Uncertainty principle using ΔpΔx ≥ ħ/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? (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 0.15 nm B) 0.29 nm C) 0.44 nm D) 0.60 nm E) 0.80 nm

Question

Quizplus · Accepted Answer

We can use the uncertainty principle: 
ΔpΔx ≥ ħ/2. 
Rearranging for Δx, we get: 
Δx ≥ ħ/2Δp. 
We need to find Δx, so we need to find Δp. 
The uncertainty in momentum Δp can be found using the percentage uncertainty in speed Δv/v: 
Δp/p = Δv/v. 
Plugging in the values: 
Δv = 0.10 × 2.0 × 10^6 m/s = 2.0 × 10^5 m/s 
p = m_electron * v = (9.11 × 10^-31 kg) * (2.0 × 10^6 m/s) = 1.82 × 10^-24 kg m/s 
Δp = (2.0 × 10^5 m/s) * (1.82 × 10^-24 kg m/s) = 3.64 × 10^-19 kg m/s 
Now we can find Δx: 
Δx ≥ ħ/2Δp = (1.055 × 10^-34 J s)/(2 * 3.64 × 10^-19 kg m/s) = 0.29 nm. 
Therefore, the minimum uncertainty in the electron's position is 0.29 nm, which is option B.

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