Quick questions on Bohr model and the Balmer-Rydberg formula: HSC Physics Module 8

By 1911 Rutherford had established that the atom has a tiny dense nucleus surrounded by electrons. The classical problem: an orbiting electron is accelerating and should radiate electromagnetic waves continuously, losing energy and spiralling into the nucleus in about $10^{-11}$ s. Atoms are stable, so classical electromagnetism cannot be the whole story.

Bohr postulated three rules to fix both problems.

Combining quantised angular momentum with the Coulomb-centripetal force balance gives the allowed orbital radii and energies. The result for hydrogen:

For a transition $n_i \to n_f$ with $n_i > n_f$:

The Balmer series is the visible band first discovered, which is why it has its own name. The full pattern lets astronomers identify hydrogen even from the most distant galaxies.

Find the wavelength of the photon emitted when an electron drops from $n = 2$ to $n = 1$ in hydrogen.

The Bohr model works astonishingly well for hydrogen (and for hydrogen-like ions such as He$^+$ and Li$^{2+}$, with $Z^2$ corrections), but it has clear limitations:

The electron in a hydrogen atom occupies certain discrete circular orbits in which it does not radiate. Each such orbit is a stationary state with a well-defined energy.

The allowed orbits are those for which the orbital angular momentum is an integer multiple of $\hbar = h/(2\pi)$:

Radiation occurs only when the electron makes a transition between two stationary states. The photon energy equals the energy difference:

For emission, $n_i > n_f$, and the formula $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ is positive. For absorption, swap them or take the magnitude.

Bound-state energies are negative ($E_n < 0$). The transition energy $\Delta E = E_i - E_f$ is positive for emission because the initial state (higher $n$) is closer to zero.

$R = 1.097 \times 10^7$ m$^{-1}$, so $1/\lambda$ is in m$^{-1}$ and $\lambda$ in m. Convert to nm at the end.

Bohr applies to hydrogen-like one-electron systems. For neutral helium use full quantum mechanics; for He$^+$ scale by $Z^2 = 4$.

It is a useful semi-classical picture, valuable for understanding spectra and ionisation energies, but the quantum-mechanical orbital picture (Schrödinger) replaces it for any serious calculation.