41.2.2 Angular Momentum and Orbit Shapes

But why is it the $3 d$ state that agrees with the Bohr atom rather than $3 s$ or $3 p$ ? All states with the same value of $n$ form a collection of “orbits” having the same energy. In FIGURE 41.8, the state with $l = n - 1$ has the largest angular momentum of the group. Consequently, the maximum- $l$ state corresponds to a circular classical orbit and matches the circular orbits of the Bohr atom.

The figure illustrates the angular momentum of different orbit shapes.

FIGURE 41.8 Full Alternative Text

States with smaller $l$ correspond to elliptical classical orbits. You can see in Figure 41.7 that the radial probability density of a $3 s$ electron has a peak close to the nucleus. The $3 s$ electron also has a good chance of being found farther from the nucleus than a $3 d$ electron, suggesting an orbit that alternately swings in near the nucleus, then moves out past the circular orbit. This distinction between circular and elliptical orbits will be important when we discuss the energy levels in multielectron atoms.

NOTE In quantum mechanics, nothing is really orbiting. However, the probability densities for the electron to be, or not to be, any given distance from the nucleus mimic certain aspects of classical orbits. They provide a useful analogy.

You can see in Figure 41.7 that the most likely distance from the nucleus of an $n = 1$ electron is approximately $a_{B}$ . The distance of an $n = 2$ electron is most likely to be between about $3 a_{B}$ and $7 a_{B}$ . An $n = 3$ electron is most likely to be found between about $8 a_{B}$ and $15 a_{B}$ . In other words, the radial probability densities give the clear impression that each value of $n$ has a fairly well-defined range of radii where the electron is most likely to be found. This is the basis of the shell model of the atom that is used in chemistry.

However, there’s one significant puzzle. In Figure 41.4, the fuzzy sphere representing the $1 s$ ground state is densest at the center, where the electron is most likely to be found. This maximum density at $r = 0$ agrees with the $1 s$ radial wave function of Figure 41.5, which is a maximum at $r = 0$ , but it seems to be in sharp disagreement with the $1 s$ graph of Figure 41.7, which is zero at the nucleus and peaks at $r = a_{B}$ .

To resolve this puzzle, we must distinguish between the probability density $| ψ (x, y, z) |^{2}$ and the radial probability density $P_{r} (r)$ . The $1 s$ wave function, and thus the $1 s$ probability density, really does peak at the nucleus. But $| ψ (x, y, z) |^{2}$ is the probability of being in a small volume $δ V$ , such as a small box with sides $δ x, δ y$ , and $δ z$ , whereas $P_{r} (r)$ is the probability of being in a spherical shell of thickness $δ r$ . Compared to $r = 0$ , the probability density $| ψ (x, y, z) |^{2}$ is smaller at any one point having $r = a_{B}$ . But the volume of all points with $r \approx a_{B}$ (i.e., the volume of the spherical shell at $r = a_{B}$ ) is so large that the radial probability density $P_{r}$ peaks at this distance.

To use a mass analogy, consider a fuzzy ball that is densest at the center. Even though the density away from the center has decreased, a spherical shell of modest radius $r$ can have more total mass than a small-radius spherical shell of the same thickness simply because it has so much more volume.

EXAMPLE 41.3 Maximum probability

Show that an electron in the $2 p$ state is most likely to be found at $r = 4 a_{B}$ .

SOLVE We can use the $2 p$ radial wave function from Equations 41.7 to write the radial probability density

\begin{array}{l} P_{r} (r) & = & 4 π r^{2} {| R_{2 p} (r) |}^{2} = 4 π r^{2} {[\frac{1}{\sqrt{24 π a_{B}^{3}}} (\frac{r}{2 a_{B}}) e^{- r / 2 a_{B}}]}^{2} \\ = & C r^{4} e^{- r / a_{B}} \end{array}

where $C = {(24 a_{B}^{5})}^{- 1}$ is a constant. This expression for $P_{r} (r)$ was graphed in Figure 41.7.

Maximum probability occurs at the point where the first derivative of $P_{r} (r)$ is zero:

\begin{array}{l} \frac{d P_{r}}{d r} & = & C (4 r^{3}) (e^{- r / a_{B}}) + C (r^{4}) (\frac{- 1}{a_{B}} e^{- r / a_{B}}) \\ = & C r^{3} (4 - \frac{r}{a_{B}}) e^{- r / a_{B}} = 0 \end{array}

This expression is zero only if $r = 4 a_{B}$ , so $P_{r} (r)$ is maximum at $r = 4 a_{B}$ . An electron in the $2 p$ state is most likely to be found at this distance from the nucleus.