Explaining Decreasing Binding Energy of Outermost Electron in Alkali Atoms

[Arcane_00]

I know this question is fairly easy, but I'm completely flabbergasted by it.
I'm sure it has to do with the Ionization energy, I'm just really confused by the negative values for the binding energies. Anyway here is the question.

The absolute value of the binding energy of the outermost electron in the alkali atoms decreases with atomic size such as;
Eb(Li)=-5.395eV
Eb(Na)=-5.142eV
Eb(K)=-4.34eV
Eb(Rb)=-4.17eV
Eb(Cs)=-3.90Ev
(Eb just stands for binding energy)
Provide an explanation for this series.
(I figure this has to do with ionization energy, given the group1 metals has only one valence electron, the binding energy of that outermost electron must be equal to the ionization energy. I just don't know why the binding energy was given to me as a negative value.)

Describe how you could measure the binding energies experimentally and describe an ansatz to calculate them. (cant think of anything for this part)

 

[Moonglum]

Bindin energy is negative, because that represents the energy required to extract the electron from its bound state and take it to a point where it has zero potential energy wrt the atom. Think of the atom as being like a valley of energy in a plain.