Uncertainty principle using ΔEΔt ≥ ħ/2: The lifetime of an excited nuclear state is 1.0 ns. What is the minimum uncertainty in the energy of this state? (ħ = 1.055 × 10^-34 J ∙ s = 6.591 × 10^-16 eV ∙ s, 1 eV = 1.60 × 10^-19 J) A) 5.0 × 10^-10 eV B) 5.0 × 10^-26 eV C) 3.3 × 10^-25 eV D) 1.6 × 10^-7 eV E) 3.3 × 10^-7 eV

Question

Quizplus · Accepted Answer

The minimum uncertainty in energy (ΔE) can be calculated using the uncertainty principle formula ΔEΔt ≥ ħ/2. Given Δt = 1.0 ns = 1.0 × 10^-9 s and ħ = 6.591 × 10^-16 eV∙s, substituting these values gives ΔE ≥ (6.591 × 10^-16 eV∙s) / (2 × 1.0 × 10^-9 s) = 3.2955 × 10^-7 eV, which rounds to 3.3 × 10^-7 eV.

Uncertainty Principle Using ΔEΔt ≥ ħ/2: the Lifetime of an Excited