The Energy Shells (The Radial Solution)

In the previous chapters, we solved the "Angular" half of the atom, which gave us the shapes of the orbitals ($s,p,d,f$). Now, we tackle the "Radial" half: the part that determines how far the electron is from the nucleus and, most importantly, its total energy.

Solving this final equation will reveal the "Energy Shells" that form the rows of the periodic table.

5.1 The Radial Equation

From our separation in Chapter 3, we have the following equation for $R(r)$:

$$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)-\frac{2m{r}^2}{{\hslash }^2}[V(r)-E]R=\ell (\ell +1)R$$

Here, $V(r)$ is the Coulomb Potential, the electrostatic "pull" of the nucleus:

$$V(r)=-\frac{Z{e}^2}{4\pi {ϵ}_{0}r}$$

Unlike the angular equation, which was universal for all spheres, the radial equation depends on the specific physical properties of the atom: the charge of the nucleus ($Z$) and the total energy ($E$).

5.2 The Strategy: Asymptotic Behavior

This equation is notoriously difficult to solve because $V(r)$ becomes infinite at $r=0$. To crack it, physicists analyze what the equation reduces to at extreme limits, then combine both into a full solution. Expanding the radial derivative and dividing by $r^2$ gives a cleaner form:

$$\frac{d^{2}R}{d{r}^2}+\frac{2}{r}\frac{dR}{dr}-\frac{2m}{{\hslash }^2}[V(r)-E]R=\frac{\ell (\ell +1)}{r^2}R$$

1. At large distances ($r\to \mathrm{\infty }$): Every term containing $\frac{1}{r}$ or $\frac{1}{r^2}$ vanishes — including $V(r)\propto \frac{1}{r}$ and the centrifugal term. The equation collapses to:

$$\frac{d^{2}R}{d{r}^2}\approx \frac{-2mE}{{\hslash }^2}R$$

For a bound electron $E<0$, so $k^2\equiv \frac{-2mE}{{\hslash }^2}>0$. The two mathematical solutions are $R\sim e^{+kr}$ and $R\sim e^{-kr}$. The first blows up to infinity as $r\to \mathrm{\infty }$ — an electron cannot have infinite probability at infinite distance — so physics discards it. Only the exponential decay survives: $R\sim e^{-kr}$.

2. At small distances ($r\to 0$): Dividing by $r^2$ makes the centrifugal term $\frac{\ell (\ell +1)}{r^2}$ grow much faster than the $\frac{1}{r}$ potential or the constant $E$. It dominates everything else, and the equation reduces to:

$$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)\approx \ell (\ell +1)R$$

Trying $R\sim r^s$ and substituting: the left side becomes $s(s+1)r^s$, so we need $s(s+1)=\ell (\ell +1)$, giving $s=\ell$ or $s=-(\ell +1)$. The second solution diverges at the origin — infinite probability density right at the nucleus is unphysical — so physics keeps $R\sim r^{\ell }$. For $\ell >0$ this forces the probability to zero at $r=0$: the centrifugal barrier acts as a wall preventing collapse into the nucleus. Only $s$-orbitals ($\ell =0$) have a non-zero probability at the origin, which has deep consequences for shielding and penetration (Section 5.5).

3. The Factoring Strategy: Knowing both limiting behaviors, we factor them out explicitly:

$$R(r)=r^{\ell }\cdot v(r)\cdot e^{-kr}$$

By stripping away the known extremes ($r^{\ell }$ near the origin, $e^{-kr}$ far away), the residual function $v(r)$ is guaranteed to be finite and well-behaved everywhere in between. Substituting this ansatz back into the full radial equation and expanding $v$ as a power series allows us to solve for the exact coefficients — which turn out to be the Associated Laguerre Polynomials of Section 5.3.

5.3 Solving for $r$: The Laguerre Polynomials

The math leads us to another set of famous functions — this time from the 19th century — the Associated Laguerre Polynomials, $L_{n-\ell -1}^{2\ell +1}(x)$:

$$L_q^p(x)=\frac{e^x{x}^{-p}}{q!}\frac{d^q}{d{x}^q}(e^{-x}{x}^{q+p})$$

For the specific values of $n$ and $\ell$ used in the Hydrogen atom, see the Mathematical Annex.

Visualize the solutions? 🤔

Explore the "onion-layer" radial node structure of these wave functions: 3D Radial Node Visualizer. For the explicit formulas, see the Mathematical Annex.

The Physical Constraint: Just like the particle in a box (Chapter 2), the math only works if we introduce an integer to keep the solution from blowing up to infinity. Remarkably, it is the very same integer we first met in the 1-D box of Chapter 2 — there it counted the standing-wave modes allowed between the walls; here it returns in three dimensions as the Principal Quantum Number:

$n$ (The Principal Quantum Number)

Together with $\ell$ and $m_{\ell }$ from Chapter 4, $n$ completes the three spatial quantum numbers of the hydrogen atom. For the math to stay stable, $n$ must be an integer, and it must always be larger than the azimuthal quantum number $\ell$. This creates the following hierarchy:

• $n=1$ ($K$ shell): $\ell =0$ ($s$).
  1 subshell ($1s$).
• $n=2$ ($L$ shell): $\ell =0,1$ ($s,p$).
  2 subshells ($2s,2p$).
• $n=3$ ($M$ shell): $\ell =0,1,2$ ($s,p,d$).
  3 subshells ($3s,3p,3d$).
• $n=4$ ($N$ shell): $\ell =0,1,2,3$ ($s,p,d,f$).
  4 subshells ($4s,4p,4d,4f$).
• $n=5$ ($O$ shell): $\ell =0,1,2,3,4$ ($s,p,d,f,g$).
  5 subshells ($5s,5p,5d,5f,5g$).
• $n=6$ ($P$ shell): $\ell =0,\dots ,5$ ($s,p,d,f,g,h$).
  6 subshells ($6s,6p,6d,6f,6g,6h$).
• $n=7$ ($Q$ shell): $\ell =0,\dots ,6$ ($s,p,d,f,g,h,i$).
  7 subshells ($7s,7p,7d,7f,7g,7h,7i$).

NOTE: Although higher shells mathematically contain many subshells, no known element in its ground state occupies orbitals beyond $7p$. For example, $5g,6f,7d$, and above are never used in ground-state chemistry.

5.4 The Result: Quantized Energy

Recall the power-series function $v(r)$ introduced in Section 5.2. For the wave function to stay finite everywhere, that series cannot run on forever — it must terminate after a finite number of terms. This termination is possible only for particular values of the energy, and that is what quantizes $E$: it is the radial counterpart of the boundary condition $\sin (kL)=0$ that quantized the particle in a box (Chapter 2).

Carrying this through, we finally derive the exact formula for the energy levels of a Hydrogen-like atom:

$$E_n=-\frac{Z^{2}m{e}^4}{32{\pi }^2{ϵ}_0^2{\hslash }^2}\frac{1}{n^2}$$

For Hydrogen ($Z=1$), this simplifies to the famous value:

$$E_n=-\frac{13.6\text{ eV}}{n^2}$$

The negative sign tells us the electron is "trapped" in the atom's potential well. To escape, it must absorb exactly enough energy to reach $E=0$.

Summary of Energy Levels (Hydrogen, $Z=1$)

$n$	Shell	Energy ($E_n$)
1	K	-13.60 eV
2	L	-3.40 eV
3	M	-1.51 eV
4	N	-0.85 eV
5	O	-0.54 eV
6	P	-0.38 eV
7	Q	-0.28 eV

5.5 The Reality: Multi-Electron Atoms

In the Hydrogen model we've discussed, the energy depends only on the principal quantum number $n$. This means a $2s$ electron and a $2p$ electron would have the exact same energy (they are "degenerate").

However, in reality, most atoms have more than one electron. This introduces three critical effects that break this degeneracy:

1. Electron-Electron Repulsion: Electrons don't just feel the "pull" of the nucleus; they also feel the "push" of other electrons.

2. Shielding (Screening): Inner electrons "shield" outer electrons from the full positive charge of the nucleus. An outer electron feels an Effective Nuclear Charge ($Z_{eff}$) that is lower than the actual $Z$.

3. Penetration: It is a common misconception that subshell shapes ($s,p,d,f$) only affect the direction of the electron. In fact, $\ell$ also changes the radial probability distribution (the likelihood of finding the electron at a certain distance).

   • The Centrifugal Barrier: High angular momentum ($\ell$) acts like an outward "force" that keeps electrons away from the center.
   • Inner Lobes: Lower $\ell$ orbitals (especially $s$) have small "peaks" of probability very close to the nucleus (see the "Secondary Peaks" in section 5.6). These "inner lobes" allow the electron to penetrate the shielding cloud of inner electrons and feel the full, unshielded pull of the nucleus.
   • Ultimate Penetration (r=0): Most importantly, $s$-orbitals are the only ones with a non-zero probability of finding the electron at the very center of the nucleus. All other orbitals ($p,d,f$) have a "node" at $r=0$, meaning they are strictly forbidden from the center.

   Because $s$-orbitals penetrate deeper than $p$-orbitals (and $p$ deeper than $d$), they are more "tightly bound" and have lower energy for the same value of $n$. This creates the subshell splitting:

$$E_{ns}<E_{np}<E_{nd}<E_{nf}$$

This splitting is what causes the "overlap" in the periodic table, such as why the $4s$ subshell fills before the $3d$ subshell.

Simplified Multi-Electron Energy

While the exact energy levels for multi-electron atoms require complex computer simulations, we can approximate them using the Effective Nuclear Charge ($Z_{eff}$):

$$E_n\approx -\frac{Z_{eff}^2\cdot 13.6\text{ eV}}{n^2}$$

Here, $Z_{eff}=Z-\sigma$, where $\sigma$ is a "shielding constant" that represents the push-back from other electrons. For example, a $2s$ electron in Lithium ($Z=3$) might feel a $Z_{eff}\approx 1.28$ because the two $1s$ electrons shield it from the nucleus.

5.6 Most Probable Radius ($r_{mp}$)

While the wavefunction $\psi$ gives the probability amplitude at a specific point, we are often interested in the probability of finding the electron at a certain distance $r$ from the nucleus, regardless of direction.

The Radial Probability Density, $P(r)$

The probability of finding the electron between $r$ and $r+dr$ is obtained by integrating $|\psi {|}^2$ over all angles. Since the spherical harmonics are normalized ($\int |Y_{\ell }^{m_{\ell }}{|}^{2}d\mathrm{\Omega }=1$), the angular part integrates to 1, leaving:

$$P(r)dr=r^2|R_{n\ell }(r){|}^{2}dr$$

The Radial Probability Density is therefore:

$$P(r)=r^2|R_{n\ell }(r){|}^2$$

Where did the $4\pi$ go? 🤔

Introductory chemistry writes radial probability as $4\pi r^2|\psi {|}^2$ for $s$-orbitals, where the $4\pi$ is the surface area of a sphere. That shortcut works only for spherically symmetric ($\ell =0$) states.

This formula is the general version. Since $\psi =R_{n\ell }(r)Y_{\ell }^{m_{\ell }}(\theta ,\varphi )$ and the spherical harmonics are normalized,

$$\int |\psi {|}^{2}d\mathrm{\Omega }=|R_{n\ell }{|}^2\underset{=1}{\underbrace{\int |Y_{\ell }^{m_{\ell }}{|}^{2}d\mathrm{\Omega }}}$$

the angular integral — including the $4\pi$ — is absorbed automatically. What remains, $P(r)=r^2|R_{n\ell }{|}^2$, holds for all orbital types ($s,p,d,f$) without adjustment.

Calculating the Maximum Likelihood

The most probable radius $r_{mp}$ (where you are most likely to find the electron) corresponds to the peaks of the $P(r)$ curve. To find these peaks, we solve for:

$$\frac{dP}{dr}=0$$

Summary of Radial Properties ($a_0=1$)

The following table summarizes the key radial characteristics of the hydrogen atom up to $n=5$.

$n$	$\ell$	Orbital	Node at $r=0$	Radial Nodes	Main $r_{mp}$	Secondary Peaks
1	0	1s	False	False	1.00	None
2	0	2s	False	True (1)	5.236	0.764
	1	2p	True	False	4.00	None
3	0	3s	False	True (2)	13.07	3.82, 0.71
	1	3p	True	True (1)	12.00	3.00
	2	3d	True	False	9.00	None
4	0	4s	False	True (3)	21.92	11.44, 5.14, 1.50
	1	4p	True	True (2)	20.92	10.68, 4.40
	2	4d	True	True (1)	18.62	9.38
	3	4f	True	False	16.00	None
5	0	5s	False	True (4)	33.00	19.55, 10.90, 5.07, 1.48
	4	5g	True	False	25.00	None

The $n=5$ row lists only the two extremes — $5s$ (the deepest-penetrating subshell) and $5g$ (the highest angular momentum, with no radial nodes); $5p$, $5d$ and $5f$ follow the same trends seen in their $n=4$ counterparts.

5.7 Summary of the Radial Logic

1. Boundary Condition (Infinity): The electron must stay near the nucleus, forcing an exponential decay.
2. Boundary Condition (Origin): Angular momentum ($\ell$) creates a "centrifugal barrier" that keeps the electron away from the center.
3. The Result: The interaction between these two forces quantizes the atom's energy, creating the structured "rows" (periods) of the periodic table.

In the final Chapter 6, we add the very last piece of the puzzle—Spin—and see how all these pieces click together to build the 118 elements.