If the uncertainty in position and momentum are equal, the uncertainty in velocity will be:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. .
Given Δx = Δp, from Heisenberg's uncertainty principle, Δx Δp ≥ h/4π, we get (Δp)^2 ≥ h/4π, so Δp ≥ √(h/4π) = (1/2)√(h/π). Since momentum p = mv, uncertainty in momentum Δp = mΔv. 1 Therefore, mΔv ≥ (1/2)√(h/π), leading to Δv ≥ (1/2m)√(h/π). The minimum uncertainty in velocity is thus (1/2m)√(h/π).