The minimum uncertainty in the speed of an electron in an one dimensional region of length 2 aₒ (Where aₒ= Bohr radius 52.9 pm) is _________ km s⁻¹.
(Given : Mass of electron = 9.1 × 10⁻³¹ kg, Planck's constant h = 6.63 × 10⁻³⁴ Js)

Solution

Chemistry ma’am · 4 steps

The box length is your Δ x, so put it into Heisenberg's relation and solve for Δ v.

1. Uncertainty in position. The electron can be anywhere in a region of length 2a₀, so Δ x = 2 × 52.9 pm = 105.8 pm = 1.058 × 10⁻¹⁰ m.

   Δ x · m Δ v ≥ h/4π ⇒ Δ vₘᵢₙ = h/(4π m Δ x)

2. Denominator first. 4π m Δ x = 12.566 × 9.1 × 10⁻³¹ × 1.058 × 10⁻¹⁰ = 1.210 × 10⁻³⁹ kg m.
3. Divide. Δ vₘᵢₙ = (6.63 × 10⁻³⁴)/(1.210 × 10⁻³⁹) = 5.48 × 10⁵ m s⁻¹.
4. So. Dividing by 1000 converts to kilometres: Δ vₘᵢₙ = 548 km s⁻¹, so the answer is 548.

The common slip: 1096. Taking Δ x as one Bohr radius instead of 2a₀ doubles the result to 1096.

The idea to keep. For minimum uncertainty, use the equality Δ x Δ p = h/4π. Whatever length the particle is confined to is Δ x, so read it carefully, and convert pm to m before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.