Particle Properties of Waves

Learning Objectives¶

By the end of this chapter, you should be able to:

• Explain the ultraviolet catastrophe and how Planck’s quantization hypothesis resolves it.

• State and apply Planck’s radiation law, Wien’s displacement law, and the Stefan–Boltzmann law.

• Describe the photoelectric effect and explain why classical (wave) theory fails to account for its observed features.

• Apply Einstein’s photon model, $E = hf$, to the photoelectric effect and interpret the stopping-potential and work-function data it predicts.

• Explain X-ray production (bremsstrahlung) as an energy-bookkeeping inverse of the photoelectric effect, and apply the Duane–Hunt relation for the minimum X-ray wavelength.

• Derive and apply the Compton scattering formula, and explain why Compton scattering demonstrates that photons carry momentum.

• Explain pair production and pair annihilation as further evidence for the particle nature of light, and use conservation of energy and momentum to find threshold conditions for each.

Introduction¶

Chapters 1–3 dismantled the ether and rebuilt kinematics and dynamics on Einstein’s postulates, but light itself — a wave, according to Maxwell’s equations, which special relativity leaves intact — remained conceptually untouched. Chapters 4–5 then developed that wave picture in full quantitative detail: interference and diffraction, both direct, precise confirmations that light obeys a wave equation. This chapter takes up a second, independent crisis of classical physics, one concerning not the kinematics of light but its very nature. Four phenomena — blackbody radiation, the photoelectric effect, X-ray production, and Compton scattering — each show, in a different experimental setting, that electromagnetic radiation exchanges energy with matter not continuously, as a wave should, but in discrete packets. The concept that emerges, the photon, carries energy $E = hf$ and momentum $p = h/\lambda$, and behaves in each of these experiments like a particle, even though light indisputably also shows the wave behavior of Chapters 4–5 in other settings. A final phenomenon, pair production, shows that a photon’s energy can be converted entirely into matter, tying the particle nature of light directly to the mass–energy equivalence of Chapter 3. Reconciling these two faces of light is the beginning of quantum mechanics.

6.1Blackbody Radiation and Planck’s Hypothesis¶

Any object at temperature $T$ emits electromagnetic radiation across a continuous range of wavelengths, with an intensity distribution depending on $T$. An idealized perfect absorber and emitter is called a blackbody; its emitted spectral distribution, the blackbody spectrum, depends only on temperature, not on the material. A good laboratory approximation is a small hole in the wall of an otherwise closed, heated cavity: radiation entering the hole is absorbed and re-emitted many times by the cavity walls before any of it can escape, so the radiation leaking out through the hole is, to excellent approximation, in thermal equilibrium with the cavity walls and independent of what the walls are made of.

The Ultraviolet Catastrophe¶

Late-nineteenth-century classical physics attempted to compute this spectrum by treating the electromagnetic field inside the cavity as a superposition of standing-wave modes, each behaving as an independent harmonic oscillator, and invoking the equipartition theorem — every such oscillator, in thermal equilibrium at temperature $T$, should carry average energy $k_BT$, independent of its frequency. Counting the number of standing-wave modes per unit volume in a wavelength interval $d\lambda$ (a purely geometric problem, fixed by the cavity’s size) and multiplying by $k_BT$ per mode gives the Rayleigh–Jeans law,

for the spectral energy density. This matches the observed spectrum reasonably well at long wavelength, but diverges as $\lambda \to 0$: since the number of short-wavelength modes grows without bound and each is assigned the same energy $k_BT$, the predicted radiated energy — obtained by integrating $u(\lambda,T)$ over all $\lambda$ — is infinite. This absurd prediction, sharply contradicted by the observed spectrum (which rises from zero at short wavelength, peaks, and falls off at long wavelength), became known as the ultraviolet catastrophe: not a small discrepancy to be patched, but a qualitative failure of classical statistical mechanics applied to the electromagnetic field.

Figure 6.1 plots both curves against each other. Raising the temperature moves the peak and lifts the total area, as the next section’s displacement and Stefan–Boltzmann laws require; the classical curve, plotted alongside, runs off the top of the graph at short wavelength no matter what temperature is chosen.

Planck’s Quantization Hypothesis¶

Max Planck resolved this in 1900 by a hypothesis with no classical justification: the oscillators making up the cavity walls cannot exchange energy with the field continuously, but only in discrete multiples of a fundamental quantum,

where $f$ is the oscillator’s frequency and $h$ is a new fundamental constant (Planck’s constant), $h = 6.626\times 10^{-34}\ \text{J}\cdot\text{s}$. With this assumption, the average energy of a mode at frequency $f$ in thermal equilibrium at temperature $T$ is not $k_BT$ (the equipartition value, independent of $f$, which causes the catastrophe) but

which suppresses high-frequency (short-wavelength) modes because at high $f$, $hf \gg k_BT$ makes a single quantum too “expensive” to be thermally excited: the Boltzmann factor $e^{-hf/k_BT}$ governing how likely a mode is to hold even one quantum falls off exponentially, cutting off the ultraviolet divergence entirely. This leads to Planck’s radiation law for the spectral energy density,

which matches the observed blackbody spectrum at all wavelengths and temperatures, and reduces to the Rayleigh–Jeans law in the limit $hf \ll k_BT$, i.e. at long wavelength, where $e^{hf/k_BT}-1 \approx hf/k_BT$ and quantization effects become negligible — exactly the regime in which the classical calculation had already succeeded. Planck himself regarded the quantization of oscillator energy as a mathematical device rather than a physical claim about light. It was Einstein who, five years later, proposed taking it literally.

Wien’s Displacement Law and the Stefan–Boltzmann Law¶

Differentiating $u(\lambda,T)$ with respect to $\lambda$ and setting the result to zero gives Wien’s displacement law for the wavelength of peak emission,

which is why hotter objects glow at shorter (bluer) wavelengths — from the deep red of a heating element to the blue-white of a hot star. Integrating $u(\lambda,T)$ over all wavelengths gives the total power radiated per unit surface area of an ideal blackbody, the Stefan–Boltzmann law,

with $\sigma$ the Stefan–Boltzmann constant. The steep $T^4$ dependence means that radiated power is extremely sensitive to temperature: doubling an object’s absolute temperature increases its radiated power per unit area sixteenfold. Real objects are not perfect blackbodies; a surface’s actual radiated power is $P/A = \varepsilon\sigma T^4$, where the emissivity $\varepsilon \le 1$ measures how closely the surface approximates an ideal absorber and emitter ($\varepsilon = 1$ for a true blackbody).

Worked Example: The Sun as a Blackbody¶

The Sun’s surface (photosphere) has an effective temperature of $T \approx 5778\ \text{K}$ and radius $R_\odot = 6.96\times10^8\ \text{m}$. Treating it as an ideal blackbody, Wien’s law gives the wavelength of peak emission,

squarely in the green part of the visible spectrum — the Sun’s spectrum actually peaks near the middle of the range of wavelengths the human eye evolved to detect, though scattering in Earth’s atmosphere and the eye’s overall spectral response make sunlight appear white or yellow rather than green. The Stefan–Boltzmann law gives the power radiated per unit area, $P/A = \sigma T^4 = (5.670\times10^{-8}\ \text{W}\cdot\text{m}^{-2}\cdot\text{K}^{-4})(5778\ \text{K})^4 \approx 6.32\times10^7\ \text{W/m}^2$. Multiplying by the Sun’s surface area, $A = 4\pi R_\odot^2 \approx 6.09\times10^{18}\ \text{m}^2$, gives a total radiated power (luminosity)

in close agreement with the Sun’s independently measured luminosity of $3.828\times10^{26}\ \text{W}$ — the same figure used in Chapter 3, Problem 6 to estimate the Sun’s rate of mass loss via $E=mc^2$. That two completely independent lines of physics (blackbody radiation here, and mass–energy equivalence there) connect so tightly to the same measured number is a testament to how well both theories describe the physical world.

Historical Context: Lenard’s Puzzle¶

Planck’s quantization hypothesis was aimed narrowly at the blackbody spectrum, and Planck himself did not initially believe that light itself came in discrete packets. Evidence pointing in that direction had, in fact, already been gathering independently. In 1902, three years before Einstein’s photon paper, Philipp Lenard studied the photoelectric effect using an arc lamp and a retarding-voltage apparatus much like the one described below, and found — to general puzzlement at the time — that the stopping voltage (and hence the photoelectrons’ maximum kinetic energy) did not increase with the lamp’s intensity, contrary to every expectation from Maxwell’s wave theory of light, in which a more intense wave carries proportionally more energy to whatever absorbs it. Lenard had no explanation; the result simply sat as an unresolved anomaly in the physics literature for three years, alongside blackbody radiation, until Einstein showed that both puzzles have the same root cause: light exchanges energy with matter in discrete quanta, not continuously.

6.2The Photoelectric Effect and X-Ray Production¶

The Photoelectric Effect¶

When light of sufficiently short wavelength strikes a metal surface, electrons are ejected — the photoelectric effect. A typical apparatus measures the photocurrent as a function of a retarding voltage $V$ applied between the emitting surface and a collector; the voltage at which the current just reaches zero, the stopping potential $V_0$, gives the maximum kinetic energy of the ejected electrons, $K_{\max} = eV_0$. Plotting the measured photocurrent against retarding voltage traces out a curve that falls smoothly to zero at $V_0$ rather than dropping abruptly, because the ejected electrons are emitted with a distribution of kinetic energies up to $K_{\max}$, not a single sharp value; $V_0$ marks where even the fastest electrons are turned back.

Three experimental features of this effect resist any explanation in terms of classical electromagnetic waves, in which energy is delivered continuously and is proportional to intensity:

1. $K_{\max}$ is independent of light intensity. Classically, a more intense wave delivers more energy per unit time to an electron and should eject electrons with more kinetic energy. Experimentally, increasing intensity increases the number of photoelectrons (the current) but not $K_{\max}$.

2. $K_{\max}$ depends linearly on frequency, and there exists a sharp threshold frequency $f_0$, characteristic of the metal, below which no photoelectrons are emitted at all, regardless of intensity or exposure time. Classically, a wave of any frequency should eventually eject electrons if given enough time to deliver sufficient energy, so a hard threshold — and one depending on frequency rather than intensity — has no classical explanation.

3. Emission is (essentially) instantaneous, with no observable time lag even at very low intensity. Classically, a dim wave should take a measurable time to deliver enough energy to an electron to free it; a rough classical estimate for a very weak source predicts delays of minutes to hours, yet no such delay is ever observed.

Einstein resolved all three in 1905 by proposing that light itself is quantized: it consists of discrete packets, photons, each carrying energy

with $h$ the same constant Planck introduced. In the photoelectric effect, a single photon transfers its entire energy to a single electron in one interaction. If $\phi$ (the metal’s work function) is the minimum energy needed to remove an electron from the metal, conservation of energy gives

the photoelectric equation. This immediately explains all three observations: $K_{\max}$ depends on $f$ (through the photon energy $hf$) but not on intensity, since intensity only changes the number of photons per second, not the energy of each one; the threshold frequency is $f_0 = \phi/h$, below which a single photon simply does not carry enough energy to free an electron, no matter how many photons arrive; and emission is instantaneous because each electron absorbs one photon’s energy all at once, not gradually. A plot of $K_{\max}$ (equivalently, $eV_0$) versus $f$ is a straight line of slope $h$ and $y$-intercept $-\phi$, and Millikan’s precise measurement of exactly this line (1916) provided both a direct experimental value of $h$ and strong confirmation of Einstein’s photon hypothesis — for which, not for relativity, Einstein received the 1921 Nobel Prize in Physics.

The three stubborn facts are best met in the order Lenard and Millikan met them, with the apparatus in front of you. In Figure 6.2 the light’s intensity, its wavelength, the target metal, and the retarding voltage are all under control, and the photocurrent and stopping potential are read off directly. Turn the intensity up at fixed wavelength: the current rises and the stopping potential does not move. Shorten the wavelength instead: the stopping potential climbs, and a plot of $eV_0$ against $f$ — the simulation will accumulate one for you — is a straight line whose slope is $h$ and whose intercept names the metal. Cross the threshold from the long-wavelength side and the current stops altogether, at full intensity.

Worked Example: The Photoelectric Effect in Cesium¶

Cesium has one of the lowest work functions of any metal, $\phi = 2.14\ \text{eV}$, which is why it is used in the photocathodes of photomultiplier tubes and early photoelectric light meters.

Ultraviolet light of wavelength $\lambda = 250\ \text{nm}$ illuminates a cesium surface. Using the convenient combination $hc = 1240\ \text{eV}\cdot\text{nm}$,

The maximum kinetic energy of the photoelectrons is $K_{\max} = E_{\text{photon}} - \phi = 4.96\ \text{eV} - 2.14\ \text{eV} = 2.82\ \text{eV}$, so the stopping potential is $V_0 = 2.82\ \text{V}$. The threshold wavelength — the longest wavelength that can still eject an electron — follows from $\phi = hc/\lambda_0$:

which falls in the visible (yellow) part of the spectrum. Cesium is therefore photoelectrically sensitive to ordinary visible light, unlike most metals, whose work functions of 4–$5\ \text{eV}$ push their threshold wavelengths into the ultraviolet.

Applications: Photomultipliers and Photovoltaic Cells¶

The photoelectric effect underlies two technologies central to modern experimental physics and everyday life. A photomultiplier tube exploits a low-work-function photocathode (often a cesium compound, as in the worked example above) to convert a single incoming photon into a single ejected photoelectron, then accelerates that electron into a series of intermediate electrodes (dynodes), each of which is struck hard enough to eject several additional electrons by secondary emission; the resulting cascade, multiplying the electron count by a factor of a few at each of perhaps ten dynode stages, converts the arrival of a single photon into a macroscopic, easily measured current pulse containing millions of electrons. This single-photon sensitivity makes photomultiplier tubes essential wherever extremely faint light must be detected and counted, including the gamma-ray detectors used in PET scanners (discussed later in this chapter) and in nuclear and particle-physics experiments generally.

A closely related but distinct effect, the photovoltaic effect, underlies solar cells: rather than ejecting an electron entirely from the material, an absorbed photon in a semiconductor promotes an electron across the material’s band gap (the semiconductor analog of the work function, examined further in solid-state contexts), and a built-in electric field at a junction between two differently doped semiconductor layers then separates the resulting electron and the vacancy (hole) it leaves behind before they can recombine, driving a current through an external circuit. As in the ordinary photoelectric effect, only photons with energy exceeding a threshold (the band-gap energy) contribute usable electrons, which is why solar-cell efficiency depends sensitively on matching the semiconductor’s band gap to the solar spectrum.

X-Ray Production: Bremsstrahlung¶

The photoelectric effect converts a photon into an energetic electron; an energetic electron can conversely give up part of its energy as radiation when it is abruptly decelerated in matter. This bremsstrahlung process is the standard laboratory and industrial method of producing X-rays. It is an energy-bookkeeping inverse of the photoelectric effect, not its literal microscopic time reverse (which would involve radiative recombination into a bound state). In an X-ray tube, electrons are accelerated from rest through a large potential difference $V$ (typically tens to hundreds of kilovolts) and strike a dense metal target (commonly tungsten). As each electron decelerates within the target — bremsstrahlung, German for “braking radiation” — it radiates part or all of its kinetic energy as one or more photons.

Because an individual electron can, in principle, lose its entire kinetic energy $eV$ in a single deceleration event (converting essentially all its kinetic energy into one photon), there is a sharply defined maximum photon energy — and correspondingly a minimum wavelength — that the tube can produce, set by

known as the Duane–Hunt limit. Most electrons instead lose their energy gradually over many collisions, radiating photons of smaller energy at each step and producing a continuous spectrum of wavelengths longer than $\lambda_{\min}$ — but no photon in the entire spectrum can have a wavelength shorter than $\lambda_{\min}$, since no single electron carries more than $eV$ of kinetic energy to begin with. Superimposed on this continuous bremsstrahlung spectrum are sharp characteristic X-ray lines, produced when an incident electron knocks an inner-shell electron out of a target atom and an outer electron falls down to fill the vacancy, emitting a photon of energy fixed by the target element’s inner-shell energy levels (Moseley’s law for these characteristic energies is developed in Chapter 11). Unlike the continuous bremsstrahlung background, the characteristic lines depend on the target material, not on the accelerating voltage (once $V$ is large enough to eject the inner-shell electron in the first place).

Bremsstrahlung is worth separating into its classical and its quantum halves, because only the second is new here. The classical half — that a charge which accelerates radiates, and that the radiation carries energy away — is already in Maxwell’s equations, and Figure 6.3 is that statement on its own: shake a charge and kinks in its field propagate outward at $c$, taking energy with them. Stopping a charge dead is an acceleration like any other, so an electron slamming into a tungsten anode must radiate. What classical physics cannot supply is the sharp edge at $\lambda_{\min}$: a continuous field theory sets no floor on the wavelength radiated in a single event, and the Duane–Hunt limit exists only because the radiated energy comes in quanta $hf$ and one electron brings only $eV$ to spend.

Worked Example: Minimum Wavelength from a Diagnostic X-Ray Tube¶

A medical diagnostic X-ray tube is operated at an accelerating voltage of $V = 80.0\ \text{kV}$. The Duane–Hunt limit gives

This is comparable to, and somewhat shorter than, typical interatomic spacings in crystals ($\sim 0.1$–$0.3\ \text{nm}$), which is why X-rays of this energy scale are useful for crystallographic diffraction (Chapter 5) as well as medical imaging, where their short wavelength (and correspondingly high photon energy) allows them to penetrate soft tissue while being partially absorbed by denser bone.

6.3Compton Scattering, Pair Production, and the Photon¶

Compton Scattering¶

Even after the photoelectric effect, one could imagine “photon-like” energy exchange as a property specific to bound electrons in a metal, without light itself consisting of localized particles carrying momentum. Arthur Compton’s 1923 experiment removed this loophole by showing that photons scattering from a free electron transfer momentum exactly as a particle collision would.

In Compton’s experiment, X-rays of a single wavelength $\lambda$ are directed at a target of loosely bound (effectively free) electrons, and the wavelength $\lambda'$ of the scattered X-rays is measured as a function of scattering angle $\theta$. Classically, an electromagnetic wave incident on a charge should simply drive that charge to oscillate at the incident frequency and re-radiate at the same frequency (Thomson scattering); no wavelength shift is expected. Compton observed a systematic increase in wavelength, $\lambda' > \lambda$, growing with scattering angle $\theta$ and independent of the target material — a signature of a two-body collision, not wave re-radiation.

Deriving the Compton Formula¶

Treat the photon as a particle with energy $E = hc/\lambda$ and momentum $p = E/c = h/\lambda$ (consistent with the massless-particle limit of the energy–momentum relation from Chapter 3), and apply conservation of relativistic energy and momentum to an elastic collision between the photon and an initially free, stationary electron of mass $m_e$. Let the photon scatter through angle $\theta$, emerging with wavelength $\lambda'$, while the electron recoils with momentum $p_e$ and (relativistic) energy $E_e$. The three momentum components entering the collision — incident photon momentum $h/\lambda$ along the initial direction, and zero for the electron — must balance the two outgoing momenta, whose vector sum (photon momentum $h/\lambda'$ at angle $\theta$, electron momentum $p_e$ at some recoil angle) forms a triangle. The law of cosines applied to that triangle gives

Conservation of energy, with the electron initially at rest ($E_e^{(0)} = m_ec^2$), gives

Squaring (ii) and using the energy–momentum invariant $E_e^2 = (p_ec)^2 + (m_ec^2)^2$ from Chapter 3 to eliminate $E_e$ in favor of $p_ec$, then substituting (i) for $(p_ec)^2$, produces (after the $\left(\frac{hc}{\lambda}\right)^2$, $\left(\frac{hc}{\lambda'}\right)^2$, and $(m_ec^2)^2$ terms cancel identically between the two sides) the much simpler relation

Dividing through by $hc$, multiplying both sides by $\lambda\lambda'/m_ec^2$, and simplifying $\lambda\left(\frac1\lambda-\frac1{\lambda'}\right)\lambda' = \lambda'-\lambda$ yields the Compton scattering formula:

The constant $h/m_ec = 2.426\times 10^{-12}\ \text{m}$ is the Compton wavelength of the electron. The formula correctly predicts zero shift at $\theta = 0$ (forward, undeflected “scattering”) and maximum shift $2h/m_ec$ at $\theta = 180°$ (photon backscattered), matches the observed angular dependence precisely, and — crucially — is independent of the incident wavelength $\lambda$ itself, matching experiment. Compton scattering is direct, quantitative confirmation that a photon carries momentum $p = h/\lambda$ and transfers it to a free electron exactly as one particle colliding with another.

Worked Example: Compton-Scattered Molybdenum X-Rays¶

X-rays of wavelength $\lambda = 0.100\ \text{nm}$ (comparable to characteristic molybdenum $K_\alpha$ X-rays used in crystallography) Compton-scatter off free electrons at $\theta = 60°$. The wavelength shift is

so the scattered wavelength is $\lambda' = 101.2\ \text{pm}$. In photon-energy terms, the incident photon carries $E = hc/\lambda = (1240\ \text{eV}\cdot\text{nm})/(0.100\ \text{nm}) = 12.40\ \text{keV}$, while the scattered photon carries $E' = hc/\lambda' = (1240\ \text{eV}\cdot\text{nm})/(0.1012\ \text{nm}) = 12.25\ \text{keV}$. The energy transferred to the recoiling electron is therefore $\Delta E = E - E' \approx 0.15\ \text{keV} = 150\ \text{eV}$ — a small but entirely measurable fraction of the incident photon’s energy, exactly the kind of energy loss Compton measured to confirm the formula.

The Compton Edge¶

In a real gamma-ray or X-ray detector, photons scatter through the full range of angles $0 \le \theta \le 180°$ available inside the detector material, and the detector records the energy $\Delta E(\theta) = E - E'(\theta)$ actually deposited by the recoiling electron for each scattering event. Because $\Delta E(\theta)$ increases monotonically with $\theta$ (least energy transfer for forward scattering, most for backscattering), there is a sharply defined maximum possible energy deposit, occurring at $\theta = 180°$ — the Compton edge — beyond which no Compton-scattered electron can deposit more energy in a single scattering event, no matter how many photons are examined. A photon that instead deposits its entire energy in one interaction (via the photoelectric effect on a bound atomic electron, rather than Compton scattering a free one) produces a separate, sharp peak at the full incident photon energy, the photopeak. Gamma-ray spectroscopists routinely distinguish these two features — a sharp photopeak plus a broad continuum of Compton-scattered energies cut off abruptly at the Compton edge — when interpreting a detector’s measured energy spectrum, since both are simultaneous, competing ways the same photon can interact with the detector material.

Pair Production and Annihilation¶

A further, still more dramatic demonstration of the particle nature of light is pair production: a sufficiently energetic photon, passing near a nucleus, can convert entirely into an electron–positron pair,

requiring a photon energy of at least $2m_ec^2 = 1.022\ \text{MeV}$ (twice the electron rest energy) — a direct manifestation of mass–energy equivalence (Chapter 3), converting a massless particle’s energy into the rest mass of two massive particles.

A nearby massive third body — typically an atomic nucleus — is required to conserve momentum. This can be seen directly from the energy–momentum four-vector formalism of Chapter 3: an isolated photon of energy $E_\gamma$ has momentum $p_\gamma = E_\gamma/c$, but a resulting electron–positron pair with the same total energy $E_\gamma$ (by energy conservation) necessarily has total momentum strictly less than $E_\gamma/c$, because each massive particle satisfies $E^2 = (pc)^2+(mc^2)^2 > (pc)^2$, so $pc < E$ for each — the pair’s combined momentum cannot match the photon’s original momentum at the same total energy. A nucleus nearby can absorb the small difference in momentum (recoiling with negligible kinetic energy, because its mass is so much larger than $m_e$) while barely affecting the energy balance, resolving the mismatch; in a vacuum with no such third body available, momentum conservation alone forbids the process outright, regardless of how much energy the lone photon carries.

The reverse process, pair annihilation, $e^- + e^+ \to 2\gamma$ (two photons are required, rather than one, to conserve momentum in the electron–positron center-of-momentum frame — see Chapter 3, Problem 3), converts rest mass entirely back into photon energy and is used, for example, in positron-emission tomography (PET) imaging, where each annihilation of an injected positron-emitting tracer with a nearby atomic electron produces two back-to-back $511\ \text{keV}$ gamma rays that a ring of detectors uses to reconstruct the tracer’s location.

Worked Example: Sharing Energy Above Pair-Production Threshold¶

A photon of energy $E_\gamma = 3.00\ \text{MeV}$, well above the $1.022\ \text{MeV}$ threshold, undergoes pair production near a heavy nucleus. Because the nucleus is far more massive than the electron or positron, it absorbs essentially none of the available kinetic energy (a large mass can supply whatever small momentum balance is needed while carrying away almost no energy, exactly analogous to how a wall barely recoils, and gains almost no kinetic energy, when it reflects a ball). The energy left over after paying the $2m_ec^2$ rest-mass cost is therefore shared, to good approximation, between the electron and positron as kinetic energy:

If the pair shares this energy symmetrically (not required by any conservation law, but a useful simplifying assumption), each particle carries about $0.989\ \text{MeV}$ of kinetic energy — comparable to its own rest energy, so each is produced at a substantial fraction of the speed of light.

The Photon: A Unified Picture¶

Four seemingly unrelated phenomena — blackbody radiation, the photoelectric effect, X-ray production, and Compton scattering — together with pair production and annihilation, are all consequences of a single underlying fact: electromagnetic radiation exchanges energy and momentum with matter in discrete, particle-like quanta, $E=hf$ and $p=h/\lambda$, rather than continuously. Blackbody radiation shows that an oscillator can only emit or absorb energy in these units; the photoelectric effect shows that a photon transfers its entire energy to a single electron in one step; X-ray production shows how an electron’s kinetic energy can be converted into photons, with a sharply bounded maximum photon energy; Compton scattering shows that a photon carries not just energy but momentum, exchanged with a free electron exactly as in a two-body collision; and pair production and annihilation show that a photon’s energy is, via $E=mc^2$, interconvertible with the rest mass of matter itself. The characteristic quantum features of all five are explained, quantitatively and without exception, by the same photon concept. Yet, as Chapters 4–5 demonstrated, light also produces interference and diffraction patterns with no possible explanation in a naive particle picture. Chapter 7 confronts this apparent contradiction directly, and shows that it is resolved not by choosing one description over the other, but by recognizing that matter, too, has a wave nature — and that wave and particle descriptions are two complementary faces of a single, more complete quantum picture.

6.4Summary¶

• The blackbody spectrum cannot be explained by the classical Rayleigh–Jeans law, $u(\lambda,T) = 8\pi k_BT/\lambda^4$, which diverges at short wavelength (the ultraviolet catastrophe); Planck’s hypothesis that oscillator energy is quantized in units of $hf$ yields a radiation law matching observation at all wavelengths and reduces to the Rayleigh–Jeans law at long wavelength.

• Wien’s displacement law, $\lambda_{\max}T = 2.898\times10^{-3}\ \text{m}\cdot\text{K}$, and the Stefan–Boltzmann law, $P/A = \sigma T^4$, follow from Planck’s radiation law and together determine both the color and total radiated power of a thermal source from its temperature alone.

• The photoelectric effect — $K_{\max}$ independent of intensity, a sharp threshold frequency, and instantaneous emission — cannot be explained by classical wave theory but follows directly from Einstein’s photon hypothesis, $E = hf$, via $K_{\max} = hf - \phi$.

• X-ray production (bremsstrahlung) is an energy-bookkeeping inverse of the photoelectric effect: decelerating electrons in an X-ray tube produce a continuous spectrum bounded by a sharp minimum wavelength, the Duane–Hunt limit, $\lambda_{\min} = hc/eV$, since no photon can carry more energy than a single electron’s kinetic energy $eV$.

• Compton scattering — the angle-dependent wavelength shift of X-rays scattered from free electrons, $\lambda' - \lambda = (h/m_ec)(1-\cos\theta)$, derivable directly from conservation of relativistic energy and momentum — shows that photons carry momentum $p = h/\lambda$ and interact with electrons as particles in an elastic collision.

• Pair production ($\gamma \to e^- + e^+$, requiring $hf \ge 2m_ec^2$ and a nearby massive body to conserve momentum) and pair annihilation ($e^-+e^+ \to 2\gamma$) further confirm that photons are quanta of energy and momentum interconvertible with massive particles.

6.5Problems¶

Solution to Exercise 6.2 #

Wien’s law gives $\lambda_{\max}=(2.898\times10^{-3}\ \text{m K})/(2900\ \text{K})=9.99\times10^{-7}\ \text{m}=999\ \text{nm}$, which is infrared. The blackbody flux is

$$\frac PA=\sigma T^4=(5.670\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4})(2900\ \text{K})^4=4.01\times10^6\ \text{W/m}^2.$$

(6.25)

Therefore, the filament peaks near $1.00\ \mu\text{m}$ in the infrared and radiates $4.01\times10^6\ \text{W/m}^2$ ideally; most of its power is invisible infrared rather than useful visible light.

Solution to Exercise 6.5 #

The frequencies are $f_1=c/(450\ \text{nm})=6.67\times10^{14}\ \text{Hz}$ and $f_2=c/(360\ \text{nm})=8.33\times10^{14}\ \text{Hz}$. Subtracting $eV_0=hf-\phi$ for the two data points gives

$$h=\frac{e(1.28-0.65)\ \text{V}}{f_2-f_1}=3.78\times10^{-15}\ \text{eV s}=6.06\times10^{-34}\ \text{J s}.$$

(6.28)

Then $\phi=hf_1-eV_1=(3.78\times10^{-15})(6.67\times10^{14})-0.65=1.87\ \text{eV}$.

Therefore, the data give $h=6.06\times10^{-34}\ \text{J s}$ and a work function of $1.87\ \text{eV}$.

Solution to Exercise 6.9 #

At $90^\circ$, $\Delta\lambda=h/(m_ec)=2.426\ \text{pm}=0.002426\ \text{nm}$. Thus

$$\lambda'=0.0711\ \text{nm}+0.002426\ \text{nm}=0.073526\ \text{nm},$$

(6.31)

$$E'=\frac{1240\ \text{eV nm}}{0.073526\ \text{nm}}=16.86\ \text{keV}.$$

(6.32)

The incident energy is $1240/0.0711=17.44\ \text{keV}$, so the electron receives $17.44-16.86=0.58\ \text{keV}$.

Therefore, the shift is $2.426\ \text{pm}$, the scattered photon has wavelength $0.07353\ \text{nm}$ and energy $16.86\ \text{keV}$, and the electron receives about $0.58\ \text{keV}$.

Solution to Exercise 6.14 #

The rest-energy cost is $2m_ec^2=1.022\ \text{MeV}$, leaving

$$K_\mathrm{available}=2.50\ \text{MeV}-1.022\ \text{MeV}=1.478\ \text{MeV}.$$

(6.39)

Equal sharing gives $K_e=K_{e^+}=0.739\ \text{MeV}$ and total energy $0.739+0.511=1.250\ \text{MeV}$ for each.

Therefore, each particle has $0.739\ \text{MeV}$ kinetic energy and $1.250\ \text{MeV}$ total energy.

Solution to Exercise 6.17 #

For $180^\circ$ scattering, $E'=E/[1+2E/(m_ec^2)]$. Hence

$$E'=\frac{0.662\ \text{MeV}}{1+2(0.662/0.511)}=0.184\ \text{MeV},$$

(6.42)

$$\Delta E_{\max}=0.662\ \text{MeV}-0.184\ \text{MeV}=0.478\ \text{MeV}.$$

(6.43)

Full photoelectric absorption deposits $0.662\ \text{MeV}$.

Therefore, the Compton edge is $0.478\ \text{MeV}$ and the photopeak is $0.662\ \text{MeV}$; both appear because photons can either scatter incompletely or be fully absorbed.