The Balmer- line of hydrogen,

PHYSICS · Multi-part · Fermi University Physics · Atomic, Molecular & Optical Physics · One-electron atoms and the origins of atomic structure · One-electron atoms and the origins of atomic structure

Deriving the Spectrum, Ch Fine structure and the Lamb shift

Problem

The Balmer-β\beta line of hydrogen, n=4→n=2n=4 \to n=2, has vacuum wavelength 486.1 nm486.1\,\mathrm{nm}. Take cRH=3.28805×106 GHzcR_{\rm H}=3.28805\times10^6\,\mathrm{GHz} and α=1/137.036\alpha=1/137.036. ((a)) From the fine-structure formula, find the 2P3/22P_{3/2}-2P1/22P_{1/2} interval and the full fine-structure spread of the n=4n=4 level, from j=12j=\tfrac12 to j=72j=\tfrac72, both in GHz. Keeping only the larger, state the predicted frequency spread of the Balmer-β\beta fine-structure pattern. ((b)) In a hydrogen discharge at 300 K300\,\mathrm{K}, the Doppler width (FWHM) of the Balmer-β\beta line is 7.6 GHz7.6\,\mathrm{GHz}. A grating spectrometer has resolving power R=λ/Δλ=1.0×105R = \lambda/\Delta\lambda = 1.0\times10^5 at this wavelength. Find the smallest frequency interval Δν\Delta\nu the spectrometer alone can resolve, and decide whether the splitting from part (a) would appear as separate lines in this discharge. ((c)) The potassium resonance doublet, the 4p 2P4p\ {}^2P-4s 2S4s\ {}^2S transition, has components at 766.49 nm766.49\,\mathrm{nm} and 769.90 nm769.90\,\mathrm{nm} (vacuum wavelengths). These two wavelengths differ only through the 4p 2P3/24p\ {}^2P_{3/2}-4p 2P1/24p\ {}^2P_{1/2} splitting, since the 4s4s level is unsplit. Calculate this measured splitting in cm−1\mathrm{cm^{-1}}, calculate the hydrogenic (one-electron, Z=1Z=1) fine-structure splitting for n=4n=4, l=1l=1 from the fine-structure formula, and give the ratio of the measured potassium splitting to the hydrogenic one.

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