Uncertainty principle using ΔxΔp ≥ ħ: 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? (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 0.29 nm B) 0.58 nm C) 0.87 nm D) 1.2 nm E) 1.6 nm

Question

Quizplus · Accepted Answer

We know that ΔxΔp ≥ ħ. We can rearrange this to solve for Δx: Δx ≥ ħ/Δp. We are given the electron's speed, but we need to find its momentum in order to calculate Δp. We can use the formula for momentum: p = mv, where m is the electron's mass. Plugging in the values given:

p = (9.11 × 10^-31 kg)(2.0 × 10^6 m/s) = 1.82 × 10^-24 kg∙m/s

Then, using the uncertainty in speed:

Δp = (0.1)(1.82 × 10^-24 kg∙m/s) = 1.82 × 10^-25 kg∙m/s

Finally, plugging into the uncertainty principle equation:

Δx ≥ ħ/Δp = (1.055 × 10^-34 J∙s)/(1.82 × 10^-25 kg∙m/s) = 0.579 nm

Rounding to two significant figures, the answer is B) 0.58 nm.

Uncertainty Principle Using ΔXΔp ≥ ħ: a Measurement of an Electron's