If the electron (9.11 $\times$ 10^-31 kg)in the hydrogen atom is traveling at about 2.2 $\times$ 10⁶ m/s ± 5%,what is the uncertainty in electron's position?

Question

Quizplus · Accepted Answer

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

Δx Δp ≥ h/4π

where Δx is the uncertainty in position, Δp is the uncertainty in momentum (which we can find using the percent uncertainty in velocity), and h is Planck's constant.

Δp/p = 5% = 0.05

Δp = (0.05)(9.11 x 10^-31 kg)(2.2 x 10^6 m/s) = 1.0 x 10^-25 kg m/s

Δx ≥ h/4πΔp

Δx ≥ (6.626 x 10^-34 J s)/(4π)(1.0 x 10^-25 kg m/s)

Δx ≥ 1.3 x 10^-10 m = 130 pm

Therefore, the uncertainty in the electron's position is ±130 pm. Answer choice C is the best option.

If the Electron (9 ×\times× 10-31 Kg)in the Hydrogen Atom Is Traveling at About 2

If the Electron (9 $\times$ 10^-31 Kg)in the Hydrogen Atom Is Traveling at About 2