Effective Nuclear Charge:

The electrostatic energy of attraction between a single
negative charge (electron) and Z units of positive charge is
given simply by -Ze^{2}/r. Here, r is the distance
between the electron and the nucleus. In the Bohr planetary
model, r is fixed. In reality, the electron is diffusely spread
over a range of r values. We'll look at a one-dimensional,
simplified "model" of this electrostatic energy of
attraction. If the electron is at a distance of 100 pm from a
nucleus with Z = 2 (like He, if this were three-dimensional), the
energy would be -Ze^{2}/100 = -e^{2}/ 50
(ignoring units since we'll be making comparisons only). If that
electron were at a distance of 200 pm, the energy of attraction
would be -Ze^{2}/200 = -e^{2}/100.

Now put a second electron into our system. If the first
electron were at r = 100 pm and the second electron were at r =
200 pm, that second electron would feel a net attraction due to
Z-1 positive charges because of screening (shielding) of the
closer negative charge. The second electron's energy of
attraction would then be -(Z-1)e^{2}/200 = -(2-1)e^{2}/200
= -e^{2}/200. Since our expression for attraction of one
electron was -Ze^{2}/r, this second electron thinks it is
attracted by a Z = 1 nucleus since for Z = 1 and r = 200 pm you
get that energy = -e^{2}/200. But the nucleus we
established has Z = 2. Hence, in this circumstance, we refer to
the effect of the nucleus and the other, first electron by saying
the *effective nuclear charge* on electron #2 is 1, or Z_{eff}
= 1.

But this is an oversimplification for electron density
distributions which are spread out. Now suppose our first
electron had an *average* distance from the nucleus of 100
pm, but that 25% of its radial distribution was near 50 pm, 50%
was near 100 pm, and 25% was near 150 pm as pictured below.

The second electron would also have a diffused radial
distribution. First, let's explore how the energy of attraction
of the second electron varies as we move that electron around. If
electron #2 is at 200 pm, we still get the energy of attraction
to be our previous value of -e^{2}/200. This value also
pertains to r for electron #2 all the way in to 150 pm.

When we move electron #2 just inside a distance of 150 pm, it
is now screened by only three-fourths of the charge distributed
by electron #1. This gives an energy of attraction of -(Z-3/4)e^{2}/r
= -(2-.75)e^{2}/150 = -e^{2}/121 (keeping the
form of the answer as was presented above). Since the electron is
at a distance of 150 pm here, we have an effective positive
charge of Z-.75 = 1.25 = Z_{eff} for this arrangement. We
can ignore any electrostatic effect on the outside of where
electron #2 is located.

If we move electron #2 down just inside 100 pm, it is screened
by 25% of electron #1's charge giving an energy of attraction of
-(Z-1/4)e^{2}/100 = -(2-.25)e^{2}/100 = -e^{2}/57.1.
Now the effective positive charge attracting the electron is 1.75
= Z_{eff}.

If we move electron #2 inside 50 pm, it is not screened at all
by negative charge between itself and the Z positive charges of
the nucleus so its energy of attraction is -(Z-0)e^{2}/50
= -e^{2}/25 and Z_{eff} = Z = 2. The animation
below moves electron #2 (outlined in black) in towards the
nucleus of charge +2 and shows how the net positive charge varies
with distance in this artificial scheme.

Our last step is to say that electron #2 also has its
distribution spread out. Let's say, for the sake of illustration,
12.5% beyond 150 pm, 37.5% just inside 150 pm, 37.5% just inside
100 pm, and 12.5% just inside 50pm. For each of these, we already
looked at the net charge attracting electron #2. From the above,
these net positive charges were 1, 1.25, 1.75, and 2. But now we
average them according to their contributions to the overall
distribution of electron #2 and get (12.5%)(1) + (37.5%)(1.25) +
(37.5%)(1.75) + (12.5%)(2) = 1.50. This is the overall effective
nuclear charge, Z_{eff}, attracting electron #2 to the
system.