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Consider the following atomic model for Helium. Helium contains 2 electrons revolving in circular orbit around the nucleus,in such a way that there is minimum possible repulsion between them.Calculate the first ionisation energy of this Helium.Will there be any difference in the ionisation energies calculated and experimentally found?

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    $\begingroup$ Bohr's model is applicable only to monoelectron species. So it may work in He+ ion. $\endgroup$ Aug 5, 2020 at 5:21
  • $\begingroup$ Now that's an interesting question. Why don't you go and compare these two numbers? $\endgroup$ Aug 5, 2020 at 5:58
  • $\begingroup$ @IvanNeretin I have calculated the value and I got 29.62 eV but the experimentally found value is 24.6 eV. Could you explain me why that difference occurs? $\endgroup$
    – user97467
    Aug 5, 2020 at 8:34
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    $\begingroup$ To be strict it doesn't mean the model is not applicable - it has just been applied after all. It means it is not very accurate. $\endgroup$
    – Ian Bush
    Aug 5, 2020 at 15:10
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    $\begingroup$ Can you show some work on how you calculated the value? It's not very helpful to ask what is wrong when we don't know what you did. $\endgroup$
    – Zhe
    Aug 5, 2020 at 16:29

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What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.

Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...

The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).

The experimental value is 1.81 (24.6 eV), which is significantly lower.

Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.

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  • $\begingroup$ I think we should go by the Slater's rules? $\endgroup$ Oct 8, 2020 at 13:49
  • $\begingroup$ The exercise at hand seems to be seeking a calculation for a thought experiment. Slater's empirical rules do not apply to that thought experiment. $\endgroup$
    – Pallas
    Oct 8, 2020 at 16:47

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