# Replacement of a nucleus with electrons by an effective nucleus

This question was asked in my test:

In a lithium atom the outer electron is in the second orbit. The interaction of this outer electron with two inner electrons can be accounted for by assuming that this electron sees a nuclear attraction of $$Z$$ protons (where $$Z < 3$$). The energy required to remove the outer electron is $$\text{5.39 eV}$$. Find the value of $$Z$$.

I started with the thought that I can replace the nucleus and the two electrons with an effective nucleus (the usual thought). I, thus, put the given values in the Bohr's formula (with values substituted)

$$5.39 = 13.6 \times \frac{Z^2}{n^2}$$

and I put $$n = 1$$; the $$Z$$ came out to be $$~ 0.63$$, which seems absurd because the combined screening effect of two electrons cannot be greater than two.

The solution followed the same process (as expected) but used a value of $$n = 2$$. This resulted in a $$Z$$ value of approximately $$1.26$$, which appears more plausible.

My way of thinking was that the nucleus and the two electrons can be considered to be an effective nucleus and the third electron can be considered to be in ground state. Rest was all in correspondence to Hydrogen atom.

Which way is correct? From values, it seems the latter one. But why not the former one?

• Commented May 9 at 17:48
• You cannot change for the 3rd electron its n=2 to n=1 just because you have replaced Li nucleus and 1s electrons by an effective equivalent charge. // BTW, Z_eff by the referred Slater rules is 3 - 2 x 0.85 = 1.3 what is in good agreement with 1.26. Commented May 11 at 11:12
• @Poutnik Your argument needs a valid reason, the thing I am searching for. Commented May 11 at 11:46
• Not more than putting n=1. // Z_eff concept does not assume hydrogen-like atom with Z_eff nucleus charge. For Li case, it does not assume 1s orbital with it's 2 electrons disappeared nor that 2s orbital is becoming 1s orbital. Commented May 11 at 13:22

There is indeed something to say for your $$n=1$$ approach. In muonic Helium, for example, it is a decent approximation to consider the nucleus + muon to form an effective nucleus (a muon is a heavy cousin of the electron), and therefore the remaining electron to be in the $$1S$$ state. In the case at hand, however, perhaps the variable-$$n$$ approach better reflects the interactions between electrons, which we ignore if we say $$n=1$$.

Nevertheless, I don't think one can say either method is completely "correct": they remain gross approximations. For example, what happens when we apply the same methods for a $$Z=92$$ atom (Uranium)? In the varying-$$n$$ method the effective charge would scale with $$n$$, yielding $$Z_{\text{eff}}=4.7$$ whereas in the $$n=1$$ method the answer would remain roughly the same at $$Z_{\text{eff}}=0.7$$. I'm not sure which of these makes more sense...

The "second orbit" mentioned in the question means "$$n=2$$".

The only $$n=1$$ orbital in any atom with a point-like nucleus can contain only one pair of electrons.
https://en.wikipedia.org/wiki/Pauli_exclusion_principle#Atoms

Things are different when some electrons are replaced by other types of particles, as Pallas mentioned in a previous answer. In such 'exotic atoms', the fermions are no longer identical. All known exotic atoms (and molecules) are unstable.

Which way is correct? From values, it seems the latter one. But why not the former one?

The latter one. It is possible to invent new concepts, of course, but then it all comes down to their usefulness.