Stability of the atom and Rutherford's model

Problem

Explain in your own words why the stability of the atom is incompatible with the classical Rutherford model, and how the quantisation of angular momentum (Bohr) or the uncertainty principle save the situation.

Solution

The classical problem. In the Rutherford model the electron orbits the nucleus like a planet. But a charge in circular motion is accelerated, and an accelerated charge radiates electromagnetic waves (Maxwell’s electrodynamics). Continuously losing energy, the electron should spiral into the nucleus in an extremely short time, of the order of $10^{-11}\;\text{s}$. The classical atom would be unstable and would collapse: a contradiction with the evidence that matter exists and is stable.

Bohr’s solution. Bohr postulates that only orbits with quantised angular momentum are allowed,

$m v r = n\hbar, \qquad n = 1, 2, 3, \dots$

On these orbits the electron does not radiate: there is a ground state ($n=1$) of minimum energy below which it cannot descend. The atom emits only by jumping between discrete levels, not continuously.

The modern reading (uncertainty). Even more deeply: confining the electron to a small region $\Delta x$ imposes, by Heisenberg’s relation, a momentum $\Delta p \gtrsim \hbar/(2\Delta x)$ and hence a kinetic energy $\sim \hbar^2/(2m\,\Delta x^2)$ that grows as $\Delta x$ decreases. “Falling into the nucleus” ($\Delta x \to 0$) would cost infinite kinetic energy: there is a trade-off between potential energy (which favours getting closer) and confinement kinetic energy (which opposes it), and the minimum gives precisely the size of the atom ($\sim 10^{-10}\;\text{m}$).

$\ev{\text{Quantisation (Bohr) and the uncertainty principle (Heisenberg) fix a stable ground state}}$

Links

Topics: Fisica quantistica Concepts: Modello di Bohr · Principio di indeterminazione Objects: Atomo di idrogeno