Physical Chemistry · Part 1 of 9

Foundations — Why Quantum Mechanics Exists

Quantum Chemistry, Part 1 · 9 sections · about 24,070 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

Quantum mechanics is usually taught as a set of equations to be trusted. This part does the opposite: it starts with the five experiments that broke classical physics, shows exactly what each one broke, and only then introduces the machinery built to replace it. By the end you should understand not just what the postulates say but why anyone would propose something so strange — which is the difference between reciting quantum mechanics and using it. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 1

  • 1Blackbody radiation and the ultraviolet catastrophe Free below
  • 2The photoelectric effect and Einstein's photon
  • 3The Compton effect
  • 4Atomic line spectra and the Rydberg formula
  • 5The Bohr model and the four things it cannot explain
  • 6The de Broglie relation, electron diffraction and wave packets
  • 7The wavefunction and the Born interpretation
  • 8Acceptable wavefunctions, normalisation and orthogonality
  • 9The postulates of quantum mechanics

Blackbody radiation and the ultraviolet catastrophe

Free extract

Section A.1 of Part 1, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

Hot things glow. The colour depends only on the temperature, not on what they are made of — and classical physics could not explain the shape of the glow at all.

Heat a poker. At 800 K it glows dull red; at 1500 K orange; at 3000 K, the temperature of a tungsten filament, yellow-white. Heat iron, graphite or a ceramic brick to the same temperature and, if the surface is a good absorber, you get the same colour. The spectrum of thermal radiation depends only on temperature, and that universality is the first clue that something deep is going on: the light is not reporting on the chemistry of the emitter.

Blackbody: an idealised object that absorbs all radiation falling on it at every wavelength, and therefore (by Kirchhoff's law, which equates emissivity and absorptivity at every wavelength and temperature) is also the best possible emitter. The standard laboratory realisation is not a black surface but a cavity: a hollow box held at temperature T with a small hole in one wall. Radiation entering the hole bounces around and is essentially certain to be absorbed before it finds its way out again, so the hole behaves as a near-perfect absorber, and the radiation leaking back out of it is a faithful sample of the thermal radiation inside.

Two experimental regularities were established before any theory existed, and both are still used every day — in astronomy, in pyrometry, in remote sensing:

λmax T = b = 2.8978 × 10−3 m K
M = σT4 ,   σ = 5.6704 × 10−8 W m−2 K−4

The first says the peak moves to shorter wavelength as things get hotter, which is why you can find a star's surface temperature from its colour. The second says the total power climbs as T4, which is why radiative loss dominates every high-temperature furnace calculation.

Cavity radiation: what is measured, and what classical physics predicted02505007501000125015001750200000.250.500.751.001.25wavelength λ / nmuλ / mJ m−3 nm−1visible6000 K5000 K4000 K3000 KRayleigh–Jeans, 5000 K→ ∞ as λ → 0Solid curves: the Planck distributionuλ = (8πhc/λ5) / (ehc/λkT − 1),sampled at 700 points each, not sketchedBlack dashes: the locus of the maxima,λmaxT = 2.8978 × 10−3 m K
The single most important graph in the history of chemistry. The four solid curves are the Planck distribution evaluated at 3000, 4000, 5000 and 6000 K; the dashed black line through their maxima is the Wien displacement law, and it is not drawn by hand — it is the set of computed peak positions. The dashed green curve is the Rayleigh–Jeans prediction at 5000 K, the best that classical electromagnetism plus classical statistical mechanics can do. It joins the true curve only at wavelengths far beyond the right-hand edge of this plot, where hc/λkT ≪ 1: even at the 2060 nm edge, hc/λkT is still 1.40 and the green line runs a factor of 2.2 above the 5000 K curve — the logarithmic figure below shows the convergence properly. As λ falls it climbs without limit. Integrated over all wavelengths it gives infinite energy in any warm box. That is the ultraviolet catastrophe — not a discrepancy to be patched but a prediction of infinity where the experiment gives a finite number.

The classical calculation, and where it goes wrong

The classical account of the cavity has two ingredients, and each is individually unimpeachable. First, count the standing electromagnetic waves that fit inside a box of volume V. Electromagnetism gives an exact answer, and the number of modes per unit volume per unit wavelength interval is 8π/λ4. There is nothing controversial here; the same counting is used today. Second, give each mode its share of thermal energy. Classical statistical mechanics has a theorem for that — the equipartition theorem — which assigns ½kT to each quadratic degree of freedom, hence kT to each oscillator with both kinetic and potential terms.

uλ = (number of modes per unit volume per unit λ) × (mean energy per mode) = (8π/λ4) × kT

Multiply the two and you get the Rayleigh–Jeans law, the green dashed curve in the figure above. It works beautifully at long wavelength. It then goes to infinity as λ → 0, which no measurement has ever done, and the total energy in the cavity, obtained by integrating over all wavelengths, is infinite. Ehrenfest named this the ultraviolet catastrophe. The name is exact and worth pausing on: the disaster is at the short-wavelength end, and it is a catastrophe rather than a discrepancy because the predicted answer is not merely wrong but unbounded.

Ultraviolet catastrophe: the classical prediction that the spectral energy density of cavity radiation diverges as λ−4 at short wavelength, so that any object at any non-zero temperature would radiate infinite energy. Its origin is the combination of (i) an unbounded number of short-wavelength modes and (ii) equipartition, which gives every one of those modes the same average energy kT.

Locate the blame carefully, because it is a favourite examination question. The mode counting is right. Maxwell's equations are right. The failure is in equipartition — that is, in the assumption that a mode may take up energy in arbitrarily small amounts and therefore always ends up with its statistical share kT.

Advanced / reference layer

Planck's quantisation, the derivation of the distribution, and both classical laws recovered as limits.

Planck's route in October 1900 was, by his own later description, an act of desperation. He already had an interpolation formula that fitted the data; what he needed was a derivation. He obtained one by treating the cavity walls as containing charged harmonic oscillators in equilibrium with the radiation, and then making one assumption whose consequences he did not at first believe:

Planck's quantum hypothesis: an oscillator of frequency ν cannot take up or give out energy continuously. Its energy is restricted to the discrete set En = nhν with n = 0, 1, 2, …, where h is a new universal constant. Energy is exchanged with the radiation field only in whole quanta of size hν.

Everything follows from evaluating the Boltzmann average with a sum instead of an integral. Classically one writes

⟨E⟩classical = ∫0∞ E e−E/kT dE ∕ ∫0∞ e−E/kT dE = kT

With the energy restricted to multiples of hν the integrals become sums:

⟨E⟩ = ∑n=0∞ nhν e−nhν/kT ∕ ∑n=0∞ e−nhν/kT

Both sums are geometric. Put x = hν/kT and s = e−x. The denominator is ∑sn = 1/(1 − s). The numerator is hν∑nsn = hν s/(1 − s)2, using the standard result obtained by differentiating the geometric series with respect to s. Dividing:

⟨E⟩ = hν · s/(1 − s) = hν / (ehν/kT − 1)

This one line is the entire quantum revolution in embryo, and the figure below is worth more than any amount of prose about it. Multiply this mean energy by the same classical mode density 8π/λ4 and, with ν = c/λ, you have the Planck distribution:

uλ(λ,T) = (8πhc / λ5) · 1 / (ehc/λkT − 1)
uν(ν,T) = (8πhν3 / c3) · 1 / (ehν/kT − 1)
Note: The two forms are not related by substituting ν = c/λ. They are densities, so the conversion carries a Jacobian: uλdλ = uνdν with |dν/dλ| = c/λ2. This is why the peak of uν and the peak of uλ do not correspond to the same colour — the frequency form peaks at x = 2.8214 and the wavelength form at x = 4.9651. A question that asks you to find λmax and then quotes a frequency-domain constant is testing exactly this point.
Where the catastrophe is cured012345600.20.40.60.81.0x = hν/kT (photon energy in units of the thermal energy)⟨E⟩ / kTclassical equipartition: ⟨E⟩ = kT, independent of νPlanck: ⟨E⟩ = hν/(ex − 1)at hν = kT the mode already holds only 0.582 kThigh-frequency modes are FROZEN OUT.There is not enough thermal energy to buyeven one quantum hν, so on average themode is empty. This is Planck's whole idea.
Equipartition is the villain, and quantisation is the fix. Classical statistical mechanics gives every oscillator, of whatever frequency, a mean energy kT (green). Since a cavity has an unbounded number of high-frequency modes, that hands out an unbounded amount of energy. Planck's assumption — that an oscillator of frequency ν can only hold energy in whole multiples of hν — replaces kT by hν/(ehν/kT − 1), the red curve, computed here at 630 points. At low frequency the two agree exactly. At high frequency the mean energy collapses exponentially, because the smallest amount of energy the mode can accept is far more than kT and thermal fluctuations essentially never supply it. The infinite sum is tamed and the catastrophe disappears.

Limit 1 — recovering Rayleigh–Jeans (long wavelength, hc ≪ λkT)

Let x = hc/λkT and take x ≪ 1, which means either long wavelength or high temperature. Expand the exponential:

ex − 1 = x + x2/2! + x3/3! + … ≈ x  (x ≪ 1)
uλ ≈ (8πhc/λ5) · (1/x) = (8πhc/λ5) · (λkT/hc) = 8πkT/λ4

Notice that h cancels completely. That is the mathematical statement of the correspondence principle for this problem: in the régime where the quantum hν is small compared with kT, the quantum result becomes independent of h and reduces to the classical one. It also explains why the ultraviolet catastrophe was not spotted decades earlier — in the infrared, where nineteenth-century detectors worked best, Rayleigh–Jeans is essentially exact.

Limit 2 — recovering Wien's law (short wavelength, hc ≫ λkT)

Now take x ≫ 1. Then ex is enormous and the −1 in the denominator is negligible:

uλ ≈ (8πhc/λ5) e−hc/λkT

This is the law Wien proposed in 1896 on thermodynamic and semi-empirical grounds, and which fitted the visible and near-ultraviolet data superbly while failing in the infrared. The Planck distribution contains it as its short-wavelength asymptote. Both limits are drawn against the exact function below, and the agreement in each region is exact rather than approximate in the appropriate limit.

One formula, two limits — and each old law is one of them100316100031601042×10410−11011031051071091011wavelength λ / nm (logarithmic)uλ / J m−4λmax = 580 nmPlanck — agrees with experiment everywhereRayleigh–Jeans — exact at large λ, diverges at small λWien — exact at small λ, wrong at large λhc/λkT ≫ 1quantum régimehc/λkT ≪ 1classical régimeT = 5000 K throughout; all three functions evaluated at 601 points from the same constants
Why Planck's formula was accepted immediately. Before 1900 there were two competing empirical laws, each right in half the spectrum and embarrassingly wrong in the other half. On logarithmic axes the relationship is unmistakable: the Wien exponential law (violet) lies exactly on the Planck curve at short wavelength and falls away at long wavelength; the Rayleigh–Jeans law (green) lies exactly on it at long wavelength and runs off to infinity at short wavelength. The Planck distribution is not a compromise between them — it is a single function that reduces to each of them in the appropriate limit of the one dimensionless group in the problem, x = hc/λkT. Both limits are derived in the advanced layer.

Limit 3 — the two empirical laws as corollaries

Wien's displacement law is now a theorem rather than an observation. Differentiate uλ with respect to λ, set the result to zero, and write x = hc/λkT:

duλ/dλ = 0  ⇒  x ex/(ex − 1) = 5  ⇒  x = 5(1 − e−x)

That transcendental equation has the non-trivial root x = 4.965114, obtained here by simple fixed-point iteration at build time. Hence

λmaxT = hc/(4.9651 k) = 2.897772 × 10−3 m K

The Stefan–Boltzmann law comes from integrating the distribution over all wavelengths. Substituting x = hc/λkT converts the integral into the standard form ∫0∞x3/(ex−1)dx = π4/15, giving

utotal = (8π5k4/15h3c3) T4 ,   M = (c/4)utotal = σT4 with σ = 2π5k4/15h3c2

Evaluating that expression from the CODATA constants gives σ = 5.670374 × 10−8 W m−2 K−4, the measured value. Two empirical laws, one spectral shape, and a single new constant — that combination is why the physics community accepted the formula long before it accepted the idea behind it.

Find the surface temperature of the Sun from its spectrum Easy

The datum. The solar spectrum peaks near λmax = 500 nm, in the green — which is why our eyes are most sensitive there.
Step 1 — use Wien's displacement law. T = b/λmax with b = 2.8978 × 10−3 m K.
Step 2 — substitute. T = 2.8978 × 10−3 / (500 × 10−9 m) = 5796 K.
Step 3 — check against the accepted value. The Sun's effective temperature is 5772 K, and 5796 K is within 0.5%. The agreement is better than the blackbody idealisation deserves — the solar photosphere is not a perfect blackbody — but it is good enough that this is how stellar temperatures are actually estimated.
The reverse direction. A blackbody at 5772 K peaks at b/5772 = 502.0 nm. Life on Earth evolved photochemistry tuned to that peak; chlorophyll, rhodopsin and every solar cell are optimised against this one number.

Show that the microwave background and a tungsten lamp obey the same formula Medium

The point. One formula spans temperatures differing by a factor of a thousand. Doing both in the same units is the fastest way to feel that.
Step 1 — cosmic microwave background, T = 2.725 K. λmax = b/T = 2.8978 × 10−3/2.725 = 1.0634 mm = 1.063 mm — microwaves, as the name says.
Step 2 — tungsten filament, T = 2800 K. λmax = 1035 nm — in the near infrared, which is precisely why an incandescent bulb is such a poor lamp: most of its output is heat you cannot see.
Step 3 — compare total emission. By Stefan–Boltzmann the ratio of emitted power per unit area is (2800/2.725)4 = 1.115 × 1012. The same equation, unchanged, covers both.
Step 4 — the peak photon energies. E = hc/λmax gives 1.1659 meV for the CMB and 1.198 eV for the filament. Note that the filament photon is of the order of a chemical bond energy while the CMB photon is about a thousandth of one (1.17 meV against a bond of a few eV); this is the origin of the rule that thermal radiation at room temperature cannot drive photochemistry.

Derive the Rayleigh–Jeans law as a limit of the Planck law, stating the condition carefully Hard

Step 1 — identify the dimensionless group. The Planck distribution contains exactly one dimensionless combination, x = hc/λkT = hν/kT. Every statement about ‘classical’ or ‘quantum’ behaviour is a statement about whether x is small or large. Never quote a limit as ‘λ large’ without saying large compared with what.
Step 2 — expand. For x ≪ 1, ex − 1 = x + x2/2 + … so 1/(ex − 1) = 1/x − 1/2 + x/12 − … . Keeping only the leading term is legitimate because the corrections are smaller by a factor x.
Step 3 — substitute. uλ = (8πhc/λ5)(1/x) = (8πhc/λ5)(λkT/hc) = 8πkT/λ4.
Step 4 — interpret the cancellation. h has vanished. That is the signature of a correspondence-principle limit and is worth stating explicitly in an answer: the classical law is not an approximation added on to the quantum one, it is what the quantum one becomes when the quantum hν is negligible beside the thermal energy kT.
Step 5 — where the boundary lies numerically. At room temperature (298 K), x = 1 corresponds to λ = hc/kT = 48.3 μm, in the mid-infrared. So for all of thermal physics at wavelengths longer than about 50 μm the classical law is perfectly adequate, and for visible light (x ≈ 88) it is hopeless.
Step 6 — the other limit, for completeness. For x ≫ 1 drop the −1 instead: uλ ≈ (8πhc/λ5)e−hc/λkT, Wien's law. Examiners frequently ask for both in one question, and the marks are for naming the condition, not for the algebra.

⚠ Common mistakes & exam traps

  • Saying the ultraviolet catastrophe means ‘the theory disagreed with experiment at short wavelength’. True but weak. The full statement is that the predicted energy density diverges, and hence the predicted total energy in any cavity at any temperature is infinite. Say ‘infinite’ and you get the mark.
  • Blaming Maxwell's equations or the mode counting. Neither is at fault. The mode density 8π/λ4 survives unchanged into the Planck derivation. The casualty is equipartition, i.e. the assumption of continuously variable energy.
  • Confusing the wavelength and frequency forms of the law. uλ and uν are densities with respect to different variables, so they peak at different places. λmaxνmax ≠ c.
  • Writing Wien's displacement law as λmax = b·T. It is a product that is constant: λmaxT = b. Hotter means shorter.
  • Believing Planck quantised the radiation field. He did not, and said so repeatedly. He quantised the energy exchange of material oscillators in the walls. Quantising the field itself — the photon — is Einstein's step in A.2, and Planck resisted it for over a decade. This distinction appears in essay questions.
Easy
A furnace is at 1500 K. At what wavelength does its emission peak, and in what region of the spectrum is that?
Show solution
λmax = b/T = 2.8978 × 10−3/1500 = 1.93 × 10−6 m = 1932 nm, in the near infrared. The visible glow you see from a 1500 K furnace is only the short-wavelength tail of the distribution — most of the energy is invisible.
Med
Calculate the ratio of the Rayleigh–Jeans prediction to the true Planck value at λ = 500 nm and T = 5000 K, and comment.
Show solution
x = hc/λkT = 1.9864 × 10−25 / (500 × 10−9 × 1.3806 × 10−23 × 5000) = 5.755. The ratio is uRJ/uPlanck = (ex − 1)/x = (315.799 − 1)/5.755 = 54.7. The classical law overestimates the energy density in the green by a factor of about fifty at the surface temperature of a cool star — and the factor grows without bound as λ falls further.
Hard
Show that the mean energy of a Planck oscillator tends to kT at high temperature and to hνe−hν/kT at low temperature, and say what each limit means physically.
Show solution
Write x = hν/kT, so ⟨E⟩ = hν/(ex − 1).
High T (x ≪ 1): ex − 1 ≈ x, so ⟨E⟩ ≈ hν/x = kT. Equipartition is recovered: when the thermal energy is much larger than one quantum, the discreteness is invisible and the mode behaves classically.
Low T (x ≫ 1): ex − 1 ≈ ex, so ⟨E⟩ ≈ hνe−hν/kT, which falls exponentially to zero. The mode is frozen out: the smallest energy it can accept is hν, the Boltzmann factor for supplying that much energy is e−hν/kT, and the mode is almost always empty. The same argument, applied to vibrations of a solid, is Einstein's and then Debye's explanation of why heat capacities fall below 3R at low temperature — a second, independent confirmation of the quantum hypothesis.
Hard
The total energy density in a cavity is u = aT4. Given that the Stefan–Boltzmann constant is σ = ac/4, and that ∫0∞x3dx/(ex−1) = π4/15, obtain σ in terms of fundamental constants and evaluate it.
Show solution
Integrate uν = (8πhν3/c3)/(ehν/kT−1) over ν. Substituting x = hν/kT gives ν = kTx/h and dν = (kT/h)dx, so u = (8πh/c3)(kT/h)4∫x3dx/(ex−1) = (8π5k4/15h3c3)T4. Hence a = 8π5k4/15h3c3 and σ = 2π5k4/15h3c2 = 5.67037 × 10−8 W m−2 K−4, which is the measured value. Note that σ is not an independent constant of nature: it is built entirely from h, c and k, which is itself strong evidence that the derivation is right.

Further reading. Levine Quantum Chemistry, Atkins & Friedman Molecular Quantum Mechanics, McQuarrie Quantum Chemistry, Szabo & Ostlund Modern Quantum Chemistry, Griffiths Introduction to Quantum Mechanics, and Pilar Elementary Quantum Chemistry.

Read the rest of Part 1

The remaining 8 sections of this part — The photoelectric effect and Einstein's photon, The Compton effect, Atomic line spectra and the Rydberg formula, The Bohr model and the four things it cannot explain — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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