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First ionisation energy?

How is this calculated?

My book just throws it in there like it was nothing, just a little fact on the side that everyone should know by default.

O(g) >>> O^+(g) + e^-[br]

First ionisation = 1314kj mo

l^1

Am I suposed to know this or is this something that is just given to me? How do I find the 1314kj/mol?

Reply 1

Plantagenet Crown

Original post by Et Tu, Brute?

How is this calculated?

My book just throws it in there like it was nothing, just a little fact on the side that everyone should know by default.

O(g) >>> O^+(g) + e^-[br]

First ionisation = 1314kj mo

l^1

Am I suposed to know this or is this something that is just given to me? How do I find the 1314kj/mol?

Are you doing A level? if so, I dont think you're expected to work out the ionisation energy, as I think it's mostly done experimentally with some theory here and there. You'll be given the values in a table most probably as it doesn't seem you've been taught how to derive it.

(edited 10 years ago)

Reply 2

Et Tu, Brute?

Original post by Plantagenet Crown

Are you doing A level? if so, I dont think you're expected to work out the ionisation energy, as I think it's mostly done experimentally. You'll be given the values in a table most probably.

undergrad year 0, so A-level content basically yeah.

k, thanks.

Reply 3

Big-Daddy

Original post by Plantagenet Crown

Are you doing A level? if so, I dont think you're expected to work out the ionisation energy, as I think it's mostly done experimentally. You'll be given the values in a table most probably.

How exactly do you propose it should be done experimentally?

Certainly it can be done theoretically, and very much something an A-level student could look into rather than just accepting.

If you have the Bohr model energies for the electrons (or any other model), ionization energy is simply

E(ionization)=E(n=infinity)-E(n)

Where E(n) is the current first quantum number value, n, for the electron. E(n) and E(n=infinity) are both found from the same Bohr model (or other) expression for E in terms of n. http://en.wikipedia.org/wiki/Bohr_model http://en.wikipedia.org/wiki/Rydberg_constant As you will be able to predict as soon as you see E in terms of n, E(n=infinity) is 0.

The calculations become more complicated when you are dealing with a molecule rather than an atom, which is why you will normally be given atomic ionization energies (because at no known level is it possible to exactly predict the ionization energies of molecules as the Schrodinger equation cannot be exactly solved for them).

Reply 4

Borek

Original post by Big-Daddy

If you have the Bohr model energies for the electrons (or any other model), ionization energy is simply

E(ionization)=E(n=infinity)-E(n)

For a one electron atom/ion. Oxygen atom doesn't fit this model.

Reply 5

Big-Daddy

Original post by Borek

For a one electron atom/ion. Oxygen atom doesn't fit this model.

It doesn't? Granted that the 1-electron Bohr model won't work to find E in terms of n but if you do find the relevant function E(n) then E(infinity)-E(current n) should be the ionization energy of the electron for which you have just specified "current n"?

Or probably we will need E(n,l,m) then E(infinity)-E(n,l,m of electron in question). But I thought decent models do exist for these? (Haven't studied them yet though.) Exact solutions impossible of course. But then if it is experimental, how do you determine it?