# What happens if the electron density around a nucleus is decreased?

Oct 19, 2015

Electron density is just the time-dependent probability of finding an electron somewhere. Wherever there is no electron density, no electrons can be observed. Hence, if you decrease electron density at a spot, you'll have a hard time finding an electron at that spot.

According to Electronic Structure Theory (first paragraph), the electron is inextricably linked to the energy of a stationary atom. Ever notice how every time you talk about energy states, relaxations, excitations, bond breaking/making, etc., you talk about the behavior of electrons? That's why.

The energy for the helium atom is approximately:

$E = K + V \left(r\right)$

$\approx \frac{1}{2} {m}_{e 1} {v}_{e 1}^{2} + \frac{1}{2} {m}_{e 2} {v}_{e 2}^{2} - {e}^{2} / \left(4 \pi {\epsilon}_{0} {\vec{r}}_{1}\right) - {e}^{2} / \left(4 \pi {\epsilon}_{0} {\vec{r}}_{2}\right) - \frac{2 {e}^{2}}{4 \pi {\epsilon}_{0} {\vec{r}}_{\text{12}}}$

where $K$ is the kinetic energy of the two electrons with respect to the nucleus, and $V$ is the overall potential energy of the two electrons with respect to the nucleus, as well as the coulombic repulsions between each electron pairwise interaction.

With less and less electrons observed around the nucleus, you simply get the convergence of the energy to $0$, because of course, if there are no electrons, there is no relationship of any electrons to another, nor is there electron movement.

$\textcolor{b l u e}{{\lim}_{\left(K , V\right) \to 0} E} = \cancel{\frac{1}{2} {m}_{e 1} {v}_{e 1}^{2} + \frac{1}{2} {m}_{e 2} {v}_{e 2}^{2}} - \cancel{{e}^{2} / \left(4 \pi {\epsilon}_{0} {\vec{r}}_{1}\right)} - \cancel{{e}^{2} / \left(4 \pi {\epsilon}_{0} {\vec{r}}_{2}\right)} - \cancel{\frac{2 {e}^{2}}{4 \pi {\epsilon}_{0} {\vec{r}}_{\text{12}}}}$

$= \textcolor{b l u e}{\text{0 J}}$