To what uncertainty,in pm,can the position of an electron traveling at 1.89 × 10⁵ m/s be measured if the uncertainty of the speed is 2.00%? A)765 B)654 C)8.33 × 10⁴ D)1.53 × 10⁵ E)1.53 × 10⁴

Question

Quizplus · Accepted Answer

We can use the Heisenberg uncertainty principle to solve this problem:

Δx Δp ≥ h/4π

where Δx is the position uncertainty, Δp is the momentum uncertainty, and h is Planck's constant.

We can relate momentum and speed using the mass of the electron:

p = mv = m(1.89 × 10^15 m/s)

The uncertainty in speed is:

Δv = 0.02 (1.89 × 10^15 m/s) = 3.78 × 10^13 m/s

We can convert this to an uncertainty in momentum:

Δp = mΔv = (9.11 × 10^-31 kg)(3.78 × 10^13 m/s) = 3.45 × 10^-17 kg m/s

Now we can solve for the position uncertainty:

Δx ≥ h/4πΔp = (6.626 × 10^-34 J s)/(4π)(3.45 × 10^-17 kg m/s) = 1.53 × 10^-14 m

We can convert this to picometers:

1.53 × 10^-14 m = 1.53 × 10^-2 pm

Therefore, the position uncertainty is 1.53 × 10^-2 pm, which corresponds to answer choice E.

To What Uncertainty,in Pm,can the Position of an Electron Traveling