Helium's ground state has two electrons

PHYSICS · Multi-part · 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

Helium's ground state has two electrons, so by the Thomas-Reiche-Kuhn sum rule the oscillator strengths of all its transitions, bound and continuum, add up to

2

. Its lowest dipole-allowed line,

1s^2\,{}^1S_0 \to 1s2p\,{}^1P_1

, lies at

58.43\,\mathrm{nm}

with

f = 0.276

. Every other transition lies higher, and the continuum begins at the ionisation energy,

24.59\,\mathrm{eV}

. The measured static polarisability is

\alpha_{\rm pol}/4\pi\epsilon_0 = 1.38a_0^3

. In atomic units the oscillator-strength sum reads

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

, where

\Delta E_{0n}

is the transition energy and

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

. ((a)) Show that for any ground state whose transitions all lie at or above an energy

\Delta E_{\rm min}

\alpha_{\rm pol}/4\pi\epsilon_0 \le \big(\sum_nf_{0n}\big)/(\Delta E_{\rm min}/E_{\rm h})^2\ a_0^3

, and evaluate this bound for helium. ((b)) Suppose all of helium's oscillator strength lay in bound lines, below the ionisation energy. Find the smallest polarisability that would then be possible, compare it with the measured value, and say what the comparison proves about where helium's oscillator strength lies. ((c)) How much of the measured polarisability does the

58.43\,\mathrm{nm}

line supply on its own, and what fraction is that?

Browse the problem library