Problem — Uncertainty of the electron in an atom

Problem

An electron is confined in an atom of radius $a_0 \approx 5\cdot 10^{-11}$ m. Estimate its minimum momentum and the corresponding kinetic energy. Compare with the binding energy of hydrogen ($13{,}6$ eV).

Solution

Step 1 — Positional uncertainty. Confining the electron within the atom means $\Delta x \sim a_0 \approx 5\cdot 10^{-11}$ m.

Step 2 — Minimum momentum. From the uncertainty principle $\Delta x\,\Delta p \sim \hbar$:

$\Delta p \sim \frac{\hbar}{a_0} = \frac{1{,}055\cdot 10^{-34}}{5\cdot 10^{-11}} \approx 2{,}1\cdot 10^{-24}\;\text{kg·m/s}$

Since the average momentum is zero, this fluctuation is effectively the typical momentum: $p \sim \Delta p$.

Step 3 — Kinetic energy.

$K \sim \frac{p^2}{2 m_e} = \frac{(2{,}1\cdot 10^{-24})^2}{2\cdot 9{,}11\cdot 10^{-31}} \approx \frac{4{,}4\cdot 10^{-48}}{1{,}82\cdot 10^{-30}} \approx 2{,}4\cdot 10^{-18}\;\text{J}$

In eV:

$K \approx \frac{2{,}4\cdot 10^{-18}}{1{,}6\cdot 10^{-19}} \approx 15\;\text{eV}$

$\ev{\Delta p \sim 2\cdot 10^{-24}\;\text{kg·m/s}, \qquad K \sim 15\;\text{eV}}$

Comparison. The estimated energy ($\sim 15$ eV) is of the same order as the real binding energy of hydrogen ($13{,}6$ eV). The uncertainty principle alone, without solving the Schrödinger equation, correctly predicts the energy scale of the atom: this is no coincidence — it is the deep reason why the atom has exactly those dimensions and energies. Compressing the electron further (reducing $a_0$) would increase $K$: there is an optimal radius that minimises the total energy, and it is indeed $\sim a_0$.

Collegamenti

Argomenti: Quantum physics Concetti: Uncertainty principle · Bohr model Competenze: Fermi estimates · Symbolic setup