Uncertainty principle using ΔpΔx ≥ h/2: An electron inside a hydrogen atom is confined to within a space of 0.110 nm. What is the minimum uncertainty in the electron's velocity? (h = 6.626 × 10^-34J ∙ s, m_el = 9.11 × 10^-31 kg) A) 3.31 × 10⁶ m/s B) 4.71 × 10⁶ m/s C) 3.30 × 10⁸ m/s D) 4.71 × 10⁸ m/s E) 3.30 × 10¹⁰ m/s

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The uncertainty principle states that the product of the uncertainty in position and the uncertainty in momentum must be greater than or equal to h/4π. In this case, the uncertainty in position is 0.110 nm, so the uncertainty in momentum must be at least:Δp ≥ h/4πΔx = (6.626 × 10^-34 J s)/(4π × 0.110 × 10^-9 m) = 4.71 × 10^-25 kg m/sThe minimum uncertainty in the electron's velocity is then:Δv ≥ Δp/m = (4.71 × 10^-25 kg m/s)/(9.11 × 10^-31 kg) = 3.31 × 10^6 m/s

Uncertainty Principle Using ΔPΔx ≥ H/2: an Electron Inside a Hydrogen