[orgo814]

Obtain expressions for (dE/dZ)n and (d^2E/dZ^2)n using the Bohr model. Using the ionization energies for isoelectronic sequences of atoms and ions from Table 1.7, and finite differences to obtain (deltaE/deltaZ)n and (deltaE^2/deltaZ^2)n, show that n=1 level is filled when the electron occupancy is 2. How would you estimate unknown ionization energies using a finite difference approximation? Estimate the electron affinities of F and O by this procedure.

I was fine up until last two questions. I attached a pic of the solutions. Numbers in parentheses represent estimated values. I am very confused on the last two questions in that paragraph so if someone can point me in right direction that would be great. Thanks.

[mjc123]

Well, it looks as if the first row of numbers is the ionisation energy of each element, and the second row is the first difference - the difference between succesive elements. The third row is the second difference - the difference between successive first differences - and these are roughly constant at about 7 eV, as shown by the low values for the third difference in the next row. So if you wanted to estimate, say, the electron affinity of F (i.e. the ionisation energy of F^-, isoelectronic with Ne, Na⁺ etc.) to a first approximation, you would take the difference between Ne and Na⁺ and subtract from it the second difference of ca. 7, giving a first difference of ca. 18.7 between F^- and Ne, and thus an estimate for IE(F^-) of ca. 2.9 eV. As you see, this is only a rough approximation, but it gives you a ball-park figure to work with.

[orgo814]

I'm pretty sure the -7.7 eV and 3.15 in first row are electron affinity though and the others are I.E

[orgo814]

What do you mean "ca."

[Enthalpy]

What do you mean "ca."

Circa or "approximately". The abbreviation doesn't look usual in English, according to my dictionary.

[Enthalpy]

Comparing nuclei with different numbers of protons surrounded with the same number of electrons, you get a simple ionization law, approximately the square of the net charge.

This is because the orbitals shrink with the net charge. It is approximate because the deeper orbitals have an effect too.

You observe the squared charge effect in X-rays too, where the K shell lets the absorption grow stepwise as photons have enough energy to ionize it.
http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html
hydrogen being an exception because of the lone electron on K.