Hydrogen-like Ions

Question

A singly ionised helium atom, He+, behaves as a one-electron system with nuclear charge twice that of hydrogen. What energy is required to excite its electron from the ground state to the first excited state?

RankGuru · Accepted Answer

For a hydrogen-like ion of nuclear charge $Z$, the energy levels scale as $E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV, so the entire spectrum is stretched by a factor of $Z^2$. For He+, $Z = 2$, hence $Z^2 = 4$. The ground state is $E_1 = -13.6 	imes 4 = -54.4$ eV and the first excited state is $E_2 = -\dfrac{13.6 	imes 4}{4} = -13.6$ eV. The excitation energy is $\Delta E = E_2 - E_1 = -13.6 - (-54.4) = 40.8$ eV. The value 10.2 eV is the corresponding $n = 1 	o 2$ gap for ordinary hydrogen and ignores the $Z^2 = 4$ scaling. The value 54.4 eV is the full ionisation energy of He+, not the first excitation energy. The value 30.6 eV results from an incorrect arithmetic combination of the levels. Because of this $Z^2$ enhancement, hydrogen-like ions emit much shorter-wavelength, more energetic radiation than neutral hydrogen for the corresponding transition. This uses the NCERT generalisation of the Bohr formula to hydrogen-like species. A plausibility check confirms the He+ excitation energy is four times the hydrogen value of 10.2 eV, exactly as the $Z^2$ dependence predicts.

Hydrogen-like Ions

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