Electrons accelerated from rest through 150 V pass through a narrow aperture of width 0.100 nm. Calculate their de Broglie wavelength and estimate the first diffraction-minimum angle using sinθ ≈ λ/a. Using Δx ≈ a and ΔxΔp_x ≥ ħ/2, estimate the minimum transverse momentum uncertainty and compare the associated angular spread Δp_x/p with the diffraction angle. Use electron mass 9.11 × 10⁻³¹ kg.

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Electron wavelength ≈ 0.100 nm; first-minimum angle is near 90° because λ/a ≈ 1.00. Minimum Δp_x ≈ 5.27 × 10⁻²⁵ kg m s⁻¹, while p ≈ 6.61 × 10⁻²⁴ kg m s⁻¹, giving minimum angular spread ≈ 0.080 rad. Electrical energy eV becomes non-relativistic kinetic energy p²/(2m), so p = √(2meV) = 6.61 × 10⁻²⁴ kg m s⁻¹. De Broglie wavelength is λ = h/p = 6.626 × 10⁻³⁴/(6.61 × 10⁻²⁴) = 1.00 × 10⁻¹⁰ m = 0.100 nm. For aperture width a = 0.100 nm, sinθ = λ/a ≈ 1.00, so the first-minimum estimate is near 90°. The small-angle approximation is invalid in this extreme case. Uncertainty gives Δp_x ≥ ħ/(2a) = 1.055 × 10⁻³⁴/[2(1.00 × 10⁻¹⁰)] = 5.27 × 10⁻²⁵ kg m s⁻¹. The minimum ratio Δp_x/p ≈ 0.080 suggests a substantial angular spread, though the inequality supplies only a lower bound and not the full aperture diffraction distribution. Diffraction provides the more specific wave-optics pattern.

Electron diffraction and uncertainty

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