Skip to content

Question

The uncertainty in the position of an electron is 1.0 × 10⁻¹⁰ m. What is the minimum uncertainty in its velocity? (h = 6.626 × 10⁻³⁴ J s, mₑ = 9.1 × 10⁻³¹ kg)

Options

Choose one · Correct answer highlighted

Explanation

Δv ≥ h/(4π m Δx) ≈ 5.8×10⁵ m s⁻¹.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.