Question

A particle has a mass of 2 × 10-30 kg and an uncertainty in its velocity of 102 m/s. What is the minimum possible uncertainty in position that the particle can have so as not to violate the uncertainty principle?

Answer 1

m = mass of the particle = 2 x 10⁻³⁰ kg

Δv = uncertainty in velocity of the particle = 10² m/s

Δx = uncertainty in position of the particle = ?

h = plank's constant = 6.63 x 10⁻³⁴ J-s

uncertainty principle is given as

m Δx Δv ≥ h/(4π)

inserting the values

(2 x 10⁻³⁰) Δx (10²) ≥ (6.63 x 10⁻³⁴)/(4(3.14))

Δx ≥ 2.64 x 10⁻⁷ m

so minimum possible uncertainty is 2.64 x 10⁻⁷ m