Electron Beam Microscopy and Lithography Tutorial Questions

Electron Beam Microscopy and Lithography Tutorial Questions

Question

1. Describe the physical origin of the proximity effect in e-beam lithography. (a) How does it manifest? (b) How can it be overcome?

Answer

Origin: electrons entering the resist undergo forward scattering (small-angle deflections in the resist) and backscattering (large-angle deflections from the substrate), exposing resist in regions adjacent to the intended pattern. (a) Manifestation: features receive extra dose from neighbouring exposed areas. Isolated features get less total dose than dense features. This causes dimensional variation — dense patterns are overexposed and isolated patterns underexposed. (b) Solutions: dose modulation (reduce dose in dense areas, increase for isolated features), proximity effect correction (PEC) software that adjusts exposure per shape, GHOST exposure (uniform low-dose background exposure to equalise), and using thinner resists or higher voltages to reduce scattering.

Question

2. Why do we say electrons have wave properties? Why particle properties?

Answer

Wave properties: electrons exhibit diffraction and interference (e.g. double-slit experiment with electrons, electron diffraction from crystals). Their behaviour is described by de Broglie wavelength λ = h/p. Particle properties: electrons have discrete mass (9.11×10⁻³¹ kg) and charge (-e), they produce localised impacts on detectors (individual clicks/spots), and their trajectories can be deflected by electric and magnetic fields. This is wave-particle duality.

Question

3. If an electron and a proton travel at the same speed, which has the shorter de Broglie wavelength?

Answer

λ = h/(mv). At the same speed, the particle with greater mass has the shorter wavelength. The proton (mₚ = 1.67×10⁻²⁷ kg) is ~1836× heavier than the electron (mₑ = 9.11×10⁻³¹ kg), so the proton has the shorter wavelength by a factor of ~1836.

Question

4. Calculate the wavelength of a 0.23 kg ball travelling at 0.10 m/s.

Answer

λ = h/(mv) = 6.626×10⁻³⁴/(0.23 × 0.10) = 6.626×10⁻³⁴/0.023 = 2.88×10⁻³² m This is absurdly small — ~10²² orders of magnitude below anything detectable. Quantum wave behaviour is completely unobservable for macroscopic objects.

Question

5. What is the wavelength of a neutron (mₙ = 1.67 × 10⁻²⁷ kg) travelling at 8.5 × 10⁴ m/s?

Answer

λ = h/(mv) = 6.626×10⁻³⁴/(1.67×10⁻²⁷ × 8.5×10⁴) = 6.626×10⁻³⁴/1.42×10⁻²² = 4.67×10⁻¹² m ≈ 0.0047 nm This is comparable to interatomic spacings, which is why thermal neutron diffraction is used for crystal structure analysis.

Question

6. Through how many volts of potential difference must an electron be accelerated to achieve a wavelength of 0.21 nm?

Answer

KE = eV = p²/(2m), and p = h/λ eV = h²/(2mλ²) V = (6.626×10⁻³⁴)²/(2 × 9.11×10⁻³¹ × (0.21×10⁻⁹)² × 1.6×10⁻¹⁹) = 4.39×10⁻⁶⁷/(2 × 9.11×10⁻³¹ × 4.41×10⁻²⁰ × 1.6×10⁻¹⁹) = 4.39×10⁻⁶⁷/(1.286×10⁻⁶⁸) ≈ 34.1 V

Question

7. What is the theoretical limit of resolution for an electron microscope with electrons accelerated through 3450 V?

Answer

λ = h/√(2meV) = 6.626×10⁻³⁴/√(2 × 9.11×10⁻³¹ × 1.6×10⁻¹⁹ × 3450) = 6.626×10⁻³⁴/√(1.006×10⁻⁴⁵) = 6.626×10⁻³⁴/3.172×10⁻²³ = 2.09×10⁻¹¹ m ≈ 0.021 nm In practice, resolution is limited by lens aberrations to ~0.1-1 nm, far above this theoretical limit.

Question

8. A potential barrier has height 14 eV and thickness 0.85 nm. Transmission coefficient T = 0.0005. What is the electron’s energy?

Answer

For a rect $ln(T)=-2\kappa d=-7.60$ $\kappa =7.6/1.7\ast 10^{-9}=4.47\ast 10^9{m}^{-1}$ $V_o-E={\hslash }^2{\kappa }^2/(2m)=1.22\ast 10^{-19}J=0.76eV;E=13.34eV$

Question

9. A 1.0 mA current of 1.6 MeV protons strikes a 2.6 MeV high barrier 2.8 × 10⁻¹³ m thick. Estimate the transmitted current.

Answer

κ = √(2mₚ(V₀-E))/ℏ = √(2 × 1.67×10⁻²⁷ × 1.0×10⁶ × 1.6×10⁻¹⁹)/(1.055×10⁻³⁴) V₀ - E = 2.6 - 1.6 = 1.0 MeV = 1.6×10⁻¹³ J κ = √(2 × 1.67×10⁻²⁷ × 1.6×10⁻¹³)/(1.055×10⁻³⁴) = √(5.344×10⁻⁴⁰)/1.055×10⁻³⁴ = 2.312×10⁻²⁰/1.055×10⁻³⁴ = 2.19×10¹⁴ m⁻¹ 2κd = 2 × 2.19×10¹⁴ × 2.8×10⁻¹³ = 122.6 T = e⁻¹²²·⁶ ≈ 10⁻⁵³ I_transmitted = 1.0×10⁻³ × 10⁻⁵³ ≈ 10⁻⁵⁶ A — effectively zero.

Question

8 (SEM question). (a) Determine wavelengths at V = 1 kV and V = 10 kV.

Answer

(a) λ = h/√(2meV) At 1 kV: λ = 1.226/√1000 nm = 1.226/31.62 = 0.0388 nm ≈ 38.8 pm At 10 kV: λ = 1.226/√10000 nm = 1.226/100 = 0.01226 nm ≈ 12.3 pm

Question

(b) Elemental composition of Sample A?

Answer

Use high voltage (10 kV) with the EDS detector. Higher accelerating voltage provides sufficient energy to excite characteristic X-rays from the different materials, and the flat surface means beam penetration depth is less of a concern.

Question

(c) Surface details of gold particles on Sample B?

Answer

Use low voltage (1 kV) with the SE detector. Low voltage reduces the interaction volume, giving better surface detail and reducing charging effects. The SE detector captures topographic information from the secondary electrons emitted from the surface.

Question

(d) Measuring the diameter of SiO₂ pillars (20 nm diameter, 220 nm tall) on Sample C?

Answer

Use moderate voltage (5 kV) with the SE detector. The SE detector gives the best edge contrast for dimensional measurements. Too high a voltage would penetrate through the thin pillars, reducing contrast. 5 kV balances resolution with appropriate interaction volume for 20 nm features.

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