Uncertainty principle using ΔpΔx ≥ h/2: A measurement of an electron's speed is 2.0 × 10⁶ m/s and has an uncertainty of 10%. What is the minimum uncertainty in its position? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 0.91 nm B) 1.8 nm C) 2.7 nm D) 2.8 nm E) 5.0 nm

Question

Quizplus · Accepted Answer

Given: 
Δv = 2.0 × 10^16 m/s
Δv/v = 10%
h = 6.626 × 10^-34 J ∙ s
m = 9.11 × 10^-31 kg

Using the uncertainty principle, we have: 
Δp Δx ≥ h/2
where 
Δp = uncertainty in momentum = mΔv = (9.11 × 10^-31 kg)(2.0 × 10^16 m/s) = 1.82 × 10^-14 kg m/s
Δx = uncertainty in position
h/2 = 3.31 × 10^-34 J ∙ s

Substituting the given values: 
(1.82 × 10^-14 kg m/s) Δx ≥ 3.31 × 10^-34 J ∙ s/2

Δx ≥ (3.31 × 10^-34 J ∙ s/2)/(1.82 × 10^-14 kg m/s) = 9.09 × 10^-21 m = 0.91 nm

Therefore, the minimum uncertainty in the electron's position is 0.91 nm, which corresponds to choice B.

Uncertainty Principle Using ΔPΔx ≥ H/2: a Measurement of an Electron's