# How to calculate electric force on valence electron?

In order to understand the periodic trends more fully, I wanted to calculate the net force exerted on the electron (with the highest principal quantum number and azimuthal quantum number) of an atom.

I decided to use the following procedure:

This is the graph that resulted from these calculations.

For example, for lithium:

• 2s electron - effective nuclear charge: $$\pu{1.279 e} \implies \pu{2.05e-19 C}$$
• atomic radius: $$\pu{167 pm} = \pu{1.67e-10 m}$$
• Coulomb's law: $$\displaystyle \frac{kq_1q_2}{r^2} = \frac{(\pu{8.99e9})(\pu{2.05e-19})(\pu{-1.602e-19})}{(\pu{1.67e-10})^2}$$
• Magnitude of force: $$\pu{-1.06e-8 N} \implies \pu{1.06e-8 N}$$

Is this a proper way to calculate the electric force exerted on the electron I specified?

If so, how does this graph relate with other atomic trends, like ionization energy and electronegativity?

• Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. There were some units missing and I'm not sure I've applied the formatting right. Please check and edit. Aug 19 '20 at 15:42
• I am not sure your calculations make sense. I may be wrong. But the force changes constantly if the distance of the electron to the nucleus changes a bit. An electron has a certain energy on a given orbital, that does not change in time. But the attractive force changes as a function of the distance to the nucleus. it is a little like the vibrating pendulum. The driving force is maximum when the extension is maximum, and equal to zero when its speed is maximum. So what you do is obtaining a sort of intermediate value of the attractive force, which does not mean too much. Aug 20 '20 at 10:29