Graduate Secondary electron in SEM

[zonnebloem]

Pretty much everywhere secondary electron has energy of 50 eV or less. I would like to know bit detail definition of secondary electron in the SEM. Why there is no SE with higher energy even beam energy is quite high like 10KeV and more. What is the difference between secondary electron and back scattered electron?

 

[Baluncore]

Welcome to PF.

In SEM, secondary electrons are defined as those with an energy below 50 eV.
https://en.wikipedia.org/wiki/Secondary_electrons

https://en.wikipedia.org/wiki/Scanning_electron_microscope#Principles_and_capacities
"Secondary electrons have very low energies on the order of 50 eV, which limits their mean free path in solid matter. Consequently, SEs can only escape from the top few nanometers of the surface of a sample. The signal from secondary electrons tends to be highly localized at the point of impact of the primary electron beam, making it possible to collect images of the sample surface with a resolution of below 1 nm."

The detector used in a SEM, for the highest resolution of surface features, is an Everhart–Thornley detector.
https://en.wikipedia.org/wiki/Everhart–Thornley_detector
"The detector consists primarily of a scintillator inside a Faraday cage inside the specimen chamber of the microscope. A low positive voltage is applied to the Faraday cage to attract the relatively low energy (less than 50 eV by definition) secondary electrons. Other electrons within the specimen chamber are not attracted by this low voltage and will only reach the detector if their direction of travel takes them to it."

 

[zonnebloem]

Thank you very much for your guide. Bit more details - From wiki "Secondary electrons are electrons generated as ionization products. They are called 'secondary' because they are generated by other radiation (the primary radiation). This radiation can be in the form of ions, electrons, or photons with sufficiently high energy, i.e. exceeding the ionization potential.", the question is when the primary radiation has high energy, I expect SE also becomes high energy. However everywhere it is mentioned that SE is around 50eV
Higher energy SE will not escape from the surface due to mean free path?

 

[Baluncore]

Higher energy SE will not escape from the surface due to mean free path?

The low energy SE may only come from the surface, so they carry most high resolution surface detail.

High energy SE can escape from deeper in the material, which is not good for high surface resolution.

 

[gentzen]

the question is when the primary radiation has high energy, I expect SE also becomes high energy. However everywhere it is mentioned that SE is around 50eV

There are also fast secondary electrons, but modeling them correctly is more important for (proximity correction for) e-beam lithography (especially if the substrate contains heavy elements like Molybdenum) than for SEM.

However, when a secondary electron is created, the probability of it being low energy always remains (much) higher, even when the primary radiation has high energy. The probability to observe them directly is a different story, because you can only directly observe slow secondary electrons created near the surface. No such restriction applies to fast secondary electrons.

 

[zonnebloem]

Let me describe my question with an example. Consider electron release from C by 10KeV e-beam and 20KeV e-beam
If the electron in C has 6 KeV binding energy, I expect the electron will have 4 KeV energy when it kicked out 10KeV e-beam. From energy aspect, this seem to be NOT SE?
If this same electron is kicked out by 20KeV e-beam, it will have 14KeV of energy. Again this is not SE?

 

[gentzen]

Let me describe my question with an example. Consider electron release from C by 10KeV e-beam and 20KeV e-beam
If the electron in C has 6 KeV binding energy, I expect the electron will have 4 KeV energy when it kicked out 10KeV e-beam. From energy aspect, this seem to be NOT SE?
If this same electron is kicked out by 20KeV e-beam, it will have 14KeV of energy. Again this is not SE?

The 4 keV or 14 keV is the energy available for both the scattered primary electron and the kicked-out secondary electron. Most often, the secondary electron would get around 10 eV, and the scattered primary electron all the rest, i.e. 3.99 keV 13.99 keV. The secondary electron can never get more than half of the energy, i.e. 2 keV or 7 keV, because the electron with more energy gets labeled as the primary electron.

 

[Baluncore]

The Everhart–Thornley detector is positioned to select only low-energy SE.

Low-energy SE can escape only from close to the surface. If you want to maximise surface resolution, you should select only SE with energy less than 50 eV.

Higher energy electrons, may escape from deeper in the sample. If they cannot be rejected by the detector, they will raise the background level, and the noise in the image.

 

[crashcat]

The Everhart–Thornley detector is positioned to select only low-energy SE.

Low-energy SE can escape only from close to the surface. If you want to maximise surface resolution, you should select only SE with energy less than 50 eV.

Higher energy electrons, may escape from deeper in the sample. If they cannot be rejected by the detector, they will raise the background level, and the noise in the image.

It seems like you're saying with a 10 kV e- beam it will eject electrons in the range 0-10kV (minus binding energy which is only tens to hundreds of eV). It is clear that the SED design is optimized for picking up only the ones <50 eV which could only have come from near the surface because otherwise they would be captured.

However, does that explain this entire figure? (I just pulled this one that has a typical shape I've seen in other figures.)

http://dx.doi.org/10.1063/1.2829783

That you get this spike at low energy just because you're going to get so many more of those out of the sample, not that the actual SE generated are all mainly 0-10 eV?

 

[Baluncore]

Your questions are not clear.

It is clear that the SED design is optimized for picking up only the ones <50 eV which could only have come from near the surface because otherwise they would be captured.

Captured by what, the sample or the detector?

However, does that explain this entire figure?

Entire has a different meaning to you than to me. You need to specify entirely, the features you need explained.

That you get this spike at low energy just because you're going to get so many more of those out of the sample, not that the actual SE generated are all mainly 0-10 eV?

The spike is the most probable SE energy detected.
The detector does not detect very low energy SE, hence the spike.

 

[zonnebloem]

The 4 keV or 14 keV is the energy available for both the scattered primary electron and the kicked-out secondary electron. Most often, the secondary electron would get around 10 eV, and the scattered primary electron all the rest, i.e. 3.99 keV 13.99 keV. The secondary electron can never get more than half of the energy, i.e. 2 keV or 7 keV, because the electron with more energy gets labeled as the primary electron.

"Most often, the secondary electron would get around 10 eV, and the scattered primary electron all the rest, i.e. 3.99 keV 13.99 keV" What is the reason for this? I expect kicked out electron(secondary electron) can have range of 4KeV to 0 KeV. However why 10eV will be most often?

 

[gentzen]

However why 10eV will be most often?

I don't know, to be honest. Maybe this is related to the plasmon peak, and only true in metals. In general, even if there is a bandgap, many more low energy electron than high energy electrons will be created, because it is much easier to create them (and doesn't cost much energy, so many of them can be created).