Question

What is the minimum uncertainty in the velocity of a 1.0 -ng particle that is at rest on the head of a \(1.0-\mathrm{mm}\) -wide pin?

Short answer

Answer: The minimum uncertainty in the velocity of the particle is approximately 1.054 × 10^{-19} m/s.

Step-by-step solution

Convert the given data to SI units

Convert the mass of the particle from nanograms (ng) to kilograms (kg) and the width of the pin from millimeters (mm) to meters (m). Mass of particle: 1.0 ng = 1.0 × 10^{-12} kg Width of pin: 1.0 mm = 1.0 × 10^{-3} m

Calculate the position uncertainty (Δx)

As we assumed that the position uncertainty Δx will be the diameter of the pin: Δx = 1.0 × 10^{-3} m

Use Heisenberg Uncertainty Principle

Heisenberg Uncertainty Principle states that Δx * Δp ≥ (ℏ/2), where ℏ is the reduced Planck constant (ℏ ≈ 1.054 × 10^{-34} Js). Since momentum (p) is given by the formula p = mass × velocity, we can write the uncertainty in momentum as Δp = mass × Δv, where Δv is the uncertainty in velocity. So, Δx × (mass × Δv) ≥ (ℏ/2) Rearranging for Δv, we get: Δv ≥ (ℏ/2) / (mass × Δx)

Plug in the values and compute Δv

Now, plug in the values of ℏ, mass, and Δx into the formula from Step 3 to calculate Δv: Δv ≥ (1.054 × 10^{-34} Js) / ( (1.0 × 10^{-12} kg) × (1.0 × 10^{-3} m)) Δv ≥ 1.054 × 10^{-19} m/s The minimum uncertainty in the velocity of the particle is approximately 1.054 × 10^{-19} m/s.