Calculating LEED Spot Angles on Si(001) Surface with 100eV Electron Beam

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SUMMARY

The discussion focuses on calculating the LEED spot angles for a 100 eV electron beam incident on an unreconstructed Si(001) surface. The wavelength of the electron beam is derived using the formula λ = 12.3/√E, resulting in a wave vector k of 5.108. The angle θ to the (1, 0) LEED spot is calculated as 12.76° using the Ewald sphere construction, while the angle to the (1, 1) spot is determined to be 17.76° by substituting √2b for b in the calculations. The methodology and results presented are confirmed as correct.

PREREQUISITES
  • Understanding of LEED (Low Energy Electron Diffraction) principles
  • Familiarity with Ewald sphere construction in crystallography
  • Knowledge of wave vector calculations in solid-state physics
  • Basic proficiency in using trigonometric functions for angle calculations
NEXT STEPS
  • Study the derivation of the Ewald sphere construction in detail
  • Learn about the implications of electron beam energy on diffraction patterns
  • Explore advanced LEED analysis techniques for surface characterization
  • Investigate the effects of surface reconstruction on LEED measurements
USEFUL FOR

Physicists, materials scientists, and students involved in surface science and electron diffraction techniques will benefit from this discussion.

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Homework Statement



A 100eV electron beam is normally incident on an unreconstructed Si(001) surface. Calculate the angle from the surface normal to the (1, 0) and (1, 1) LEED spots (the spots are indexed using the surface unit cell)

Si cubic latice parameter is 5.43072 angstroms



Homework Equations



Can calculate the wavelength in angstroms of the electron beam using λ = \frac{12.3}{√E} where E is in eV. And hence wave vector k = \frac{2π}{λ}



The Attempt at a Solution



I have used the Ewald sphere construction to calculate the angle, θ, to the (1, 0) LEED spot in terms of the wave vector, k , and the reciprocal lattice constant, b. b = \frac{2π}{5.43072} = 1.157 k = \frac{2π}{1.23} = 5.108

Hence θ = arctan \frac{b}{k} = arctan \frac{1.157}{5.108} = 12.76°

And then to the (1, 1) spot I used √2b instead of just b in the first part to find the angle = 17.76 degrees.

I'm very uncertain of my method and the answers. Are they correct? Thanks in advance
 
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