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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?

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

The formula for the 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 (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"