Question

Calculate the uncertainty in the position of an electron if it has a speed of 500 m/s with an uncertainty of 0.02%

Answer 1

Speed of the electron = v = 500 m/s.
uncertainity in speed is 0.02 %
∆v = 0.02 % of 500 m/s
= 0.02100×500 = 0.1 m/s
According to Heisenberg uncertainity principle, 

∆x.∆p ≥h4π
where ∆x = uncertainity in position
∆p = uncertainity in momentum
h = Planck's constant = 6.63×10-34 kg m2/s
to calculate ∆p =​ m∆v
m = mass of electron = 9.1×10-31 kg
∆v = uncertainity in speed = 0.1 m/s
∆p =​ m∆v​ = 9.1×10-31 kg​ × 0.1 m/s = 9.1×10-32 kg​ m/s​
∆x.∆p ≥h4π
∆x≥h4π1∆p
∆x≥6.63×10−344×3.1419.1×10−32
= 5.8×10-4 m
Uncertainity in position = ​5.8×10-4 m


Hope this will help u