Uncertainty principle using ΔpΔx ≥ ħ/2: A nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. What is the kinetic energy of the proton if its speed is equal to the minimum uncertainty possible in its speed? (1 eV = 1.60 × 10^-19 J, ħ = 1.055 × 10^-34 J ∙ s, m_proton = 1.67 × 10^-27 kg) A) 0.13 eV B) 1.3 eV C) 13 eV D) 130 eV E) 1300 eV

Question

Quizplus · Accepted Answer

We can use the uncertainty principle equation to find the minimum uncertainty in the proton's momentum:
ΔpΔx ≥ ħ/2
Δp ≥ ħ/2Δx
Δp ≥ (1.055 × 10^-34 J ∙ s)/(2(2.0 × 10^-12 m))
Δp ≥ 2.64 × 10^-23 kg m/s

We can use the minimum uncertainty in momentum to find the minimum uncertainty in speed:
Δp = mΔv
Δv = Δp/m
Δv = (2.64 × 10^-23 kg m/s)/(1.67 × 10^-27 kg)
Δv = 1.58 × 10^4 m/s

Since the minimum uncertainty in speed is equal to the proton's speed, we can use the kinetic energy formula to find the proton's kinetic energy:
KE = (1/2)mv^2
KE = (1/2)(1.67 × 10^-27 kg)(1.58 × 10^4 m/s)^2
KE ≈ 1.32 × 10^-17 J
KE ≈ 0.082 eV

Therefore, the answer is B) 1.3 eV

Uncertainty Principle Using ΔPΔx ≥ ħ/2: a Nonrelativistic Proton Is