You measure the velocity of an electron to an accuracy of ±22.6 m/s.What is the minimum uncertainty in its position?
A)8.05 μm
B)4.03 μm
C)16.1 μm
D)32.2 μm

Question

Quizplus · Accepted Answer

The minimum uncertainty in position is given by the Heisenberg uncertainty principle: ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant. The uncertainty in momentum is given by Δp = mΔv, where m is the mass of the electron and Δv is the uncertainty in velocity. Substituting these expressions into the Heisenberg uncertainty principle, we get: ΔxΔv ≥ h/4πm. Plugging in the given values, we get: Δx(22.6 m/s) ≥ (6.626 × 10^-34 J s)/(4π × 9.109 × 10^-31 kg) = 8.05 μm.

You Measure the Velocity of an Electron to an Accuracy