Question

An electron has a speed of 500 m/s with an uncertainty of 0.02%. What is the uncertainty in locating its position?​

Answer 1

Explanation:

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^−34*4×3.1419.1×10^−32

= 5.8×10^-4 m

Uncertainity in position = 5.8×10^-4 m