energies, and Zack responded that the binding energies are not that different at all, but that is has more to do with the "magic" shells number, or as Andrew subsequently confirmed:

Quote
Zack is right, it's because of the nuclear closed shells for both protons and neutrons, at the numbers that he gave:  2, 8, 20, 28, 56, 82, 126, ...

This why there are s-process abundance peaks at 56Ni (which decays to 56Fe), Ba, Pb, and r-process peaks at Sn and Pt (for these there is pile-up at Z=56 and 82, then the beta-decay after the r-process is over shifts the peak down by a few elements.

He also adds:

Quote
In the case of K and Ar, it's not because of binding energy, but just because 40Ar is the most abundant isotope of Ar in the Earth's atmosphere and 39K is the most abundant isotope of K.  The reversal is not universal, it's just because most of the Ar in the atmosphere comes from the decay of 40K.   If you used the average atomic weight of the elements based on the *cosmic* abundances, e.g., in interstellar gas, you would not see a reversal -- that is, the *cosmic* atomic weight of Ar is close to 36, not 40.  So the sequence starting from S would be 32, ~35.4, 36, 39, ...

So the atomic weight is not a universal number, but varies according to the isotopic composition of the material that you're considering.

I have to say, I'm still not sure I understand completely, but I'm thinking about it.
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