A particle of mass 4.00 x 10- 9 kg moves with speed 4.00 m/sec. If its speed is uncertain by 0.2 percent (i.e. Δv/v = 0.2/100), what is the minimum uncertainty in its position to 3 significant figures and in units of metres? Note: use the appropriate formula given on the information sheet.

The problem involves the Heisenberg Uncertainty Principle, which states that the more precisely the position of a particle is determined, the less precisely its momentum can be known, and vice versa. The formula for this principle is:

Δx * Δp ≥ h/4π

where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant (6.626 x 10^-34 Js).

First, we need to calculate the uncertainty in momentum (Δp). The momentum of a particle is given by the product of its mass (m) and velocity (v). So, Δp can be calculated as:

Δp = m * Δv

Given that the uncertainty in velocity (Δv) is 0.2% of the velocity, we can calculate Δv as:

Δv = 0.2/100 * v Δv = 0.002 * 4.00 m/s Δv = 0.008 m/s

Now, we can calculate Δp:

Δp = m * Δv Δp = 4.00 x 10^-9 kg * 0.008 m/s Δp = 3.2 x 10^-11 kg m/s

Now, we can substitute Δp into the Heisenberg Uncertainty Principle to find Δx:

Δx * Δp ≥ h/4π Δx * 3.2 x 10^-11 kg m/s ≥ 6.626 x 10^-34 Js / 4π Δx ≥ (6.626 x 10^-34 Js / 4π) / 3.2 x 10^-11 kg m/s Δx ≥ 5.27 x 10^-24 m

So, the minimum uncertainty in the position of the particle is 5.27 x 10^-24 m.