Question

An electron in a particle accelerator has an uncertainty in position of 5.00 pm. Calculate the Uncertainty in the momentum of the electron.​

Answer 1

Answer:

EXAMPLE 1. HEISENBERG UNCERTAINTY PRINCIPLE IN POSITION AND MOMENTUM FOR AN ATOM

If the position of an electron in an atom is measured to an accuracy of 0.0100 nm, what is the electron’s uncertainty in velocity?

If the electron has this velocity, what is its kinetic energy in eV?

Strategy

The uncertainty in position is the accuracy of the measurement, or Δx = 0.0100 nm. Thus the smallest uncertainty in momentum Δp can be calculated using ΔxΔp≥h4πΔxΔp≥h4π. Once the uncertainty in momentum Δp is found, the uncertainty in velocity can be found from Δp = mΔv.

Solution for Part 1

Using the equals sign in the uncertainty principle to express the minimum uncertainty, we have

ΔxΔp=h4πΔxΔp=h4π

Solving for Δp and substituting known values gives

Δp=h4πΔx=6.63×10−34 J⋅ s4π(1.00×10−11 m)=5.28×10−24 kg⋅ m/sΔp=h4πΔx=6.63×10−34 J⋅ s4π(1.00×10−11 m)=5.28×10−24 kg⋅ m/s

Thus, Δp = 5.28 × 10−24 kg · m/s = mΔv.

Solving for Δv and substituting the mass of an electron gives

Δv=Δpm=5.28×10−24 kg⋅ m/s9.11×10−31 kg=5.79×106 m/sΔv=Δpm=5.28×10−24 kg⋅ m/s9.11×10−31 kg=5.79×106 m/s

Solution for Part 2

Although large, this velocity is not highly relativistic, and so the electron’s kinetic energy is