Q1(d). Uncertainty in the position of a honeybee

Given: m=0.68 g =6.8×10⁻⁴ kg; Δ v=0.1 m s⁻¹; h=6.627×10⁻³⁴ J s.

1. Principle: position and momentum along one axis cannot both be known exactly: Δ x Δ p≥(h)/(4π).
2. Δ p=m Δ v=(6.8×10⁻⁴)(0.1)=6.8×10⁻⁵ kg m s⁻¹
3. Δ x=(h)/(4π m Δ v)=6.627×10⁻³⁴4×3.1416×6.8×10⁻⁵, Δ x≥7.76×10⁻³¹ m
4. The velocity 0.85 m s⁻¹ does not enter: the principle limits the spreads, not the values. (With the form Δ x Δ p≥ h/2π the answer is 1.55×10⁻³⁰ m; the conclusion is the same.)
5. Scale: 7.76×10⁻³¹ m is 10¹⁵ times smaller than a proton. No instrument can detect it, so the bee has a well-defined trajectory.
6. Contrast: an electron (m=9.11×10⁻³¹ kg) with the same Δ v has Δ x=5.8×10⁻⁴ m, far larger than an atom. For electrons in atoms the principle replaces orbits by orbitals (probability distributions).

Application. the same relation sets the zero-point energy of a particle in a box and of a vibrating bond: confining a particle to a region Δ x forces a momentum spread and hence a minimum kinetic energy.