An electron is allowed to move freely in a closed cubic box of length 10 cm. The minimum uncertainty in its velocity will be observed as:

Explanation

We are given:The electron is in a closed cubic box of length L = 10 cm = 0.1 mWe are to find the minimum uncertainty in its velocity, which means we will use Heisenberg's uncertainty principleStep 1: Use Heisenberg's uncertainty principle:Δx × Δp ≥ h / (4π)Where:Δx = uncertainty in position ≈ size of the box = 0.1 mΔp = uncertainty in momentum = m × Δvh = Planck's constant = 6.626 × 10⁻³⁴ J·sm = mass of electron = 9.1 × 10⁻³¹ kgStep 2: Rearranging the formula:Δv ≥ h / (4π × m × Δx)Now plug in the values:Δv ≥ (6.626 × 10⁻³⁴) / (4 × 3.1416 × 9.1 × 10⁻³¹ × 0.1)First calculate the denominator:4 × 3.1416 × 9.1 × 10⁻³¹ × 0.1 ≈ 1.143 × 10⁻³⁰Now divide:Δv ≥ (6.626 × 10⁻³⁴) ÷ (1.143 × 10⁻³⁰) ≈ 5.8 × 10⁻⁴ m/s