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? ,h = 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

The minimum uncertainty in speed is given by:
Δv ≥ ħ/(2mΔx)

where ħ is the reduced Planck constant, m is the mass of the particle (proton in this case), and Δx is the length of confinement.

Substituting the values, we get:
Δv ≥ (1.055 × 10^-34 J•s)/(2 * 1.67 × 10^-27 kg * 2 × 10^-12 m)
Δv ≥ 3.15 × 10^4 m/s

The kinetic energy of the proton is given by:
K = (1/2)mv^2

Substituting the minimum uncertainty in speed, we get:
K = (1/2) * 1.67 × 10^-27 kg * (3.15 × 10^4 m/s)^2
K = 0.127 eV

Therefore, the kinetic energy of the proton is approximately 0.13 eV, which is closest to option B (1.3 eV).

A Nonrelativistic Proton Is Confined to a Length of 2