Lithium's resonance line (the transition) lies at a vacuum wavelength of and carries…

PHYSICS · Subjective · Fermi University Physics · Atomic, Molecular & Optical Physics · Radiative transitions, external fields and collisions · Radiative transitions, external fields and collisions

Deriving the Spectrum, Ch Atoms in electric fields

Problem

Lithium's resonance line (the

2s\to2p

transition) lies at a vacuum wavelength of

670.8\,\mathrm{nm}

and carries an absorption oscillator strength

f = 0.75

. Treating this single line as if it carried all of lithium's oscillator strength, estimate the atom's static polarisability

\alpha_{\rm pol}/4\pi\epsilon_0

using

\frac{\alpha_{\rm pol}}{4\pi\epsilon_0}\Big|_{a_0^3} = \sum_n \frac{f_{0n}}{(\Delta E_{0n}/E_{\rm h})^2},

where

\Delta E_{0n}

is the transition energy and

E_{\rm h} = 27.211\,\mathrm{eV}

is the Hartree energy. Give the result in units of

a_0^3

and in

10^{-30}\,\mathrm{m^3}

, and compare with the measured value,

164\,a_0^3

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