If the position of an electron (m = 9.11 × 10−³¹ kg) could be measured to within 10−³⁰ m, the uncertainty in the magnitude of its speed could be as much as A) 6 × 10−³⁴ m/s. B) 6 × 10²⁵ m/s. C) 6 × 10³⁰ m/s. D) 10³¹ m/s. E) 10⁶¹ m/s.

Question

Quizplus · Accepted Answer

According to Heisenberg’s uncertainty principle, ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck’s constant. We can use the fact that momentum equals mass times velocity (p = mv) to find the uncertainty in the magnitude of the velocity. Rearranging the uncertainty principle equation to solve for Δp, we get Δp ≥ h/(4πΔx). Using the known values of h, m, and Δx, we find Δp ≥ 5.27 × 10^-25 kg·m/s. Then, using Δp = mΔv, we can solve for Δv, the uncertainty in velocity magnitude. We get Δv = Δp/m ≥ 5.78 × 10^5 m/s. Since this represents the maximum possible uncertainty, we can round up to get the final answer of 6 × 10^5 m/s, which is choice B.

If the Position of an Electron (M = 9