Uncertainty principle using ΔpΔx ≥ ħ/2: A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J) A) 0.00038 eV B) 0.0038 eV C) 0.038 eV D) 0.38 eV E) 3.8 eV

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The uncertainty in position Δx is 500 pm = 500 × 10^-12 m. Using the uncertainty principle ΔpΔx ≥ ħ/2, we find Δp = ħ/(2Δx). The minimum uncertainty in speed Δv is Δp/m. The kinetic energy is (1/2)m(Δv)^2. Converting to eV, the result is approximately 0.038 eV.

Uncertainty Principle Using ΔPΔx ≥ ħ/2: a Nonrelativistic Electron Is