Uncertainty principle using ΔxΔp ≥ ħ: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 (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 1.05 × 10⁶ m/s B) 1.50 × 10⁶ m/s C) 1.05 × 10⁸ m/s D) 1.50 × 10⁸ m/s E) 1.05 × 10¹⁰ m/s

Question

Quizplus · Accepted Answer

Using the uncertainty principle:
ΔxΔp ≥ ħ
Δx = 0.110 nm = 1.10 × 10^-10 m
We need to solve for Δp (minimum uncertainty in momentum) to find the minimum uncertainty in velocity:
Δp = ħ/Δx
Δp = (1.055 × 10^-34 J ∙ s)/(1.10 × 10^-10 m)
Δp = 9.59 × 10^-25 kg ∙ m/s
Minimum uncertainty in velocity, Δv:
Δv = Δp/m
Δv = (9.59 × 10^-25 kg ∙ m/s)/(9.11 × 10^-31 kg)
Δv = 1.05 × 10^6 m/s
Therefore, the minimum uncertainty in the electron's velocity is 1.05 × 10^6 m/s, which is answer choice A.

Uncertainty Principle Using ΔXΔp ≥ ħ:An Electron Inside a Hydrogen