A measurement of an electron's speed is 2.0 × 10⁶ m/s and has an uncertainty of 10%.What is the minimum uncertainty in its position? ( h = 1.055 × 10^-34 J • s,m_el = 9.11 × 10^-31 kg) A) 0.15 nm B) 0.29 nm C) 0.44 nm D) 0.60 nm E) 0.80 nm

Question

Quizplus · Accepted Answer

The uncertainty principle states that the minimum uncertainty in position (Δx) can be calculated using the following equation:
Δx Δp ≥ h/4π
where Δp is the uncertainty in momentum, h is Planck's constant.
Δp can be calculated by multiplying the speed of the electron by its mass and the percentage uncertainty:
Δp = (m_e)*(v)*(Δv)/100
Substituting the given values into the equation:
Δp = (9.11 × 10^-31 kg) * (2.0 × 10^6 m/s) * (10%)
Δp = 1.82 × 10^-25 kg m/s
Now, substituting both Δp and h in the uncertainty principle equation, we get:
Δx * (1.82 × 10^-25 kg m/s) ≥ (1.055 × 10^-34 J s)/(4π)
Δx ≥ (1.055 × 10^-34 J s)/(4π*(1.82 × 10^-25 kg m/s))
Δx ≥ 0.286 nm ≈ 0.29 nm (to two significant figures)
Therefore, the minimum uncertainty in position is 0.29 nm.

A Measurement of an Electron's Speed Is 2