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Does an electron in an orbital have exactly the same energy as the orbital's shell?

Physics Asked on January 3, 2021

Solving the Schrödinger equation gives a wave function for each electron in an atom of any element. The wave functions under the atom can be squared to yield probability distribution maps, or orbitals, for electrons in that atom.

Since an electron moves around the nucleus, it possesses at least kinetic energy, gravitational potential energy, and electrical potential energy. When we talk about the energy of the electron, we usually mean the sum of all forms of its energy.

According to the quantum-mechanical model of atoms, each electron orbital falls under a shell, which has a very clearly defined energy value. For example, the $2s$, $2p_x$, $2p_y$, and $2p_z$ orbitals are all deemed to possess the energy value at $n = 2$. When an electron is assigned to an orbital, does it possess exactly the same amount of energy as the orbital’s shell?

My intuitive answer is no, since an orbital covers a sizable portion of atomic space. An electron cannot possess exactly the same total energy at any position within this orbital space. But if an electron can possess more or less total energy than its orbital’s shell depending on its position, what rule determines that it must belong to the orbital in question, instead of an orbital with lower or higher shell energy?

One Answer

In hydrogen

For single-electron problems, the "energy of the electron" is exactly the energy of the orbital. There is no "energy of the electron" separate from the energy eigenvalue of its wavefunction in the time-independent Schrödinger equation (which is the specific technical quantity often summarized as the 'energy of the orbital').

In particular, when you say

an orbital covers a sizable portion of atomic space. An electron cannot possess exactly the same total energy at any position within this orbital space.

that's just your classical intuition leading you astray. The electron doesn't have a position (unless and until you perform a projective measurement of position), and it is instead described by a delocalized wave which covers an extended region of space. Some of its energy comes from a position-based potential energy, but the kinetic energy (which is proportional to the Laplacian) is a direct measure of how wavy and localized the wave gets. That's what energy is like in QM; the rest is just classical baggage that needs to be discarded.


In multi-electron atoms

Some words of warning:

  • In multi-electron atoms, the potential energy includes the electron-electron electrostatic repulsion. This energy belongs to the system as a whole, and cannot be assigned to any individual electron.

  • Moreover, in multi-electron atoms orbitals are not well-defined physical quantities, and they are only uniquely determined within the Hartree-Fock method which is, at best, an approximation.

Answered by Emilio Pisanty on January 3, 2021

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