Uncertainty principle using ΔpΔx ≥ h/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? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J) A) 0.015 eV B) 0.15 eV C) 1.5 eV D) 15 eV E) 150 eV

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The uncertainty in position Δx is 500 pm = 500 × 10^-12 m. Using the uncertainty principle ΔpΔx ≥ h/2, we find Δp ≥ h/(2Δx). The minimum uncertainty in momentum Δp corresponds to the minimum uncertainty in speed Δv, where Δp = m_el * Δv. Solving for Δv gives Δv = h/(2m_el * Δx). The kinetic energy KE = 0.5 * m_el * (Δv)^2. Substituting the values and converting to eV gives approximately 1.5 eV.

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