Question

Calculate the minimum uncertainty in the momentum of a 4He atom confined to 0.40 nm.

(a) 2.02 X 10^-25 kg m/s

(b) 2.53 X 10^-25 kg m/s

(c) 2.64 X 10^-25 kg m/s

(d) 2.89 X 10^-25 kg m/s

This question was posed to me at a job interview.

This intriguing question originated from Electron Probability Density topic in division Quantum Physics of Engineering Physics I

Answer 1

Right choice is (c) 2.64 X 10^-25 kg m/s

The explanation: We know that ^4He atom is somewhere in the 0.40 nm region, therefore, Δx = 0.40 nm.

Using, Δpx ≥ ℏ/Δ*x

For minimum uncertainty, Δpx = 6.626 X 10^-34 Js/2π X 0.40 X 10^-9

= 2.64 X 10^-25 kg m/s.