Why is the electron-nucleus potential energy negative?

The one-electron energy is negative because it is based on one negative and one positive charge, describing an attractive interaction between one electron and the positively-charged nucleus.

If one were examining the two-electron repulsion energy, it would be positive because the charges are the same sign, indicating a repulsive interaction.

From Coulomb's law,

#V = (q_1q_2)/(4piepsilon_0r_(12))#

where #q# is a charge, #epsilon = 8.854 xx 10^(-12) "F"cdot"m"^(-1)# is the vacuum permittivity, and #r_(12)# is the radial distance between two point charges.

Since electrons have negative charges, #q_1 = -1.602 xx 10^(-19) "C"#, and similarly, protons have #q_2 = 1.602 xx 10^(-19) "C"#.

For hydrogen atom (a proton plus an electron) for simplicity, we thus have:

#V = -e^2/(4piepsilon_0 r)#

• Here, the first charge is the electron, and the other is the proton in hydrogen's nucleus.
• #r# is now the radial distance of the electron from the nucleus (assumed fixed), where #e = q_2# and #-e = q_1#, with #e > 0#.

It is clear that #V < 0#, a negative value for an attractive potential energy of the electron with the nucleus, since #e^2 > 0#, #epsilon_0 > 0#, and #r >= 0#.