An electron is located on a pinpoint having a diameter of 2.04 µm. What is the minimum uncertainty in the speed of the electron?

Answer 1

Δx• Δp ≥ ħ / 2
Δx •m •Δv ≥ ħ / 2
Δv ≥ ħ / 2• m •Δx
The minimum uncertainty in speed
Δv = ħ / 2• m •Δx =
=( h / 2 π ) / 2• m •Δx =
= h / 4• π •m •Δx .
Using the diameter of the pinpoint for the uncertainty in position
Δv =6.63•10^-34/4•π•9.1•10^-31•2.04•10 ^-6 =
= 28.37 m/s.