Question

From the Heisenberg uncertainty principle how do you calculate Δx for each of the following: (a) an electron with Δv = 0.340 m/s (b) a baseball (mass = 145 g) with Δv = 0.200 m/s?

Answer 1

(a) `Δx ≥ "0.170 mm"`; (b) `Δx ≥ 1.82 × 10^"-33"color(white)(l) "m"`.

Explanation:

The formula for the Heisenberg uncertainty principle is

`color(blue)(|bar(ul(color(white)(a/a) ΔpΔx≥h/(4π) color(white)(a/a)|)))" "`

where

• `Δp` is the uncertainty in the momentum
• `Δx` is the uncertainty in the position
• `h` is Planck's constant (`6.626 × 10^"-34" color(white)(l)"kg·m"^2"s"^"-1"`)

The momentum `p = mv`, and the mass `m` is a constant, so

`Δp =Δ(mv) = mΔv`

The uncertainty principle then becomes

`mΔvΔx≥h/(4π)`

or

`Δx ≥h/(4πmΔv)`

(a) `Δx` for an electron

`Δx ≥h/(4πmΔv) = (6.626 × 10^"-34" color(red)(cancel(color(black)("kg")))·stackrel("m")(color(red)(cancel(color(black)("m"^2))))color(red)(cancel(color(black)("s"^"-1"))))/(4π × 9.109 × 10^"-31" color(red)(cancel(color(black)("kg"))) × 0.340 color(red)(cancel(color(black)("m·s"^"-1")))) = 1.70 × 10^"-4"color(white)(l) "m" = "0.170 mm"`

(b) `Δx` for a baseball

`Δx ≥h/(4πmΔv) = (6.626 × 10^"-34" color(red)(cancel(color(black)("kg")))·stackrel("m")(color(red)(cancel(color(black)("m"^2))))color(red)(cancel(color(black)("s"^"-1"))))/(4π × 0.145 color(red)(cancel(color(black)("kg"))) × 0.200 color(red)(cancel(color(black)("m·s"^"-1")))) = 1.82 × 10^"-33"color(white)(l) "m"`