Exercise 1 solutions
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Exercise 1 solutions

Exercise 1 solutions

The right (high-energy) end of the graphs is from the surface of the Pt film. At this end the backscattered have maximum possible energy, as they have lost none in interactions with electrons while traveling through the sample. Therefore the energy loss of the ions originates solely from the backscattering.

Pt surface


Exercise 1 solutions1

Exercise 1 solutions

(Not required to explain)

The scattering probability depends on inversely on the ion energy. The lower energy, the higher scattering probability. Thus, atoms deeper in the sample give higher intensity.

Yield ~ 1/E


Exercise 1 solutions2

Exercise 1 solutions

The low-energy side of the graph originates from the Si-Pt interface, where the scattered ions have lost energy while traveling through the Pt film and back. The location of the low-energy side can be used to determine film thickness.

Pt/Si interface

  • The interface is moving to lower E

    • Platinum silicide forms at the interface

    • Longer diffusion depth at higher T


Exercise 1 solutions3

Exercise 1 solutions

Dent at Pt/Si interface

  • Pt diffuses into Si (and vice versa) at the interface

  • -> less Pt neat the interface


Exercise 1 solutions4

Exercise 1 solutions

Intensity peak decreases with annealing

  • The amount of Pt is constant, and with annealing it spreads to Si (and Si diffuses to Pt)-> lower intensity over the film


Exercise 2 solutions

Exercise 2 solutions

  • (i) Electron ejects electron: AES

  • (ii) Electron ejects X-ray: EMP-EDS/WDS

  • (iii) X-ray ejects electron: XPS

  • (iv) X-ray ejects X-ray: TXRF

  • (v) Ion ejects ion: SIMS (or scattering: RBS)


Exercise 2 solutions1

Exercise 2 solutions

  • b) X-rays are non-destructive and they penetrate deep into the sample: XRF is suitable.

    (X-rays can also escape from the depths of the sample)


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