Image formation and interpretation by 4t1kE0o

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									          Surface Emissions
        Pole Piece,
        etc           SE3



X-rays
Cathodoluminescence



                                            ≈ 1 nm for
                                            metals up
                                           to 10 nm
                                           for insulator
                            Specimen current
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:

• Rk-o = (0.0276 A E01.67)/(Z0.89ρ) μm

  – For E0 in keV, ρ in gm/cm3, A in gm/mole and
    Z = atomic number
Where are electrons coming from?

 Material A       Z    ρ    E0 R (μm)

 PMMA     12.01       6 1.16 20 8.63295

 Fe       55.26   26 7.87 20 1.58759

 Fe       55.26   26 7.87    1 0.01066

      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%!




            Nearly independent
                                 10%
Why do secondary electrons give
 great topographic information?
Image formation and
   interpretation
              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 kV                 500 V
Everhart-
Thornley
detector (1960)
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...300 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 = 4Ip/π2αp2 dp2
• dp = (2WD/πrA)sqrt(Ip/ β)
• FEI: dp = (2*5 mm/π*.015 mm)sqrt(10-10
  Acm2sr/108A) = 0.5 nm!
• Leo: (2*8 mm/π*.01 mm)sqrt(10-10 A
  cm2sr/105A) = 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?
• dp = (2WD/πrA)sqrt(Ip/ 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?

								
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