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Start here — today's preview. Watch “Why Real Atoms Don't Look Like This - Quantum Numbers to Understand Atomic Structure by Parth G” from Parth G before (or after) this session. The lab will wait. Mostly.
Why Real Atoms Don't Look Like This - Quantum Numbers to Understand Atomic Structure by Parth G
Parth G explains why the tidy shell picture of the atom misleads, then introduces the principal quantum number n and the orbital quantum number l that label real atomic states, with a short look at the other two and the exclusion principle. Watch for his account of subshells and orbital angular momentum, then let this session sharpen it: for l = 1 the full angular momentum is √2 times ħ, while its component along any one axis is at most ħ.
Watch on YouTubeMore from the video vault
We finally understood orbital shapes intuitively! (My mind is blown)
A patient, intuition-first rebuild of the orbitals: a one-dimensional standing-wave analogy, probability density versus radial probability, why p orbitals look like dumbbells and d orbitals like double dumbbells, radial versus angular nodes, and where the quantum numbers and the counts of three, five and seven orbitals come from. Watch it after Parts 3 and 4; it is the long-form companion to this session.
Watch on YouTube6. Hydrogen Atom Wavefunctions (Orbitals)
Lecture 6 of MIT's 5.111 Principles of Chemical Science (Fall 2014), taught by Catherine Drennan: where the electron is likely to be found, radial probability distributions and the hydrogen wavefunctions, with a guest segment on spin and magnetic resonance imaging that makes a fine bridge to the next chapter.
Watch on YouTube- Why Real Atoms Don't Look Like This - Quantum Numbers to Understand Atomic Structure by Parth G — Parth G — https://www.youtube.com/watch?v=EhIbCrl1pGw
- We finally understood orbital shapes intuitively! (My mind is blown) — FloatHeadPhysics — https://www.youtube.com/watch?v=M--6_0F62pQ
- 6. Hydrogen Atom Wavefunctions (Orbitals) — MIT OpenCourseWare — https://www.youtube.com/watch?v=V-RPM3e8Ws0
Today's mission
Dr. Q · cold openPsigor, draw the blinds and fetch the hydrogen! Last session we learned that a measurement throws the quantum dice. Tonight we throw them a billion times at the simplest atom in the universe, one proton and one electron, and plot where the electron turns up. Bohr promised us neat little racetracks. What we will actually find are spheres nested inside spheres, dumbbells, cloverleaves and stranger shapes still, clouds with holes in them where the electron is never found, every one labelled by a handful of whole numbers that will one day build the periodic table. No orbit survives tonight's experiment. Not one! Back to the blackboard!
By the end of this session you will be able to:
- Describe how solving Schrödinger's equation for an electron in the proton's Coulomb potential produces the energies Eₙ = −13.6 eV/n² and three quantum numbers
- State the allowed values of n, l, mₗ and mₛ, and use them to list and count the states of any level, including the n² and 2n² degeneracies
- Name the s, p, d and f orbitals, describe their shapes, and count their radial and angular nodes
- Interpret an orbital as a probability cloud rather than a path or a container, and show that the 1s electron's most probable distance is the Bohr radius, 0.0529 nm
- Explain how fine structure and the Zeeman effect exposed relativity and electron spin, and calculate a simple Zeeman splitting
- Connect hydrogen's shells and subshells to the structure of the periodic table
Read it — or listen along
7 parts. Press play on any part and the paragraph being read lights up as Dr. Q speaks.
From Racetracks to Standing Waves
Chapter 5 left hydrogen with the right energies and the wrong picture: Bohr's electron circled the proton on fixed racetracks that the uncertainty principle of chapter 8 forbids. Tonight Schrödinger's equation, as chapter 10 promised, supplies the right picture. Hydrogen is the ideal patient: one proton, one electron and the electrical attraction between them. Its potential energy, V(r) = −e²/(4πε₀r), depends only on the distance r between the particles, not on direction; a potential like this is called central. The proton is 1,836 times heavier than the electron, so we treat it as fixed; letting it move shifts every energy by only about 0.05 percent.
With no preferred direction, the natural coordinates are spherical: the distance r from the nucleus, a polar angle θ measured down from a chosen z axis, and an azimuthal angle φ around that axis, like latitude and longitude on a globe. In these coordinates Schrödinger's equation separates: every solution factors into a radial part depending only on r and an angular part depending only on θ and φ. The angular parts are spherical harmonics, patterns Pierre-Simon Laplace and Adrien-Marie Legendre studied in the 1780s to describe the gravity of planets; the radial parts are polynomials times an exponential that fades with distance.
Quantization then arrives as it did for the particle in a box of chapter 11, from the demand that the wavefunction be well behaved. Going once around the z axis must bring ψ back to its starting value, so its phase can turn only a whole number of times per circuit; that whole number is mₗ. Staying finite at the poles restricts the angular pattern to a whole number l at least as large as the size of mₗ. And the radial part must fade away far from the nucleus rather than blow up, which happens only at special energies, Eₙ = −13.6 eV/n², with n a whole number greater than l. Bohr's ladder returns, derived rather than postulated.
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Three Numbers for Three Dimensions, Plus One
Each solution is labelled by the whole numbers its boundary conditions produced, its quantum numbers. The principal quantum number n = 1, 2, 3 and so on fixes the energy: −13.6 eV for n = 1, then −3.40, −1.51 and −0.85 eV, crowding toward zero, where the electron breaks free. It also sets the cloud's overall size, which swells roughly as n², and its total number of nodes, n − 1, the surfaces where the wavefunction vanishes. In hydrogen the energy depends on n alone; the other numbers change the cloud's shape and orientation but not its energy.
The orbital quantum number l runs from 0 up to n − 1 and fixes the magnitude of the orbital angular momentum, the rotational momentum of a whirling object: L = √(l(l + 1))·ħ, which is 0, about 1.41ħ, 2.45ħ and 3.46ħ for l = 0, 1, 2 and 3. Here is the first nail in the coffin of the orbit. In the ground state, n = 1 forces l = 0, so the electron has no orbital angular momentum whatsoever, where Bohr had given it exactly ħ. Anything circling a centre carries angular momentum. The 1s electron carries none, so it is not circling anything.
The magnetic quantum number mₗ takes the 2l + 1 whole-number values from −l to +l and fixes the component of angular momentum along the z axis, exactly mₗħ. For any l above zero, even the largest component, lħ, is shorter than the full length √(l(l + 1))·ħ, so the angular momentum never points exactly along any axis: its components are incompatible observables, as chapter 14's commutators warned. Finally, nature adds a fourth label that Schrödinger's equation knows nothing about: spin, an intrinsic angular momentum whose measured component along any axis is always +ħ/2 or −ħ/2, recorded by the spin quantum number mₛ = ±½.
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Clouds, Not Racetracks
What, then, is an orbital? It is one of these solutions: a single-electron wavefunction labelled by its n, l and mₗ, a three-dimensional standing wave. By the Born rule of chapter 9, |ψ|² is its probability density, the probability per unit volume of finding the electron at a point. The chemist Robert Mulliken coined the word in 1932, short for one-electron orbital wave function, and its echo of orbit has misled students ever since. Textbook shapes are boundary surfaces enclosing, by convention, about 90 percent of the probability; the cloud itself has no edge.
Take the ground state, 1s. Its wavefunction is spherically symmetric and largest at the nucleus, shrinking by a factor of e, about 2.718, for every Bohr radius of distance; the Bohr radius, a₀, is about 0.0529 nm. So the most probable point is the nucleus itself, yet the most probable distance is a₀, because there is little room near the centre and a great deal farther out. The chance of finding the electron in a thin spherical shell of radius r, per unit of the shell's thickness, is the density times the shell's area, 4πr² times |ψ|², a quantity called the radial probability. Rising area and falling density compete, and their product peaks at exactly r = a₀: Bohr's radius survives, not as a track but as the summit of a probability distribution.
Now retire the first myth. Nothing in the 1s cloud goes round: it is a stationary state with zero angular momentum and no preferred direction to circulate in. The cloud is not a long-exposure photograph of a tiny planet but a map of the odds for where a position measurement will find the electron. Each measurement yields one point, and the cloud emerges only from many measurements on identically prepared atoms, as single dots built the double-slit fringes of chapter 7. Interpretations differ about what, if anything, the electron does between measurements, a debate for chapter 22, but all of them predict this same cloud.
Now the second. An orbital is not a container: it has no walls and holds nothing, because it is the electron's state itself. Saying an electron is in a 2p orbital describes its wavefunction, just as saying a guitar string is in its second harmonic describes how it vibrates, not a box it sits in. An empty orbital is simply an allowed state, a note the string could play.
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s, p, d, f: Shapes Carved by Nodes
The values of l carry letters inherited from the spectroscopy of the late 1800s and early 1900s: s, p, d and f for l = 0, 1, 2 and 3, from the sharp, principal, diffuse and fundamental series of lines in alkali-metal spectra; beyond f the letters run on alphabetically, g, h, i, k, skipping j. A full label puts n in front, so 3d means n = 3 and l = 2. Every s orbital is spherical. A p subshell, meaning all the orbitals sharing one n and l, holds three orbitals, usually drawn as dumbbells along the x, y and z axes. A d subshell holds five: four four-leaf clovers and one, labelled z², a dumbbell threaded through a doughnut. An f subshell holds seven. The count is always 2l + 1, one per value of mₗ.
The shapes are carved by nodes, surfaces on which ψ, and so the probability density, is exactly zero. Angular nodes are planes or cones through the nucleus, exactly l of them: none for s, one plane for each p orbital, two planes or cones for each d orbital. Radial nodes are spheres centred on the nucleus, n − l − 1 of them, so the total is n − 1, just as the nth standing wave in chapter 11's box had n − 1 nodes. The 2s orbital has one radial node, a sphere of radius 2a₀, about 0.106 nm; 3s has two, near 1.90a₀ and 7.10a₀; the z² orbital's angular nodes are two cones at 54.7 degrees from the z axis. Only s orbitals have nonzero density at the nucleus.
Which picture is real? Label the states by mₗ and the 2p states with mₗ = +1 and −1 turn out to be not dumbbells but doughnuts circling the z axis, identical at every angle around it, because their wavefunctions only rotate in phase as you travel round. The familiar x and y dumbbells are superpositions, the sum and the difference, of those two states. A doughnut has no nodal plane, only a nodal line along the z axis, so the rule of l angular nodes counts the nodes of the real, dumbbell-style orbitals. Because the pair share one energy, every superposition is an equally valid stationary state, and choosing dumbbells or doughnuts is choosing a basis, exactly as in chapter 15. Chemists prefer the dumbbells, which point along bonds; physicists prefer definite mₗ when a magnetic field singles out the z axis.
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Counting the Seats: Degeneracy
When different states share one energy, the level is called degenerate, and hydrogen is spectacularly so. In level n, l runs from 0 to n − 1 and each l contributes 2l + 1 values of mₗ, so the total is 1 + 3 + 5 and so on up to 2n − 1, which always adds up to n². Level 1 has one state, 1s; level 2 has four, 2s plus three 2p; level 3 has nine and level 4 sixteen. Spin doubles every count, because each spatial state can hold the electron with mₛ = +½ or −½, giving 2n²: 2, 8, 18 and 32.
Degeneracy is the fingerprint of symmetry. The mₗ degeneracy follows from rotational symmetry: space has no preferred direction, so tilting an orbital cannot change its energy, in any central potential. The l degeneracy, which lets 2s and 2p share an energy, requires the exact 1/r Coulomb potential. It is often called accidental, but it springs from the hidden symmetry that keeps planetary orbits closed ellipses, the one Pauli exploited; in 1935 Vladimir Fock showed it to be a rotational symmetry in four dimensions. Distort the 1/r law, as the extra electrons of heavier atoms do, and the l degeneracy breaks; apply a magnetic field, which picks out a direction, and the mₗ degeneracy breaks too. Broken degeneracies appear as split spectral lines.
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Cracks in the Pattern: Fine Structure and the Zeeman Effect
Eₙ = −13.6 eV/n² ignores spin and relativity, and nature is not quite that simple. In 1887 Albert Michelson and Edward Morley, examining hydrogen's red line with an interferometer, found that it is a close doublet. In 1916 Arnold Sommerfeld reproduced the splitting by adding relativity to Bohr's orbits, introducing the fine-structure constant α ≈ 1/137, the strength of electromagnetism, though partly by luck, since spin was unknown. Fine-structure splittings are smaller than the level energies by roughly α², about 5×10⁻⁵. They have two main sources: relativity, since the ground-state electron's typical speed is about 1/137 of the speed of light, and spin-orbit coupling, the interaction between the electron's spin magnetism and the magnetic field it feels from the proton's relative motion. Paul Dirac's relativistic equation of 1928 contains both.
A magnetic field cracks the pattern more visibly. In 1896 at Leiden, Pieter Zeeman put a sodium flame between the poles of an electromagnet of about one tesla and saw its yellow lines broaden; sharper sources resolved the broadening into separate components. Hendrik Lorentz explained the effect with his theory of electrons, and the two shared the 1902 Nobel Prize in Physics. In quantum terms, an electron with orbital angular momentum is a tiny magnet. A field B along z shifts each state's energy by mₗ times μ_B times B, where the Bohr magneton, μ_B = eħ/2mₑ, is about 5.79×10⁻⁵ electron-volts per tesla, so a level with a given l splits into 2l + 1 equally spaced sublevels: hence the name magnetic quantum number.
Yet many lines refused to play along. Beginning with Thomas Preston in Dublin in 1897, spectroscopists found more complicated patterns; sodium's two yellow lines split into four and six components, not the three that orbital magnetism alone allows. This anomalous Zeeman effect began to make sense in 1925, when two Leiden graduate students, George Uhlenbeck and Samuel Goudsmit, proposed that the electron carries intrinsic angular momentum, spin, with components of ±ħ/2 and twice the magnetism its angular momentum alone would suggest. Spin explained the anomalous patterns and, with relativity, the fine structure. Hydrogen's cracks were clues.
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From One Electron to the Periodic Table
Hydrogen is the only neutral atom whose Schrödinger equation can be solved exactly, yet its solutions organize every atom. In a many-electron atom each electron feels the nucleus plus the averaged charge of the others, which is still roughly spherical, so the angular story carries over unchanged: the same quantum numbers, shapes, orientations and node counts. What changes is the radial story. Inner electrons shield the nucleus, so the potential is no longer a pure 1/r, the l degeneracy is lost, and within a shell s lies below p, which lies below d.
Now add the Pauli exclusion principle of chapter 18: no two electrons in an atom may share all four quantum numbers. Each spatial orbital then takes at most two electrons, one of each spin, and a shell can hold 2n² electrons: 2, 8, 18 and 32. Those numbers are written across the periodic table, whose rows hold 2, 8, 8, 18, 18, 32 and 32 elements; the repeats and reshuffles come from the broken l degeneracy, which lets 4s fill before 3d. Hydrogen supplies the seating chart; chapter 18 supplies the seating rules. Chemistry, from the angles of molecules to the colours of gemstones, is a negotiation among these shapes.
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Go deeper on video
18 more videos for this chapter, 12 of them from universities, research institutes and Nobel lectures. The clips under “Watch this part” in the lesson start at the moment that matters.
Key equations
Schrödinger's equation for hydrogen
The time-independent Schrödinger equation in three dimensions, with the proton's Coulomb attraction as the potential. The first term is kinetic energy, measured by the curvature of psi in all three directions; the second is electrical potential energy, which depends only on the distance r. Well-behaved solutions exist only at special energies.
| Symbol | Meaning | Unit |
|---|---|---|
| ψ | the electron's wavefunction, a function of r, θ and φ | m⁻³ᐟ² (so that |ψ|² is a probability per cubic metre) |
| ∇² | the Laplacian: total curvature of ψ along x, y and z | 1/m² |
| mₑ | electron mass, 9.109×10⁻³¹ | kg |
| e | elementary charge, 1.602×10⁻¹⁹ | C |
| ε₀ | electric constant, 8.854×10⁻¹² | F/m |
| r | distance between electron and proton | m |
| E | energy of the stationary state | J or eV |
Say it: “minus h-bar squared over two m e, times the Laplacian of psi, minus e squared over four pi epsilon nought r, times psi, equals E times psi”
Hydrogen energy levels
Solving the equation reproduces Bohr's ladder from constants of nature alone. In hydrogen the energy depends only on the principal quantum number n, not on l or m l, which is the source of its large degeneracy.
| Symbol | Meaning | Unit |
|---|---|---|
| Eₙ | energy of level n, negative because the electron is bound | eV |
| n | principal quantum number, 1, 2, 3 … | dimensionless |
| h | Planck constant, 6.626×10⁻³⁴ | J·s |
| 13.6 eV | hydrogen's ionization energy, 2.18×10⁻¹⁸ J | eV |
Say it: “E n equals minus m e times e to the fourth, over eight epsilon nought squared h squared, times one over n squared, which is approximately minus thirteen point six electron-volts over n squared, for n equals one, two, three and so on”
Quantum numbers and angular momentum
The rules that label every hydrogen state. The number l fixes the length of the orbital angular momentum and m l its projection on the z axis; because the projection is always shorter than the length, the angular momentum never points exactly along an axis. Spin, with m s equal to plus or minus one half, is a fourth label added by nature, not by Schrödinger's equation.
| Symbol | Meaning | Unit |
|---|---|---|
| l | orbital quantum number (s, p, d, f for 0, 1, 2, 3) | dimensionless |
| mₗ | magnetic quantum number, 2l + 1 values | dimensionless |
| mₛ | spin quantum number, +½ or −½ | dimensionless |
| |L| | magnitude of the orbital angular momentum | J·s |
| L_z | component of the orbital angular momentum along z | J·s |
| ħ | reduced Planck constant, 1.055×10⁻³⁴ | J·s |
Say it: “l runs from zero to n minus one, m l runs from minus l to plus l, and m s is plus or minus one half. The orbital angular momentum has magnitude equal to the square root of l times l plus one, times h-bar, and its z component equals m l times h-bar”
The 1s cloud and the Bohr radius
The radial probability of the ground state: the chance per unit distance of finding the electron at distance r from the nucleus. The r squared factor, from the growing area of a spherical shell, competes with the exponentially falling density, and their product peaks at exactly r equals a nought.
| Symbol | Meaning | Unit |
|---|---|---|
| P(r) | radial probability per unit distance | 1/m |
| ψ₁₀₀ | the 1s wavefunction (n = 1, l = 0, mₗ = 0) | m⁻³ᐟ² |
| a₀ | Bohr radius, 5.29×10⁻¹¹ | m |
Say it: “P of r equals four pi r squared times the squared magnitude of psi one zero zero, which equals four r squared over a nought cubed, times e to the minus two r over a nought. The Bohr radius, a nought, equals four pi epsilon nought h-bar squared over m e times e squared, about zero point zero five two nine nanometres”
Counting states and nodes
Every level of hydrogen holds n squared spatial states sharing one energy, doubled by spin. Every orbital has n minus one nodes in total: l of them angular planes or cones through the nucleus, and the rest spherical radial nodes.
| Symbol | Meaning | Unit |
|---|---|---|
| n² | number of spatial states (orbitals) in level n | dimensionless |
| N_radial | number of spherical nodes | dimensionless |
| N_angular | number of nodal planes or cones | dimensionless |
Say it: “the sum over l from zero to n minus one of two l plus one equals n squared, or two n squared with spin. The number of radial nodes is n minus l minus one, and the number of angular nodes is l”
The normal Zeeman effect
Ignoring spin, a magnetic field along z shifts each state by an amount proportional to m l, splitting a level of given l into 2l + 1 equally spaced sublevels. Spin adds magnetism of its own, which produces the anomalous patterns.
| Symbol | Meaning | Unit |
|---|---|---|
| ΔE | energy shift of the state | J or eV |
| μ_B | Bohr magneton, 9.274×10⁻²⁴ J/T | J/T |
| B | magnetic field strength along z | T |
Say it: “the energy shift equals m l times the Bohr magneton times B, where the Bohr magneton equals e h-bar over two m e, about five point seven nine times ten to the minus five electron-volts per tesla”
Real numbers, step by step
Cover the steps, try the problem yourself, then reveal the solution one step at a time.
A census of the n = 3 level
List every orbital in hydrogen's n = 3 level. For each subshell give the number of orbitals, the radial and angular nodes and the magnitude of the orbital angular momentum. How many states are there with and without spin, and what is their energy?
Step 1: Allowed l values are 0, 1 and 2, so the subshells are 3s, 3p and 3d.
Step 2: 3s: l = 0 and mₗ = 0, one orbital. Radial nodes n − l − 1 = 2, angular nodes 0. Angular momentum √(0 × 1)·ħ = 0.
Step 3: 3p: l = 1 and mₗ = −1, 0 or +1, three orbitals. Radial nodes 1, angular nodes 1. Angular momentum √2·ħ ≈ 1.41ħ ≈ 1.49×10⁻³⁴ J·s.
Step 4: 3d: l = 2 and mₗ = −2 to +2, five orbitals. Radial nodes 0, angular nodes 2. Angular momentum √6·ħ ≈ 2.45ħ ≈ 2.58×10⁻³⁴ J·s.
Step 5: Total: 1 + 3 + 5 = 9 = 3² spatial states, 18 with spin, all at E₃ = −13.6 eV ÷ 9 = −1.51 eV.
Nine orbitals (one 3s, three 3p, five 3d) and eighteen states with spin, all at −1.51 eV; every orbital has n − 1 = 2 nodes in total
Why it matters: The node budget is fixed by n and merely redistributed by l: 3s spends both nodes on spheres, 3d spends both on angular surfaces. Same energy, utterly different shapes.
Locating the 1s electron
Using the 1s radial probability of equation e4, find the most probable distance of the electron from the nucleus, the probability of finding it within a₀ and beyond 2a₀, its average distance, and how much denser the cloud is at the nucleus than at a₀. The probability of finding it within a distance R is 1 minus the product of (1 + 2x + 2x²) and e to the power −2x, where x = R/a₀.
Step 1: Most probable distance: P(r) is proportional to r² times e to the power −2r/a₀. Its slope is zero when 2r − 2r²/a₀ = 0, so r = a₀ = 0.0529 nm.
Step 2: Within a₀, x = 1: 1 − 5e⁻² = 1 − 5 × 0.1353 = 0.323.
Step 3: Within 2a₀, x = 2: 1 − 13e⁻⁴ = 1 − 13 × 0.01832 = 0.762, so the probability beyond 2a₀ is 0.238.
Step 4: Average distance: ⟨r⟩ = 1.5a₀ = 0.0794 nm, larger than the peak because the distribution has a long outer tail.
Step 5: Density: |ψ|² falls as e to the power −2r/a₀, so the density at the nucleus is e², about 7.39, times the density at a₀.
Most probable distance a₀ = 0.0529 nm; about 32 percent within a₀ and 24 percent beyond 2a₀; average distance 0.0794 nm; density 7.4 times higher at the nucleus than at a₀
Why it matters: The most probable point (the nucleus), the most probable distance (a₀) and the average distance (1.5a₀) are three different numbers. The cloud has no edge, only a long, fading tail.
2s versus 2p: nodes, peaks and Bohr's ghost
With x = r/a₀, the 2s radial probability is proportional to x²(2 − x)² times e to the power −x, and the 2p radial probability to x⁴ times e to the power −x. Find the 2s radial node, the radii of its two peaks and the most probable radius of 2p, and compare with Bohr's n = 2 orbit.
Step 1: 2s node: the factor (2 − x) vanishes at x = 2, a spherical node at r = 2a₀ = 0.106 nm.
Step 2: 2s peaks: setting the slope of x²(2 − x)² e to the power −x to zero gives 2/x − 2/(2 − x) − 1 = 0, which rearranges to x² − 6x + 4 = 0, so x = 3 ± √5. The peaks sit at 0.764a₀ = 0.0404 nm and 5.24a₀ = 0.277 nm, and the outer peak is about 3.7 times taller.
Step 3: 2p peak: the slope of x⁴ times e to the power −x is zero when 4/x − 1 = 0, so x = 4 and r = 4a₀ = 0.212 nm.
Step 4: Bohr's n = 2 orbit had radius n²a₀ = 4a₀ = 0.212 nm, exactly the 2p peak.
2s node at 0.106 nm; 2s peaks at 0.0404 nm and 0.277 nm; 2p peaks at 0.212 nm, Bohr's n = 2 radius
Why it matters: For the orbital with the largest l in each level (1s, 2p, 3d, 4f) the most probable radius is exactly n²a₀, so Bohr's radii survive as peaks of clouds. The 2s electron's small inner hump puts about 5 percent of its probability inside 2a₀, close to the nucleus: a habit of penetration that chapter 18 cashes in.
Splitting Lyman-alpha with a magnet
Hydrogen's Lyman-alpha line, at 121.6 nm, comes from the 2p → 1s transition. Ignoring spin, find how far the 2p sublevels shift in a 2.00 tesla field, the frequency and wavelength spacing of the resulting three lines, and compare the shift with the 2p fine-structure splitting of 4.5×10⁻⁵ electron-volts.
Step 1: Shift per unit of mₗ: ΔE = μ_B × B = 9.274×10⁻²⁴ joules per tesla × 2.00 tesla = 1.855×10⁻²³ joules, which is 1.16×10⁻⁴ electron-volts.
Step 2: The 2p level splits into mₗ = −1, 0 and +1, shifted by −1.16×10⁻⁴ electron-volts, 0 and +1.16×10⁻⁴ electron-volts. The 1s level has l = 0, so it does not shift. Three lines appear.
Step 3: Frequency spacing: Δf = ΔE ÷ h = 1.855×10⁻²³ joules ÷ 6.626×10⁻³⁴ J·s = 2.80×10¹⁰ hertz, or 28.0 GHz.
Step 4: Wavelength spacing: Δλ = λ² × Δf ÷ c = (121.6×10⁻⁹ metres)² × 2.80×10¹⁰ hertz ÷ 2.998×10⁸ m/s = 1.38×10⁻¹² metres, or 0.00138 nm.
Step 5: Comparison: 1.16×10⁻⁴ ÷ 4.5×10⁻⁵ ≈ 2.6, so the magnetic shift is about 2.6 times the fine-structure splitting, yet only about one part in 88,000 of the photon's 10.2 eV.
Shifts of ±1.16×10⁻⁴ electron-volts; three lines 28.0 GHz, or 0.00138 nm, apart; the magnetic shift is about 2.6 times the fine-structure splitting
Why it matters: Tiny, but easy for a good spectrometer. Because the field's effect here is comparable to the fine-structure coupling, the real pattern is intermediate; much stronger fields push it toward this simple triplet, while in weak fields spin's extra magnetism produces the anomalous patterns that first revealed spin.
What everyone gets wrong
Read the myth, decide whether you believed it, then bust it.
- Myth #1
Electrons orbit the nucleus in fixed circles, like planets around the Sun.
TruthHydrogen's electron occupies orbitals, stationary standing-wave states with no trajectory. The ground state has zero orbital angular momentum, so nothing circulates, and it is spherically symmetric, so there is no direction to circulate in. Bohr's radius survives only as the most probable distance of a probability cloud, about 0.0529 nm, and the electron can be found nearer or farther. A charge truly circling the nucleus would also radiate its energy away within about 10⁻¹¹ seconds.
- Myth #2
An orbital is a container, a fuzzy box or bubble that the electron sits inside.
TruthAn orbital is the electron's state itself, a wavefunction, not a place with walls. The drawn boundary is a convention enclosing about 90 percent of the probability; the cloud has no edge. Saying an electron is in a 2p orbital describes how it exists, as saying a string is in its second harmonic describes how it vibrates. Empty orbitals are simply allowed states that no electron occupies.
- Myth #3
The most likely place to find the 1s electron is somewhere on a sphere at the Bohr radius.
TruthThe probability density |ψ|² is greatest at the nucleus itself, about 7.4 times its value at a₀. What peaks at a₀ is the radial probability, the chance of finding the electron at a given distance, because a thin shell at a₀ has far more volume than one near the centre.
- Myth #4
To get from one lobe of a p orbital to the other, the electron must somehow sneak through the nodal plane where it can never be found.
TruthThe puzzle assumes a path, and there is none. The two lobes are two parts of one standing wave, like the two halves of a guitar string vibrating with a node at its centre: both halves move, yet nothing travels through the still point. Measurements find the electron in one lobe or the other, with zero probability on the plane.
Your turn at the controls
Welcome to the Orbital Explorer, where we map the electron cloud one slice at a time. Choose any hydrogen orbital up to n = 4, from 1s to the seven 4f orbitals, and the lab draws a slice through its probability density, plots its radial probability, counts its radial and angular nodes, and reads out its energy. Then the lights go down and a mystery orbital appears, and you must identify it from its density map alone.
Your tasks
0/4How it connects to the theory
Every image is the Born-rule probability density |ψ|² of a genuine hydrogen solution, a radial function multiplied by an angular one, sliced through the nucleus. The radial plot is the shell-weighted probability of equation e4, extended to every orbital, so you can find a₀, 2a₀ and 4a₀ on it. The node counts follow the budget from Part 2, n − 1 in all, split into l angular and n − l − 1 radial nodes, which is exactly how you decode a mystery map. The energy readout depends on n alone, which is why orbitals of different shapes share one value. The lab draws the real-valued orbitals, the dumbbells and cloverleaves of Part 4, whose angular nodes are the planes and cones you count; the doughnut-shaped states of definite mₗ share the same energies but hide some of those nodes.
Orbital Explorer: Map the Electron Cloud loads as you scroll down.
Make it stick
The Four S's, then Walls and Balls. Name each quantum number by what it sets: n is Size, and with it the energy; l is Shape; mₗ is Slant, the orientation; mₛ is Spin. Chant the limits: l stays under n, m sweeps from minus l to plus l, spin flips two ways. Then the node rhyme: 'n minus one in all; l are walls, the rest are balls.' Walls are the angular nodes, flat planes or cones cutting through the nucleus; balls are the spherical radial nodes. To use it on any label such as 4d: Size n = 4, so −13.6 ÷ 16 = −0.85 eV; Shape d, so l = 2; Slant gives five orientations; Spin doubles them to ten states; nodes: three in all, two walls and one ball.
An orbital is a ringing pattern, like the figures Ernst Chladni made visible in 1787 by sprinkling sand on vibrating metal plates: the sand gathers along nodal lines where the surface stays still, higher notes show more nodes, and each pattern is a standing wave fixed by the shape of the object. Hydrogen's orbitals are the ringing patterns of the electron's wave in the bowl of the proton's attraction, with n setting the pitch, l the number of nodal surfaces through the centre and mₗ their orientation. Where it breaks: a plate's vibration is real motion of metal that you can see all at once, whereas ψ is a probability amplitude, and a measurement finds the electron at a single point. And in hydrogen, patterns with different numbers of nodal surfaces through the centre ring at exactly the same pitch, a coincidence that a round plate or a bell does not share; it is the signature of the special 1/r force.
The whole chapter in one lineHydrogen's electron is a 3D standing wave labelled n, l, mₗ, mₛ: energy −13.6 eV/n², shapes set by l; a cloud, never an orbit.
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Dr. Q's Quantum Quiz Show
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Prove it on paper
Answer every question, check your work, and read the explanations — the stretch challenge is for the truly mad.
Work through every question with the lesson closed, then check yourself. Use these facts: Eₙ = −13.6 eV/n²; l runs from 0 to n − 1 and mₗ from −l to +l; mₛ = ±½; an orbital has l angular and n − l − 1 radial nodes; a₀ = 0.0529 nm; the probability of finding a 1s electron within distance R of the nucleus is 1 minus the product of (1 + 2x + 2x²) and e to the power −2x, with x = R/a₀; the Bohr magneton μ_B = 9.274×10⁻²⁴ J/T = 5.79×10⁻⁵ eV/T; ħ = 1.055×10⁻³⁴ J·s; h = 6.626×10⁻³⁴ J·s. Show every step in the calculation questions.
Part A
Vocabulary
- Orbital
- A single-electron wavefunction ψₙₗₘ, a stationary standing-wave state whose squared magnitude gives the probability density; not a path and not a container.
- Principal quantum number
- n = 1, 2, 3 …; sets the energy, −13.6 eV/n² in hydrogen, the overall size of the cloud and the total number of nodes, n − 1.
- Orbital quantum number
- l = 0 to n − 1; sets the orbital's shape (s, p, d, f) and the magnitude of its orbital angular momentum, √(l(l + 1))·ħ.
- Magnetic quantum number
- mₗ = −l to +l; sets the orbital's orientation and the z-component of its angular momentum, mₗħ, and so its shift in a magnetic field.
- Spin quantum number
- mₛ = +½ or −½; labels the component of the electron's intrinsic angular momentum, +ħ/2 or −ħ/2, along a chosen axis.
- Subshell
- All the orbitals that share the same n and l, such as the three 2p orbitals; it contains 2l + 1 orbitals.
- Node
- A surface where the wavefunction, and so the probability density, is zero: angular nodes are planes or cones through the nucleus, radial nodes are spheres.
- Radial probability
- The probability per unit distance of finding the electron at distance r from the nucleus, 4πr² times |ψ|² for an s orbital; it peaks at a₀ for 1s.
- Bohr radius
- a₀ = 4πε₀ħ²/(mₑe²) ≈ 0.0529 nm, the most probable distance of hydrogen's 1s electron from the nucleus.
- Degeneracy
- The sharing of one energy by several distinct states; hydrogen's level n holds n² spatial states, 2n² with spin.
- Fine structure
- The small splitting of hydrogen's levels, roughly α² ≈ 5×10⁻⁵ of their energy, caused by relativity and spin-orbit coupling.
- Zeeman effect
- The splitting of spectral lines by a magnetic field, which separates states with different mₗ and, through its anomalous form, revealed electron spin.
Part B
Questions
Auto-checked except short answers, which you mark yourself.
- Question 1.Multiple choice
Which set of quantum numbers (n, l, mₗ, mₛ) describes an allowed state of hydrogen's electron?
- Question 2.Multiple choice
How many radial and angular nodes does a 4d orbital have?
- Question 3.Multiple choice
Ignoring fine structure, how do the energies of hydrogen's 3s, 3p and 3d orbitals compare?
- Question 4.True or false
The probability density |ψ|² of hydrogen's 1s state is greatest at a distance a₀ from the nucleus.
- Question 5.True or false
For a 2p orbital pointing along the z axis, the probability density is zero everywhere on the plane through the nucleus perpendicular to z.
- Question 6.True or false
Hydrogen's ground-state electron has orbital angular momentum ħ, as Bohr proposed in 1913.
- Question 7.Fill in the blank
The quantum number mₗ is called the quantum number, because a field of that kind splits states with different mₗ.
- Question 8.Fill in the blank
A surface on which an orbital's wavefunction is exactly zero is called a .
- Question 9.Short answer
Using the idea of a stationary state, explain why it is wrong to picture the 1s electron circling the nucleus.
- Question 10.Short answer
A classmate says: 'An orbital is a bubble that the electron lives inside.' Write two or three sentences correcting this.
- Question 11.Calculation
For hydrogen's n = 5 level, list the subshells, count the spatial states with and without spin, and find the energy.
This one asks for more than one value — give each, with its units.
- Question 12.Calculation
Find the probability that a 1s electron is found within 0.5a₀ of the nucleus, and the probability that it is found farther than 3a₀ away.
This one asks for more than one value — give each, with its units.
- Question 13.Calculation
A hydrogen atom in a 3d state sits in a 1.50 tesla magnetic field. Ignoring spin, into how many sublevels does the 3d level split, what is the energy spacing between neighbouring sublevels in eV, what is the total spread from the lowest to the highest, and what frequency corresponds to one spacing?
This one asks for more than one value — give each, with its units.
- Question 14.Calculation
For an electron in a d orbital, find the magnitude of the orbital angular momentum in J·s, its largest possible z-component, and the smallest possible angle between the angular momentum and the z axis.
This one asks for more than one value — give each, with its units.
Part C · optional, dangerous, extremely rewarding
Mad scientist challenge
The 1s electron meets the proton. The probability of finding a 1s electron within distance R of the nucleus is 1 minus the product of (1 + 2x + 2x²) and e to the power −2x, with x = R/a₀. (a) Check the formula: show that its rate of change with x equals 4x² times e to the power −2x, the 1s radial probability written in terms of x, and that it tends to 1 for large R. (b) Show that for very small x the probability is approximately (4/3)x³, and derive the same result directly from the density at the nucleus, |ψ(0)|² = 1/(πa₀³). (c) Taking the proton's radius as 0.84 fm, estimate the probability that the electron is found inside the proton. (d) Explain why the corresponding probability for a 2p electron is far smaller still, and why the tiny overlap of s electrons with the nucleus matters in real physics.
Recap & sign-off
Key takeaways
- Solving Schrödinger's equation with the Coulomb potential gives Eₙ = −13.6 eV/n², deriving Bohr's levels from boundary conditions
- Three dimensions give three quantum numbers: n for energy and size, l from 0 to n − 1 for shape and angular momentum √(l(l + 1))·ħ, and mₗ from −l to +l for orientation; spin adds mₛ = ±½
- Orbitals are probability clouds, not paths: the 1s density peaks at the nucleus, but the most probable distance is the Bohr radius a₀, about 0.0529 nm
- An orbital is the electron's state, not a container; its drawn boundary is a 90 percent convention
- s, p, d and f mean l = 0, 1, 2 and 3; every orbital has l angular nodes and n − l − 1 radial nodes, n − 1 in all
- Each level holds n² spatial states, 2n² with spin, all at one energy in hydrogen
- Fine structure and the Zeeman effect split these levels and exposed relativity and electron spin
- With shielding and the Pauli principle, hydrogen's shells and subshells become the periodic table
Dr. Q signs offI'm mad, not wrong! Tonight we evicted Bohr's electron from its racetrack and found it living as a cloud, sculpted by n, l and mₗ into spheres, dumbbells and cloverleaves, with nodes where it is never found. But hydrogen left a riddle in its cracks: lines splitting into even numbers of pieces, and a fourth quantum number with no home in Schrödinger's equation. Next session we fire silver atoms through a magnet in Frankfurt, 1922, and watch spin split the beam in two. Bring a compass.
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Sources & further reading
- HyperPhysics (Georgia State University): Hydrogen Wavefunctions · hyperphysics.phy-astr.gsu.edu
- HyperPhysics: Hydrogen Fine Structure · hyperphysics.phy-astr.gsu.edu
- NIST Physical Measurement Laboratory: Energy Levels of Hydrogen and Deuterium · nist.gov
- NobelPrize.org: The Nobel Prize in Physics 1902 (Lorentz and Zeeman) · nobelprize.org
- Physics World (2013): Quantum microscope peers into the hydrogen atom, on Stodolna et al., Phys. Rev. Lett. 110, 213001 · physicsworld.com