Confined electron and uncertainty

Problem

An electron is confined to a region of size $\Delta x \approx 10^{-10}\;\text{m}$ (of the order of the size of an atom). Estimate, using Heisenberg’s uncertainty principle, the minimum uncertainty on its momentum and the corresponding typical kinetic energy in eV.

Solution

Step 1 — Minimum uncertainty on momentum. From the uncertainty principle $\Delta x\,\Delta p \geq \hbar/2$:

$\Delta p \geq \frac{\hbar}{2\,\Delta x} = \frac{1{,}055\cdot 10^{-34}}{2\cdot 10^{-10}} \approx 5{,}3\cdot 10^{-25}\;\text{kg·m/s}$

Step 2 — Typical kinetic energy. The momentum is of the order of its own uncertainty, $p \sim \Delta p$, so

$K \sim \frac{(\Delta p)^2}{2 m_e} = \frac{(5{,}3\cdot 10^{-25})^2}{2\cdot 9{,}11\cdot 10^{-31}} \approx \frac{2{,}8\cdot 10^{-49}}{1{,}82\cdot 10^{-30}} \approx 1{,}5\cdot 10^{-19}\;\text{J}$

Step 3 — In electronvolts.

$\ev{K \sim \frac{1{,}5\cdot 10^{-19}}{1{,}6\cdot 10^{-19}} \approx 1\;\text{eV}}$

A confined electron on the atomic scale therefore has a kinetic energy of a few eV, precisely the order of magnitude of atomic binding energies. The confinement itself “generates” the kinetic energy: this is the quantum reason why the electron does not fall onto the nucleus.

Links

Topics: Fisica quantistica Concepts: Principio di indeterminazione Objects: Atomo di idrogeno