An object is moving along a straight line, and the uncertainty in its position is 2.40 m. (a) Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass = 0.0450 kg) and (c) an electron.

Answer 1

(a) The minimum uncertainty in the momentum p can be found by employing Heisenberg's Uncertainty Principle:

(Δx)(Δp) = ħ/2

We are given the uncertainty in position Δx and want to find the uncertainty in momentum Δp. Rearranging the equation to solve for Δp gives:

Δp = ħ/(2Δx)

Where ħ is the reduced Planck constant, ħ = 1.054571818x10^(-34) J s. We are given Δx = 2.40 m. Plugging it into the equation:
Δp = 1.054571818x10^(-34) J s / (2 * 2.40 m)
Δp = 2.19411412x10^(-35) kg m/s

The uncertainty in the object's momentum is 2.19x10^(-35) kg m/s.

(b) To find the minimum uncertainty in the velocity of a golf ball (mass = 0.0450 kg), we use the formula:

Δv = Δp/m

Where Δp = 2.19x10^(-35) kg m/s is the uncertainty in the momentum and m = 0.0450 kg is the mass of the golf ball.
Δv = 2.19x10^(-35) kg m/s / 0.0450 kg

Δv = 4.87x10^(-34) m/s

The uncertainty in the golf ball's velocity is 4.87x10^(-34) m/s.

(c) To find the minimum uncertainty in the velocity of an electron, we use the same formula for Δv:
Δv = Δp/m

Now, the mass of the electron, m = 9.11x10^(-31) kg. Plugging that into the equation:

Δv = 2.19x10^(-35) kg m/s / 9.11x10^(-31) kg
Δv = 2.40x10^(4) m/s

The uncertainty in the electron's velocity is 2.40x10^(4) m/s.