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Where Bohr's model fails: many-electron atoms and shielding, the ground state's angular momentum, line intensities, fine structure and the Zeeman effect.
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By the end of this lesson you will be able to identify Bohr's model's failures, quantify shielding with effective charges, and compute Zeeman splittings.
From the last two lessons you can use Bohr's model to predict hydrogen's levels and lines, and scale them to one-electron ions. From magnetism you know magnetic moments and their energy in a field. This lesson collects what the model cannot do, which is most of atomic physics beyond hydrogen, and why a new theory had to replace it.
| Term | What it means |
|---|---|
| Shielding | Inner electrons partly cancelling the nucleus's charge seen by outer ones. |
| Effective nuclear charge | $Z_{\text{eff}}$, the net charge an electron behaves as if it feels. |
| Fine structure | Small splittings of lines from spin and relativistic effects, about $\alpha^2$ of the line energy. |
| Zeeman effect | The splitting of spectral lines in a magnetic field. |
| Bohr magneton | $\mu_B = 9.274 \times 10^{-24}$ J/T $= 57.88$ μeV/T, the natural unit of atomic magnetic moment. |
| Electron spin | An intrinsic angular momentum of $\hbar/2$, absent from Bohr's model. |
| Semiclassical model | Classical motion with quantum rules added, like Bohr's. |
Bohr's model reproduces hydrogen's energy levels exactly, but it fails in several ways that no adjustment could fix:
Measuring these failures turns them into tools: effective charges reveal shielding, and Zeeman splittings reveal magnetic fields, even on the Sun.
Another way: picture
Picture a map of a city drawn with every street as a perfect circle around the town hall. For a village with one ring road it would be accurate, and you could find every address. For a real city, with crossing streets and neighborhoods crowded together, the map would fail, however carefully you adjusted the circles. Bohr's model is that map: exactly right for hydrogen, wrong in kind for everything else.
Another way: steps
In helium, each electron is attracted by a nucleus of charge $2e$ and repelled by the other electron. Bohr's model offers no way to handle the repulsion: an orbit is defined by one electron circling one center. Treating each electron as if the other were absent predicts that removing one takes $54.4$ eV; the measured value is $24.6$ eV.
Physicists tried for a decade to extend the model to helium, with elaborate arrangements of orbits, and failed. The first successful calculation came in 1929, with quantum mechanics, by Egil Hylleraas, who matched the measured energy to a part in ten thousand. Every atom beyond hydrogen needs the new theory.
An outer electron in sodium spends most of its time outside the ten inner electrons, which cancel ten of the nucleus's eleven charges. It should feel about $+1e$ and behave almost like hydrogen's electron in level $n = 3$, with an ionization energy of $13.6/9 = 1.51$ eV. The measured value is $5.14$ eV, corresponding to $Z_{\text{eff}} = 1.84$.
The extra charge comes from the outer electron's wave dipping inside the inner shells part of the time, where it feels more of the nucleus. Bohr's circular orbit cannot dip anywhere. Quantum mechanics describes this penetration, and it explains why, within one shell, $s$ electrons are more tightly bound than $p$ or $d$ electrons.
Bohr's rule gives the ground state angular momentum $\hbar$. Quantum mechanics says hydrogen's ground state is an $s$ state with orbital angular momentum exactly zero: the electron's probability cloud is spherical, with no preferred direction of circulation. The Stern–Gerlach experiment and hydrogen's magnetic behavior confirm zero.
That Bohr's energies are nonetheless right is something of a coincidence. The exact solution depends on the principal quantum number $n$ alone, for the pure Coulomb force, and Bohr's rule happens to produce the same $n$-dependence. Change the force slightly, as in atoms with shielding, and the coincidence disappears.
Look closely at hydrogen's red line with a high-resolution spectrograph and it is a pair, separated by about $0.016$ nm. Similar splittings appear throughout the spectrum. Arnold Sommerfeld added elliptical orbits and relativity to Bohr's model and got the right size, about $\alpha^2 \approx 5 \times 10^{-5}$ of the line energy, but for partly wrong reasons.
The full explanation needs electron spin, discovered in 1925, and Dirac's relativistic quantum mechanics of 1928. Sodium's famous yellow line is a pair, at $589.0$ and $589.6$ nm, split by the spin of the outer electron interacting with its orbital motion.
In a magnetic field, an atom's energy depends on how its magnetic moment is oriented. A level with orbital quantum number $l$ splits into $2l + 1$ sublevels, shifted by $m_l\mu_BB$, where $\mu_B = 57.88$ μeV/T is the Bohr magneton. A spectral line splits correspondingly, most simply into three.
Pieter Zeeman saw the effect in 1896, and Hendrik Lorentz explained the simplest pattern with classical electrons. But most lines split into more complicated patterns, the anomalous Zeeman effect, which only electron spin explains. Bohr's model, with its planar orbits and no orientation quantum number, cannot account for either.
Checking an answer. Effective charges must lie between 1 and $Z$. Zeeman shifts for laboratory fields must be tens of μeV, a few parts in a hundred thousand of visible photon energies. Wavelength shifts must be picometers. And any claim that Bohr's model gives a many-electron level directly is a warning sign.
Fitting an effective charge to a measured energy is a description, not a prediction: it summarizes shielding in one number. Its trend across the periodic table is meaningful, but computing it from first principles requires quantum mechanics.
The Zeeman formula $\Delta E = m\mu_BB$ is the normal Zeeman effect, valid when spin plays no role or when fields are strong enough to overwhelm spin–orbit coupling. It gives the right order of magnitude for any atom, which is all that estimates of magnetic fields from line splittings need.
In 1925 Werner Heisenberg, and in 1926 Erwin Schrödinger, built quantum mechanics, in which an electron is described by a wave function rather than an orbit. For hydrogen it reproduces Bohr's energies and adds the angular momentum quantum numbers, the correct zero ground-state angular momentum, line intensities, and, with spin and relativity, fine structure.
For many-electron atoms it gives the shell structure behind the periodic table, worked out by Wolfgang Pauli, Douglas Hartree and others. Bohr's model survives as a first picture and a quick calculator for one-electron systems, but the next lesson's quantum numbers come from the new theory.
The Zeeman effect lets astronomers measure magnetic fields they can never visit. George Ellery Hale at Mount Wilson Observatory near Los Angeles found in 1908 that lines in sunspot light were split, with components polarized in the pattern the Zeeman effect predicts. Sunspots, he concluded, hold fields of thousands of gauss, tenths of a tesla.
Today NASA's Solar Dynamics Observatory maps the Sun's magnetic field every $45$ seconds from Zeeman splittings of an iron line, and the National Solar Observatory's Daniel K. Inouye telescope in Hawaii resolves field structures a few tens of kilometers across. Forecasts of solar flares, which disrupt satellites and power grids, rest on these maps.
Bohr's model says which photons an atom can emit but not how often it emits each one. Yet line strengths carry as much information as their wavelengths. Astronomers infer temperatures and compositions of stars from the relative brightness of lines, and lasers work only on transitions strong enough to amplify light.
Quantum mechanics computes the rate of each jump from the overlap of the two levels' wave functions with the light's electric field. Some jumps, like hydrogen's $2p$ to $1s$, happen in about $1.6$ ns; others are forbidden and take seconds or even millions of years. The $2s$ level of hydrogen lives about a tenth of a second because a single-photon drop to $1s$ is forbidden. The famous $21$ cm radio line of hydrogen, used to map our galaxy, comes from a transition so slow that a given atom emits it about once every $11$ million years, detectable only because the galaxy contains so much hydrogen. None of this can be read from a circular orbit.
Since Hale's discovery at Mount Wilson in 1908, solar physicists have measured the Sun's magnetic fields through the Zeeman effect. NASA's Solar Dynamics Observatory, operated from Goddard Space Flight Center in Maryland, carries the Helioseismic and Magnetic Imager, which measures the splitting of an iron line at $617.3$ nm across the whole solar disk.
Its maps show fields up to a few tenths of a tesla in sunspots, twisting and reconnecting before flares. The Space Weather Prediction Center in Boulder, Colorado, uses such data to warn airlines, satellite operators and power companies of solar storms, all from picometer shifts in spectral lines.
Magnetic resonance imaging relies on the same physics as the Zeeman effect, applied to protons instead of electrons. In a $1.5$ T scanner, a proton's two spin orientations differ in energy by a tiny amount, corresponding to a radio frequency of $64$ MHz. Radio pulses at that frequency flip spins, and the signal as they relax builds the image.
Because the proton's magnetic moment is about $660$ times smaller than the electron's, the splitting is much smaller than an atom's Zeeman shift in the same field. The idea, energy levels split in proportion to the field, is what Bohr's model left out and what makes the scanner work.
It is tempting to see Bohr's model as a slightly rough version of the truth, like Newton's mechanics compared with relativity. But its central picture, electrons on definite circular orbits, is wrong: electrons in atoms have no trajectories, and hydrogen's ground state has no orbital angular momentum at all. Its correct energies for hydrogen are partly a coincidence of the pure Coulomb force.
A related error is to apply $-13.6Z^2/n^2$ to atoms with several electrons. It works only for one electron; otherwise shielding must be included.
Sodium's outer electron is in level $n = 3$ and needs $5.14$ eV to remove. Find the unshielded Bohr energy with $Z = 11$.
$\dfrac{13.6 \times 121}{9} = 183\ \text{eV}$
Ignoring the other electrons.
Compare with the measured value.
$\dfrac{5.14}{183} = 0.028$
Off by a factor of 35.
Try full shielding, $Z_{\text{eff}} = 1$.
$\dfrac{13.6}{9} = 1.51\ \text{eV}$
Too small by a factor of 3.4.
Find the effective charge that fits.
$Z_{\text{eff}} = 3\sqrt{\dfrac{5.14}{13.6}} = 1.84$
Between the two extremes.
Interpret the effective charge.
$\text{the outer electron penetrates the inner shells}$
Feeling more than one charge part of the time.
A hydrogen lamp sits in a $1.5$ T field. Find the energy shift for $m_l = \pm 1$.
$\Delta E = 57.88 \times 1.5 = 86.8\ \mu\text{eV}$
$\mu_BB$.
Find the photon energy of the red line.
$E = 1.889\ \text{eV}$
At $656.5$ nm.
Find the fractional shift.
$\dfrac{\Delta E}{E} = \dfrac{86.8 \times 10^{-6}}{1.889} = 4.6 \times 10^{-5}$
Tiny.
Find the wavelength shift.
$\Delta\lambda = 656.5 \times 4.6 \times 10^{-5} = 0.030\ \text{nm}$
Thirty picometers.
Find the frequency shift.
$\Delta f = 14.0 \times 1.5 = 21\ \text{GHz}$
$\mu_B/h = 14.0$ GHz/T.
Describe what a spectrograph shows.
$\text{three lines } 0.030\ \text{nm apart}$
Needs a high-resolution instrument.
An iron line at $525.0$ nm splits by $\pm 0.020$ nm in a sunspot. Find the energy shift.
$\Delta E = \dfrac{1240 \times 0.020}{525.0^2} = 9.0 \times 10^{-5}\ \text{eV}$
$hc\,\Delta\lambda/\lambda^2$.
Divide by the Bohr magneton.
$B = \dfrac{9.0 \times 10^{-5}}{5.788 \times 10^{-5}} = 1.56\ \text{T}$
Normal Zeeman effect.
Adjust for the line's actual splitting factor of $3$.
$B = \dfrac{1.56}{3} = 0.52\ \text{T}$
Spin makes this line split three times as much.
Convert to gauss.
$0.52\ \text{T} = 5200\ \text{gauss}$
$1$ T $= 10^4$ gauss.
Compare with Earth's field.
$\dfrac{0.52}{5 \times 10^{-5}} \approx 10^4$
Ten thousand times stronger.
Explain why spots are dark.
$\text{strong fields slow the upflow of hot gas}$
So the spot is cooler than its surroundings.
Note Bohr's model's role.
$\text{none: it has no magnetic sublevels}$
The measurement needs quantum mechanics.
Write the formula.
$Z_{\text{eff}} = n\sqrt{\dfrac{E}{13.6}}$
Bohr's energy solved for the charge.
Substitute the values.
$Z_{\text{eff}} = 2\sqrt{\dfrac{5.39}{13.6}}$
Measured energy.
Evaluate the effective charge.
Bohr's model gives He⁺, with one electron, an ionization energy of $54.4$ eV, matching experiment. Student group $2$ tries it on neutral helium, with two electrons, and gets nothing close to the measured $24.6$ eV for removing the first. What is the main reason?
Complete the worked solution: hydrogen's red line, $656.5$ nm and $1.889$ eV, is observed in a $1$ T magnetic field. With $\mu_B = 57.88$ μeV/T, find the energy shift of the $m_l = \pm 1$ components in μeV, the photon energy of the unshifted line in eV, and the wavelength shift of the components in pm.
Multiply the Bohr magneton by the field.
$\Delta E = \mu_BB =$ e
In μeV.
State the unshifted photon energy.
$E =$ p
In eV.
Scale the wavelength by the fractional shift.
$\Delta\lambda = \lambda\,\dfrac{\Delta E}{E} =$ l
In pm.
Describe the pattern.
$\text{three lines: unshifted and } \pm\Delta\lambda$
The normal Zeeman effect.
Match each failure of Bohr's model to its cause or the correct result.
| repulsion and shielding | really zero, not ħ | not predicted | spin and magnetic energy | |
|---|---|---|---|---|
| many-electron atoms | ||||
| ground-state angular momentum | ||||
| line brightness | ||||
| fine and Zeeman structure |
sodium has nuclear charge $Z = 11$ and one outer electron in level $n = 3$, measured to need $5.14$ eV to remove. Fill in the unshielded Bohr energy $13.6Z^2/n^2$ in eV, the effective charge $n\sqrt{E/13.6}$ that the outer electron feels, and the ratio of the measured to the unshielded energy.
| value | |
|---|---|
| unshielded Bohr energy (eV) | |
| effective charge | |
| measured ÷ unshielded |
In a magnetic field $B$ (T), an atomic level with magnetic quantum number $m_l = 3$ shifts in energy by $m_l\mu_BB$. With $\mu_B/h = 14.0$ GHz/T, write the shift in the frequency of light from that level, in GHz, as a formula in $B$.
Answer:
It takes $5.39$ eV to remove the outer electron of lithium, which sits in level $n = 2$. Fitting Bohr's formula $E = 13.6Z_{\text{eff}}^2/n^2$, what effective nuclear charge does that electron feel?
Answer: effective charge
At Mount Wilson Observatory in California in 1908, George Ellery Hale found spectral lines split in sunspot light. An iron line at $630.25$ nm shows components shifted by $40$ pm. Treating it as the normal Zeeman effect, $\Delta E = \mu_BB$ with $\mu_B = 5.788 \times 10^{-5}$ eV/T and $hc = 1240$ eV·nm, how strong is the sunspot's field, in T?
Answer: T
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In a magnetic field $B$ (T), an atomic level with magnetic quantum number $m_l = 2$ shifts in energy by $m_l\mu_BB$. With $\mu_B/h = 14.0$ GHz/T, write the shift in the frequency of light from that level, in GHz, as a formula in $B$.
Answer:
You can judge Bohr's model. Explain to someone why it cannot predict helium's ionization energy.
21. Your turn: lithium's outer electron, in $n = 2$, needs $5.39$ eV to remove. Find its effective charge., step 3
$Z_{\text{eff}} = 1.26$
Well below lithium's 3.