If light of wavelength 600 nm is used to locate an electron to within one wavelength of the light, what is the minimum uncertainty in the electron's momentum? A) 1.1 × 10⁻²⁷ kg·m/s B) 1.8 × 10⁻²⁸ kg·m/s C) 8.8 × 10⁻²⁹ kg·m/s D) 1.2 × 10⁻³² kg·m/s E) 1.8 × 10⁻³² kg·m/s

Question

Quizplus · Accepted Answer

The minimum uncertainty in the electron's momentum can be calculated using Heisenberg's uncertainty principle, which states Δx * Δp ≥ ħ/2, where Δx is the uncertainty in position (600 nm or 600 × 10^-9 m), Δp is the uncertainty in momentum, and ħ (h-bar) is the reduced Planck's constant (approximately 1.055 × 10^-34 m^2 kg / s). Rearranging for Δp gives Δp ≥ ħ/(2Δx). Substituting the given values, we find Δp ≥ 1.055 × 10^-34 / (2 * 600 × 10^-9) = 8.8 × 10^-25 kg·m/s.

If Light of Wavelength 600 Nm Is Used to Locate