Question

A proton has velocity v=( 1500± 0.30 )m/s . What is the uncertainty in its position?

Answer 1

By using Heisenberg's uncertainty relation between momentum and position

∆x ∆p ≥ h

here ∆p is the uncertainty in momentum, ∆x the uncertainty in position and h is Plank's constant (6.63x10^-34J).

So, from the information given we have,

∆x (M_p ∆v) ≥ h

where M_p is the mass of the proton (1.7x10^-27kg) and ∆v the uncertainty in velocity(. This gives,

∆x ≥ h / (M_p ∆v) in meters

ΔV = (1500.3 -1499.7) =0.6m/s

∆x ≥ (6.63x 10 ^-34 ) / (1.7 x 10^-27 x 0.6)

the uncertinity in position is ∆x ≥ 6.5 10^-4 m