The Schrödinger-equation solution for the hydrogen atom matches the experimental evidence, but so did the Bohr hydrogen atom. The real test of Schrödinger’s theory is how well it works for multielectron atoms. A neutral multielectron atom consists of $Z$ electrons surrounding a nucleus with $Z$ protons and charge $+Ze$. Z, the atomic number, is the order in which elements are listed in the periodic table. Hydrogen is $Z=1$, helium $Z=2$, lithium $Z=3$, and so on.

The potential-energy function of a multielectron atom is that of $Z$ electrons interacting with the nucleus and Z electrons interacting with each other. The electron-electron interaction makes the atomic-structure problem more difficult than the solar-system problem, and it proved to be the downfall of the simple Bohr model. The planets in the solar system do exert attractive gravitational forces on each other, but their masses are so much less than that of the sun that these planet-planet forces are insignificant for all but the most precise calculations. Not so in an atom. The electron charge is the same as the proton charge, so the electron-electron repulsion is just as important to atomic structure as is the electron-nucleus attraction.

The potential energy due to electron-electron interactions fluctuates rapidly in value as the electrons move and the distances between them change. Rather than treat this interaction in detail, we can reasonably consider each electron to be moving in an average potential due to all the other electrons. That is, electron $i$ has potential energy

where the first term is the electron’s interaction with the $Z$ protons in the nucleus and $U_{\text{elec}}$ is the average potential energy due to all the other electrons. Because each electron is treated independently of the other electrons, this approach is called the independent particle approximation, or IPA. This approximation allows the Schrödinger equation for the atom to be broken into $Z$ separate equations, one for each electron.

A major consequence of the IPA is that each electron can be described by a wave function having the same four quantum numbers $\mathit{n}\mathit{\text{,}}\mathit{l}\mathit{\text{,}}\mathit{m}\mathit{\text{,}}\mathbf{\text{and}}{\mathit{m}}_{\mathit{s}}$ used to describe the single electron of hydrogen. Because $m$ and $m_s$ do not affect the energy, we can still refer to electrons by their $n$ and $l$ quantum numbers, using the same labeling scheme that we used for hydrogen.

A major difference, however, is that the energy of an electron in a multielectron atom depends on both n and l. Whereas the $2s$ and $2p$ states in hydrogen had the same energy, their energies are different in a multielectron atom. The difference arises from the electron-electron interactions that do not exist in a single-electron hydrogen atom.

FIGURE 41.14 shows an energy-level diagram for the electrons in a multielectron atom. For comparison, the hydrogen-atom energies are shown on the right edge of the figure. The comparison is quite interesting. States in a multielectron atom that have small values of $l$ are significantly lower in energy than the corresponding state in hydrogen. For each $n$, the energy increases as $l$ increases until the maximum-$l$ state has an energy very nearly that of the same $n$ in hydrogen. Can we understand this pattern?

FIGURE 41.14 An energy-level diagram for electrons in a multielectron atom.

Indeed we can. Recall that states of lower $l$ correspond to elliptical classical orbits and the highest-$l$ state corresponds to a circular orbit. Except for the smallest values of $n$, an electron in a circular orbit spends most of its time outside the electron cloud of the remaining electrons. This is illustrated in FIGURE 41.15. The outer electron is orbiting a ball of charge consisting of $Z$ protons and $(Z-1)$ electrons. This ball of charge has net charge $q_{\text{net}}=+e$, so the outer electron “thinks” it is orbiting a proton. An electron in a maximum-$l$ state is nearly indistinguishable from an electron in the hydrogen atom; thus its energy is very nearly that of hydrogen.

FIGURE 41.15 High-$l$ and low-$l$ orbitals in a multielectron atom.

The low-$l$ states correspond to elliptical orbits. A low-$l$ electron penetrates in very close to the nucleus, which is no longer shielded by the other electrons. The electron’s interaction with the $Z$ protons in the nucleus is much stronger than the interaction it would have with the single proton in a hydrogen nucleus. This strong interaction lowers its energy in comparison to the same state in hydrogen.

As we noted earlier, a quantum electron does not really orbit. Even so, the probability density of a $3s$ electron has in-close peaks that are missing in the probability density of a $3d$ electron, as you should confirm by looking back at Figure 41.7. Thus a low-$l$ electron really does have a likelihood of being at small $r$, where its interaction with the $Z$ protons is strong, whereas a high-$l$ electron is most likely to be farther from the nucleus.