Physicists talk about the electron in a hydrogen atom being in an “energy well”. Explain why the difference in energy, ∆E, is negative. What does a negative ∆E represent?

Apr 3, 2018

$\Delta E$ might refer to the electrostatic potential energy of the electron in the potential well surrounding the atomic nucleus.

Explanation:

Taking a nucleus as a point charge of electrostatic charge ${q}_{1}$, the strength of its electrostatic interaction on an electron of charge $- {q}_{2}$ would be proportional to the product of their charge, ${q}_{1} \cdot {q}_{2}$ while being inversely proportional to the square of their separation $r$. This relationship is known as the Coulomb's law.
$F = {k}_{e} \cdot \frac{{q}_{1} \cdot \left(- {q}_{2}\right)}{r} ^ 2$

Note that $F$ is negative here since it resembles attraction.

The electrostatic potential energy of the electron is defined as the energy it takes to bring them apart to an infinite separation, such that there no longer any interaction between the two particles. [2] The relationship between this energy, the charges of the two particles, and their initial separation ${r}_{0}$ can be modeled with the equation

${E}_{P} = - {k}_{e} \cdot \frac{{q}_{1} \cdot {q}_{2}}{r} _ 0$

${E}_{P}$ is also negative here since opposite charges attract each other, so it takes energy to split them apart.

Plotting (in two dimensions) the electrostatic potential energy against the separation between the two charges shall give something that looks like half of the cross-section of a well- the smaller the separation, the lower the electrostatic potential energy. And that's why the electrostatic interaction behaves in a way similar to gravity acting on an object in a well.

graph{-1/x [-1.315, 8.685, -4.8, 0.2]}
color(grey)("Graph resembling the shape of "E_P" plotted against separation r"

FYI: With some knowledge of calculus it is possible to derive this expression by integrating the magnitude of their separation with reference to the separation, from the initial distance to the final separation of infinity
${E}_{P} = {\int}_{{r}_{0}}^{\infty} F \cdot \mathrm{dr} = {\int}_{{r}_{0}}^{\infty} \frac{{q}_{1} \cdot \left(- {q}_{2}\right)}{r} ^ 2$
$= {\left[{k}_{e} \cdot \frac{{q}_{1} \cdot {q}_{2}}{r} _ 0\right]}_{{r}_{0}}^{\infty} = {k}_{e} \cdot \frac{{q}_{1} \cdot {q}_{2}}{\infty} - {k}_{e} \cdot \frac{{q}_{1} \cdot {q}_{2}}{r} _ 0$
$= - {k}_{e} \cdot \frac{{q}_{1} \cdot {q}_{2}}{r} _ 0$