Problem — Why the atom does not collapse (standing wave)

Problem

An electron orbiting the nucleus, treated as a classical accelerated charge, should radiate energy and fall onto the nucleus in an extremely short time. Why is the atom actually stable? Argue using the quantisation of levels and de Broglie’s standing-wave condition.

Solution

The classical paradox. According to classical electromagnetism an accelerated charge radiates electromagnetic waves. The electron on a circular orbit is accelerated (centripetal acceleration), so it would continuously lose energy and spiral into the nucleus in about $10^{-11}$ s. Matter would be unstable: a contradiction with the very existence of atoms.

The wave solution. The electron has an associated de Broglie wave of wavelength $\lambda = h/p$. For the wave on a closed orbit not to cancel itself through destructive interference, it must close in phase: the circumference must contain a whole number of wavelengths.

$n\lambda = 2\pi r, \qquad n = 1, 2, 3, \dots$

Substituting $\lambda = h/(mv)$ gives Bohr’s angular-momentum quantisation:

$n\,\frac{h}{mv} = 2\pi r \quad\Longrightarrow\quad mvr = n\,\frac{h}{2\pi} = n\hbar$

Why the atom is stable. Only the radii (and hence energies) that satisfy this condition give standing waves: these are the allowed stationary states. In a stationary state the electron does not radiate, exactly as a standing wave on a string carries no net energy. There is also a state of minimum energy (ground state, $n=1$): below it no standing waves are possible, so the electron cannot “fall” any further. The atom is stable because the wave nature of the electron imposes discrete energy levels with an insurmountable minimum.

Collegamenti

Argomenti: Quantum physics Concetti: Bohr model · De Broglie wavelength · Standing waves · Wave-particle duality