Uncertainty principle using ΔxΔp ≥ ħ: A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? (ħ = 1.055 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J) A) 0.0015 eV B) 0.015 eV C) 0.15 eV D) 1.5 eV E) 15 eV

Question

Quizplus · Accepted Answer

We are given Δx = 500 pm = $5	imes 10^{-10}$ m and want to find the kinetic energy of the electron. We are also told that the speed of the electron is equal to the minimum uncertainty in its speed:
$$\Delta v = \frac{\hbar}{2m\Delta x}$$
Substituting the values given, we have:
$$\Delta v = \frac{(1.055 	imes 10^{-34} 	ext{ J} \cdot 	ext{s})}{2(9.11 	imes 10^{-31} 	ext{ kg})(5 	imes 10^{-10} 	ext{ m})} \approx 1.15	imes10^{6}	ext{ m/s}$$
The kinetic energy of the electron can be found using the formula:
$$K = \frac{1}{2}mv^2$$
Substituting the values given, we have:
$$K = \frac{1}{2}(9.11 	imes 10^{-31} 	ext{ kg}) (1.15	imes10^{6}	ext{ m/s})^2 \approx 6 	imes 10^{-20} 	ext{ J}$$
Converting to electronvolts (eV):
$$K \approx \frac{(6 	imes 10^{-20} 	ext{ J})}{(1.60 	imes 10^{-19} 	ext{ J/eV})} \approx 0.15 	ext{ eV}$$
Therefore, the best choice is (C), 0.15 eV.

Uncertainty Principle Using ΔXΔp ≥ ħ: a Nonrelativistic Electron Is