# Kinetic energy (KE) in atomic orbital

Within an atomic orbital, electrons must obviously have relative differences between points in space due to potential gradient. But there is kinetic energy as well. If we choose a particular point as zero potential, the electron will have a base level KE here. What determines this KE? In other words, there must be relative differences in KE, but adding a constant everywhere maintains this difference. What determines this constant?

• "But there is kinetic energy as well." - [citation needed] The orbital states have a definite energy. Whether that is "kinetic" energy is meaningless, since quantum objects do not move in the classical sense. – ACuriousMind Jan 5 '15 at 0:35
• In the Bohr model, electrostatic forces must balance inertial forces. This is why an electron replaced with a heavier particle produces a smaller atomic radius. Within this model, the negative potential energy is equal to 2 times the kinetic energy. I'm sorry to answer your question in a comment, but I've answered this same effective question multiple times on this site already. – Alan Rominger Jan 5 '15 at 0:37

Your saying that "electrons must obviously have relative differences between points in space due to potential gradient" is not rigorous, because an electron is not a localized object inside an atom, is doesn't occupy a definite point in space. Neither has it a well-defined velocity, nor a well-defined kinetic energy. To put it clearly, the electron is not a Newtonian particle that follows a trajectory and has at every time a position and velocity. There is an uncertainty in its location, and so in its linear momentum.

An interesting discussion on this issue was done by R. Feynman, see the question

Quantum mechanics,and how the law $ΔxΔp≥ℏ/2$ explains the paradox regarding atoms

About its kinetic energy, it's the wave-function that can tell you, not the kinetic energy because it doesn't have a precise value, but its mean value. If you want to see a calculus, for the electron on the $n$-th level and zero angular momentum ($l = 0$), in the hydrogen atom

$⟨E_K⟩ = \frac {\hbar ^2}{2m_e n^2 a_0^2}$,

(where $a_0$ is the Bohr radius).

Since you speak of choosing a particular point as zero potential, that's quite difficult as in the Coulombian field the potential increases as $1/r$ when approaching the nucleus.

Could it be that you make a confusion? We often refer to $n=1$ level as ground level, and to its total energy as ground energy. Thus, for an electron to jump from the ground level to a higher level, the atom needs to absorb the energy difference between the higher level and the ground energy. By the way, total energy is already not an average, but a constant value, typical to each level.