Question

What is the uncertainty in the velocity of an electron whose position is known to within 2 × 10–8 meters? If the electron is moving at a speed of 5.0 × 105 m·s–1, what fraction of this speed does the uncertainty represent?

Answer #1

According to heisenberg uncertainty principle,

m Δx ΔV = h / (4π)

Where

m = mass of electron =
9.1x10^{-31} kg

Δx = uncertainity in position = 2 x 10^-8

ΔV = uncertainity in velocity = ??

h = planck's constant = = 6.625x
10^{-34} J.s

m Δx ΔV = h / (4π)

9.1x10^{-31} x 2 x 10^-8 x ΔV =
6.625x 10^{-34} / 4 x 3.14

ΔV = 2832.8

**uncertainty in the velocity =
2.8 x 10^3 m /s**

electron is moving at a speed = 5.0 × 10^5 m/s

uncertainity = (2.8 x 10^3 / 5 x 10^5 ) x 100

**uncertainity = 0.56
%**

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