Chemistry, Finding Ionization Energies

[andii2121]

1. The ionization energies of Rb and Ag are 4.28 eV and 7.57 eV, respectively. Calculate the ionization energies of an H atom with its electron in the same orbitals as the highest energy electrons in these two atoms. Another way of thinking of this is: what would the ionization energies of RB and Ag be if they each only had one proton?




I wasn't sure on what equation to use for this problem or how to go about it.
I tried the E=-hcR∞(z^2/n^2) but got a negative number and that didn't seem right

 

[epenguin]

1. The ionization energies of Rb and Ag are 4.28 eV and 7.57 eV, respectively. Calculate the ionization energies of an H atom with its electron in the same orbitals as the highest energy electrons in these two atoms. Another way of thinking of this is: what would the ionization energies of RB and Ag be if they each only had one proton?




I wasn't sure on what equation to use for this problem or how to go about it.
I tried the E=-hcR∞(z^2/n^2) but got a negative number and that didn't seem right

Well it sounds to me: you are supposed to be able to know from knowledge of atomic structure what orbital the outermost electron in Rb and in Ag are in. And then from the theory of the H atom (Schrödinger) you know what the energy of the corresponding orbitals in H are. You also know from same theory the ionisation energy of H (you remember that corresponds to where the spectral lines crowd together at a limit). Well these last things are also known from experiment, the spectra, i.e. the theory works well. But it might be easier to do the calculation than find the data! I don't know. So the difference between these two energies is the ionisation energy of an electron in that orbital in H.

The comparison with the other elements I guess gives a measure of the screening effect of all the other electrons in Rb and Ag. I had never heard or thought of this method of measuring the screening effect which they talk about in chemistry due to my stunted chemical education. I will be interested if you come back and tell us the conclusions.

Perhaps a proper chemist will come and confirm if this is the question, but it sounds like an interesting one to find the answer to, and useful revision in any case.