(5-18) ① We can estimate the ionisation energy as the energy difference between having the electron at the predicted mean distance we have calculated and at a very large distance (effectively, removed completely from the atom). So we take our energy expression (equation V-16), substitute in <r> = 4.0 x 10-11m and <r> = ∞, and find the difference in the calculated energies.
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Using the values of the constants given, we get, for <r> = 4.0 x 10-11m, E = -2.9 x 10-18 J. When <r> = ∞, both terms go to zero, so E = 0 (as it should since this is in effect our reference energy). Therefore the predicted ionisation energy is:
0 - - 2.9 x 10-18 = +2.9 x 10-18 J.
② Ionisation energy / eV = +2.9 x 10-18 J / 1.602 x 10-19 J = +18.1 eV. This is not too far away, given the simplicity of the model, from the measured value of +13.6 eV.
③ This is the energy for 1 atom. So for a mole of atoms, the energy required is given by multiplying by Avogadro’s number:
2.9 x 10-18 x 6.02 x 1023 ≈ 1.7 x 106 Jmol-1 = +1700 kJmol-1
The experimental value is +1312 kJmol-1.
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