Uncertainty principle using ΔpΔx ≥ ħ/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? (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 5.26 × 10⁵ m/s B) 7.50 × 10⁵ m/s C) 5.26 × 10⁷ m/s D) 7.50 × 10⁷ m/s E) 5.26 × 10⁹ m/s

Question

Quizplus · Accepted Answer

We are given Δx = 0.110 nm = 1.10 × 10^-10 m and we need to find Δp (minimum uncertainty in momentum) to calculate the minimum uncertainty in velocity.
Using the uncertainty principle,
ΔpΔx ≥ ħ/2
Δp ≥ ħ/2Δx
Δp ≥ (1.055 × 10^-34 J ∙ s)/(2 × 1.10 × 10^-10 m)
Δp ≥ 4.80 × 10^-25 kg m/s
Now, we can use the uncertainty principle again to find the minimum uncertainty in velocity:
Δp = mΔv
Δv = Δp/m
Δv = (4.80 × 10^-25 kg m/s)/(9.11 × 10^-31 kg)
Δv = 5.26 × 10^5 m/s
Therefore, the minimum uncertainty in the electron's velocity is 5.26 × 10^5 m/s or 5.26 × 10^17 nm/s, which is closest to option A.

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