If the x-component of the velocity of an electron can be measured only to a precision of 5 × 10^-2 m/s,what is the minimum uncertainty of the position of the electron in the x-direction? ( A)9.5 × 10³²m B)1 × 10^-3m C)1.1 × 10^-33m D)8.6 × 10²m E)2 × 10¹m

Question

Quizplus · Accepted Answer

According to Heisenberg's Uncertainty Principle, 
ΔxΔp ≥ h/4π
where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.

Since we are given the uncertainty in velocity, we can use the relationship between momentum and velocity for an electron:
p = mv
where m is the mass of the electron and v is the velocity.

Therefore, Δp = Δv * m
and Δx ≥ h/4πΔp

Substituting the values given:
Δv = 5 × 10^-12 m/s
m = 9.1 × 10^-31 kg
h = 6.63 × 10^-34 J.s

Δp = (5 × 10^-12 m/s) * (9.1 × 10^-31 kg) 
≈ 4.55 × 10^-42 kg.m/s

Δx ≥ (6.63 × 10^-34 J.s)/(4π * 4.55 × 10^-42 kg.m/s)
≈ 1 × 10^-13 m
≈ 1 × 10^-1^3 S1S1P0m

Therefore, the minimum uncertainty in position is 1 × 10^-13 m, or option (B).

If the X-Component of the Velocity of an Electron Can