Question

(b) A 0.13−kg baseball thrown at 100 mph has a momentum of 5.9 kg · m/s. If the uncertainty in measuring the mass is 1.0 × 10−7 of the mass, calculate the uncertainty in the baseball's position.

Answer 1

Rearrange the inequality that describes the Heisenberg uncertainty principle (Equation 6.4.7) to solve for the minimum uncertainty in the position of an object (Δx).

Find Δv by converting the velocity of the baseball to the appropriate SI units: meters per second.

Substitute the appropriate values into the expression for the inequality and solve for Δx.

Solution:

A The Heisenberg uncertainty principle tells us that (Δx)[Δ(mv)] = h/4π. Rearranging the inequality gives

Δx≥(h4π)(1Δ(mv))

B We know that h = 6.626 × 10−34 J•s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ(mv) = mΔv and Δv = ±1 mi/h. We have

Δν=(1mih)(1h60min)(1min60s)(1.609kmmi)(1000mkm)=0.4469m/s

C Therefore,

Δx⩾(6.626×10−34J⋅s4(3.1416))(1(0.149kg)(0.4469m⋅s−1))

Inserting the definition of a joule (1 J = 1 kg•m2/s2) gives

Δx⩾⎛⎝⎜6.626×10−34kg⋅m2⋅s4(3.1416)(s2)⎞⎠⎟⎛⎝⎜1s(0.149kg)(0.4469m)⎞⎠⎟

Δx⩾7.92±×10−34m

This is equal to 3.12 × 10−32 inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.

Step-by-step explanation:

Answer 2

Δp.ΔX
`\geq`h/4
`\pi`

Δp = 1.0 × 10−7× 5.9 kg · m/s

= 5,9×
`10^(-7)`kg · m/s

Δx
`\geq` h/ 4×
`\pi`×m×Δv

Δx
`\geq` 6.626×
`10^(-34)` J.s / 4
`\pi`×5,9×
`10^(-7)`kg · m/s

`\geq` 8.94×
`10^(-29)` m

Step-by-step explanation:

using the Heisenbergs uncertainty principle which states that we cannot simultaneously measure with great precision both momentum and position of the particle.