The brightness when using back-scattered electrons depends on the intensity of signal from a given area.  The intensity of back-scattered electrons depends on the average atomic number. The atomic number of aluminum is 13.  The average atomic number of alumina is 10.  Thus, the aluminum phase should appear brighter than the alumina phase when using back-scattered electrons to image the sample because aluminum has a higher atomic number.  The extent to which the aluminum phase is brighter (white or grey) simply depends on the specific contrast and brightness settings that were used to capture the image.

Al matrix will be darker than Al2O3, because Al is electric conductive. High resistivity of Al2O3 lead to white or light gray color doe to charging. Shade of gray depend on SEM setting - mainly voltage. Lower boltage means darker color of resistive phase. Colors of SEM images are not calibrated. Calibration is not possible. If you abalyse the SEM image by image analysis, only way is threshold phases for every image individually (try to keep SEM setting for all images).

dear Radovan bures

thanks alot .

I think it's not very important, because it depends on SEM analysis kind which can be either "back scatter" or "secondary'...

and if you want to make sure about a particle material, you can use EDS analysis (Energy Dispersive Spectroscopy).

good luck

dear Amir 

thanks alot

I used BSE for image analysis. 

The brightness when using back-scattered electrons depends on the intensity of signal from a given area.  The intensity of back-scattered electrons depends on the average atomic number. The atomic number of aluminum is 13.  The average atomic number of alumina is 10.  Thus, the aluminum phase should appear brighter than the alumina phase when using back-scattered electrons to image the sample because aluminum has a higher atomic number.  The extent to which the aluminum phase is brighter (white or grey) simply depends on the specific contrast and brightness settings that were used to capture the image.

Dear james

thank you 

It depends on atomic number as oxygen having an effective atomic number lower than the aluminium matrix, therefore it should be black in color.

Dear Esmaeil

Please note that there is not any mandatory difference between the color of alumina particles and Al grains, although as Mr. Radovan Bures kindly said, Al matrix should be darker (it completely depends on the intensity of electron beam and the kind of sample). anyway, the most effective way is morphological investigations.

In a polished and Etched section of a Sample, you may find Alumina Particles as  fine grains distributed in a semi-homogenized matrix (Al grains), high magnification will help you check the boundaries of these grains morphologically and be sure about presence of alumina particle, regarding to its coarse boundaries (EDS analysis is the best way to check the chemical composition of particle, after finding it via morphological investigations- a final check).

In a fracture surface (specially in castings), you may face with Alumina Layers (bifilms) and/or Alumina particles. bifilms are naturally folded and deformed in shape (surface), and seem to be brighter than the other phases, and Alumina Particles may be easily indicated, regarding to their semi-porous surfaces (coarse) and 3D shapes, without any adhesion to other phases.

in the case of needing more information and examples, please contact me directly via [email protected].

Regards   

Hi, Esmaeil,

I think the easiest explanation is that the article images were SE images, and yours BSE images. That your BSE image gave you a grey Al2O3 color it’s easy to understand and James explained it best.

As for the article images situation, my shot at an explanation of how you could get a white Al2O3 particle in a grey Al matrix in SE mode is that these particles experienced the edge effect, meaning the edges appear brighter than their surroundings (the edges reject more electrons and therefore the SEM detector gets more signal). If, on top of that, the contrast/ brightness ratio is unbalanced (poor), you may see small particles as white. There is one caveat to this, though: this happens mostly on specimens with some topography (for example, fracture surfaces). In a metallographic sample, for instance, you don’t typically get the edge effect.

If a BSE detector was used for the article images, then I would venture to say that either there were negative images (which it may happen, depending on how the instrument/ image was set); or the oxide was more complex than a simple Al2O3 and contained a lot of other, heavier, elements that made it appeared brighter than the aluminum matrix.

However, interpreting the shades of grey in the literature images is sometimes difficult - image post-processing may throw you off. Better trust your own EDS analysis, if your microscope has the capability to do that.

Dear behzad

thanks alot 

sure.

Dear Roxana 

thank you 

Usually it has white or whit gray relative to the dark gray observed in the case of the SiC. Moreover the EDS or EDX for particle is misleading unless you make in the interface between the matrix and the particle. Usually use area EDS not point it is more accurate

Dear Mohammad Ibrahim

thanks alot 

By answering in a simple way: the difference in the colors of various ceramic particles (not only SiC or Al2O3 ) depends on the microsection preparation, the conductivity of the same ceramics and research opportunities hardware, ie the microscopic settings.

Dear Anna

thanks

Some of the responses provided here can be misleading. However James Kelly's explanation is correct for your question since you have indeed confirmed that you used BSE. On the other hand, it is difficult to assign colour to the different phases when using SE since it depends on several factors as most of the respondents have indicated. Roxana's answer shows how difficult this can be.

Dear Thomas

thanks

To follow up on Roxana's comments if the Al2O3 particles are small (i.e. less than about 1 µm in diameter) there is a greater chance that they will appear completely bright in an SE2 image. This depends on how the sample surface was prepared. Al2O2 is significantly harder than Al and will therefore polish slower than the matrix. In most cases (unless the sample has been prepared carefull) the Al2O3 particles will stick slightly out of the sample surface after conventional mechanical polishing. If the sample has been electropolished to remove the final mechanical polishing damage layer then the Al2O3 particles will most likely stick even more out of the surface as the oxides are usually chemically inert to acids/electrolytes.

If the particles are small they do not need to stick out of the surface by very much (less than about 1µm) to create significant edge effects. On large particles the edge effect will be pronounced only on one side of the particles on the side that points to the SE2 detector.. However if the particles are small then the interaction volume (usually larger for oxides than their parent metals) will be larger and it will be possible for the whole particle to be inside the interaction volume of the electron beam and thus the whole particle will emit many more electrons from the particle edges. Depending on the brightness and contrast settings then the particle may saturate the detector and appear uniformly white and it will not be possible to see if the edges are brighter only on one side.

Lastly, if the particles are stick significantly out of the surface then one can tilt the sample to see if they cast a shadow. As Roxana said, if your alloy is complex then EDS is necesssary.

Dear Jacob

yes, I must do this

thanks alot