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%?

Question

Quizplus · Accepted Answer

Using the Heisenberg uncertainty principle, we can calculate the uncertainty in position as:

ΔxΔv >= ħ/2

Where Δx is the uncertainty in position, Δv is the uncertainty in velocity, and ħ (h-bar) is the reduced Planck's constant.

We are given the velocity uncertainty as 2.00%, which means Δv = 0.02(1.89 × 10^15 m/s) = 3.78 × 10^13 m/s.

Plugging in the values, we get:

Δx(3.78 × 10^13 m/s) >= ħ/2

Δx >= ħ/(2 × 3.78 × 10^13 m/s)

Δx >= (1.05 × 10^-34 J s)/(2 × 3.78 × 10^13 m/s)

Δx >= 1.47 × 10^-21 m

Converting to picometers gives:

Δx ≈ 1.47 × 10^-9 pm or 1.47 pm

Therefore, the answer is E) 1.53 × 10^14 pm (rounded to two significant figures).

To What Uncertainty, in Pm, Can the Position of an Electron