Surface Emissions Specimen current X-rays Cathodoluminescence Pole Piece, etc SE3  ≈ 1 nm for metals up to 10 nm for insulators

Dependence of backscatter coefficient on backscattered electron energy

Backscattering increases strongly with Z independent of beam energy

Backscatter dependence on tilt and Z

Where are electrons coming from? Kanaya – Okayama equation for electron range in a material: R k-o = ( A E )/(Z 0.89 ρ) μm –For E 0 in keV, ρ in gm/cm 3, A in gm/mole and Z = atomic number

Where are electrons coming from? MaterialAZρE0E0 R (μm) PMMA Fe Fe Homework 2.2

Depth of origination of backscattered electrons Backscattered electrons can Investigate deep into the sample

Distribution of scattered electron energies Note that the distinction between SE and BSE is a definition There are only “scattered electrons”

Why do backscattered electrons give the most information about chemical composition? Why don’t secondary electrons give much information about chemical composition?

Dependence of scattered electron yield on Z 50%! 10% Nearly independent

Why do secondary electrons give great topographic information?

Image formation and interpretation

Images from exactly the same area of the same sample taken with different detectors.

In case you thought the second image was just taken at higher contrast…

And it’s not just your detector choice that can impact your image… 20 kV500 V

Everhart- Thornley detector 

Scanning and Data Collection

Transfer of image from sample to screen Works for both topographic and elemental information

Pixels (Picture elements) NEC monitor: 380 x 300 mm; 1280 x 1024 pixels. Hence pixel size on monitor size is 297 x 293 microns microns. Typical file size used in FEI Nova NanoSEM is 1024 x 884 pixels. Pixel size on sample is pixel size on monitor divided by magnification, about 15 microns (20X) to 0.6 nm (500kX).

So what; who cares? Example: 10 keV beam at 100 pA viewing at 100 X (neural array was taken at 118 X) saved into 1024 x 884 file using Leo and FEI. β = 4I p /π 2 α p 2 d p 2 d p = (2WD/πr A )sqrt(I p / β) FEI: d p = (2*5 mm/π*.015 mm)sqrt( Acm 2 sr/10 8 A) = 0.5 nm! Leo: (2*8 mm/π*.01 mm)sqrt( A cm 2 sr/10 5 A) = 40 nm!

So what; who cares? Cont’d So –Your pixel size varies from 0.6 nm to 15 microns –Your beam diameter can vary from 0.5 nm to 40 microns, at the smallest –Your interaction volume varies from 10 microns to 10 nm (BSE, SE2; last homework)

Implications If probe size is too small –You are wasting resolution: topography can change between sampling points (Nyquist Theorem!) –Resolution regained by sampling and saving more points –You are wasting signal to noise –You are wasting contrast

How to make the spot larger? d p = (2WD/πr A )sqrt(I p / β) –Increase working distance –Go to a smaller aperture –Increase probe current –Decrease accelerating potential Is this the dominant effect of decreasing the accelerating potential?