# 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”##

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