A measurement of an electron's speed is 2.0 × 10⁶ m/s and has an uncertainty of 10%. What is the minimum uncertainty in its position? ( = 1.055 × 10^-34 J ∙ s, ^mel = 9.11 × 10^-31 kg) A) 0.15 nm B) 0.29 nm C) 0.44 nm D) 0.60 nm E) 0.80 nm

Question

Quizplus · Accepted Answer

We can use the Heisenberg uncertainty principle, which states that:

Δx * Δv >= h/(4π)

where Δx is the uncertainty in position, Δv is the uncertainty in velocity, and h is Planck's constant.

We can rearrange this equation to solve for Δx:

Δx >= h/(4πΔv)

Plugging in the given values, we get:

Δx >= (6.63 × 10^-34 J ∙ s)/(4π * 2.0 × 10^16 m/s * 0.1)

Δx >= 2.95 × 10^-11 meters

Converting this to nanometers, we get:

Δx >= 0.29 nm

Therefore, the minimum uncertainty in the electron's position is 0.29 nm, which corresponds to choice B.

A Measurement of an Electron's Speed Is 2