Dual Nature of Matter and Radiation

Shine a very dim red lamp on a caesium surface for an hour. Then shine an extremely faint blue flash on it for a nanosecond. Which one ejects electrons?

The blue flash — instantly. The red lamp, never, however long you wait.

Classical wave theory predicts precisely the opposite. A wave spreads its energy continuously over the surface, so a faint source should simply take longer to accumulate the escape energy. Estimate the lag: for a lamp delivering $10^{-10}$ W m${}^{-2}$, an atom of cross-section $10^{-20}$ m${}^2$ collects energy at $10^{-30}$ W, and a work function of $2$ eV — about $3\times 10^{-19}$ J — would take

$t\approx \frac{3\times 10^{-19}}{10^{-30}}=3\times 10^{11}\text{s}$

roughly ten thousand years. The measured delay is under a nanosecond.

One assumption fixes everything. Light arrives in indivisible packets of energy $h\nu$. A packet either has enough energy to free an electron or it does not, and no amount of waiting lets two red packets combine.

1. Four failures, and what each one rules out

The photoelectric effect defeats classical theory in four separate ways, and Advanced questions test each of them individually.

There is a threshold frequency. Below it, no emission occurs at any intensity. A wave theory has no mechanism for this at all, since intensity and frequency are independent.

Emission is instantaneous. The delay is under $10^{-9}$ s even for the faintest source, against the classical prediction of hours or years.

Intensity controls the current, not the energy. Doubling the brightness doubles the number of electrons but leaves the maximum kinetic energy exactly unchanged.

Maximum kinetic energy is linear in frequency, with a slope that turns out to be the same constant $h$ for every metal.

Each of these follows immediately if light is quantised, and none of them follows from a wave picture.

Illustration 1

Light of wavelength $400$ nm falls on a metal of work function $2.5$ eV. Find whether emission occurs, and the maximum kinetic energy if it does.

Photon energy $=\frac{1240}{400}=3.10$ eV, using $hc=1240$ eV nm.

Since $3.10>2.5$, emission occurs.

$K_{max}=3.10-2.50=0.60$ eV

Keep $hc=1240$ eV nm memorised. It converts wavelength in nanometres to photon energy in electronvolts in one division and saves a chain of SI conversions in every question of this type.

2. Einstein's equation and the four graphs

$h\nu =\varphi +K_{max},e{V}_0=h\nu -\varphi ,{\lambda }_0=\frac{hc}{\varphi }$

where $V_0$ is the stopping potential — the reverse voltage that just prevents the fastest electron from reaching the collector. Four graphs carry most of the marks, and each slope means something specific.

The plot of stopping potential against frequency is the most informative of all. It is a straight line

$V_0=\frac{h}{e}\nu -\frac{\varphi }{e}$

whose slope is $h/e$ for every metal, and whose intercepts differ. That universality is the strongest single piece of evidence for the photon, since it says one constant governs all materials.

Illustration 2

Stopping potentials of $1.85$ V and $0.82$ V are measured at wavelengths $300$ nm and $400$ nm. Find $h$ and the work function.

$E_1=\frac{1240}{300}=4.13$ eV, $E_2=\frac{1240}{400}=3.10$ eV

$\Delta E=1.03$ eV for $\Delta V_0=1.03$ V, confirming $e{V}_0=h\nu -\varphi$ with unit slope in electronvolts.

$\varphi =4.13-1.85=2.28$ eV, and the same from the second pair: $3.10-0.82=2.28$ eV.

Two measurements determine both constants, which is exactly how Millikan measured $h$ — reluctantly, having set out to disprove Einstein's equation.

Illustration 3

A metal has threshold wavelength $600$ nm. Find the stopping potential for incident light of $400$ nm.

$\varphi =\frac{1240}{600}=2.07$ eV

$E=\frac{1240}{400}=3.10$ eV

$V_0=3.10-2.07=1.03$ V

Doubling the frequency does not double $V_0$. The work function is subtracted first, so the stopping potential grows faster than proportionally — which is exactly what makes the graph a line with a negative intercept rather than one through the origin.

3. Photons: counting them, and their momentum

A source of power $P$ at wavelength $\lambda$ emits

$N=\frac{P}{h\nu }=\frac{P\lambda }{hc}\text{photons per second}$

Each carries momentum

$p=\frac{h}{\lambda }=\frac{E}{c}$

which is what produces radiation pressure. The enormous photon numbers from ordinary sources are why the granularity is invisible in everyday life.

Illustration 4

A $100$ W sodium lamp emits at $590$ nm with $10\%$ efficiency. Find the number of photons emitted per second.

Photon energy $=\frac{1240}{590}=2.10$ eV $=3.36\times 10^{-19}$ J

Radiated power $=10$ W

$N=\frac{10}{3.36\times 10^{-19}}=3.0\times 10^{19}$ per second

Thirty billion billion per second. Detecting a single one requires a photomultiplier precisely because the ordinary flux is so overwhelming that individual arrivals are impossible to distinguish.

Illustration 5

Find the momentum of a $500$ nm photon and compare it with that of an electron of the same wavelength.

$p=\frac{h}{\lambda }=\frac{6.63\times 10^{-34}}{5\times 10^{-7}}=1.33\times 10^{-27}$ kg m s${}^{-1}$

The electron has the same momentum, since $p=h/\lambda$ applies to both.

Their energies differ enormously: the photon has $E=pc=2.48$ eV, while the electron has $E=\frac{p^2}{2m}=6\times 10^{-6}$ eV.

Same wavelength does not mean same energy. For a photon $E\propto 1/\lambda$; for a slow particle $E\propto 1/{\lambda }^2$, which is why matter waves of useful wavelength need so little accelerating voltage.

4. de Broglie: a wavelength for matter

If radiation carries momentum $h/\lambda$, de Broglie proposed the converse — that a particle of momentum $p$ has a wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}$

For a charge $q$ accelerated through a potential $V$, $p=\sqrt{2mqV}$, so

$\lambda =\frac{h}{\sqrt{2mqV}},{\lambda }_{electron}\left(\text{A˚}\right)=\frac{12.27}{\sqrt{V\left(\text{volts}\right)}}$

and for a particle in thermal equilibrium, $\lambda =\frac{h}{\sqrt{3mkT}}$.

The reason matter waves went unnoticed for so long is arithmetic. A cricket ball has a wavelength around $10^{-34}$ m — smaller than any aperture that exists — while a $100$ V electron has one of about $1.2$ Å, comfortably the spacing between atoms in a crystal.

Illustration 6

Find the de Broglie wavelength of an electron accelerated through $150$ V, and of a $150$ g ball moving at $30$ m s${}^{-1}$.

Electron: $\lambda =\frac{12.27}{\sqrt{150}}=\frac{12.27}{12.25}=1.00$ Å

Ball: $\lambda =\frac{6.63\times 10^{-34}}{0.15\times 30}=1.5\times 10^{-34}$ m

Twenty-four orders of magnitude apart. No aperture remotely that small exists, so the ball's wave nature is not merely hard to detect — it is unobservable in principle with any conceivable apparatus.

Illustration 7

A proton and an electron are accelerated through the same potential difference. Compare their de Broglie wavelengths.

$\lambda \propto \frac{1}{\sqrt{m}}$ at fixed $V$ and $q$.

$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{1836}=42.8$

The electron's wavelength is about forty-three times longer. This is exactly why electron microscopes rather than proton microscopes are built: for the same voltage the heavier particle gives a shorter wavelength, but it is far harder to produce and steer.

Illustration 8

Find the de Broglie wavelength of a neutron in thermal equilibrium at $300$ K, given $m=1.67\times 10^{-27}$ kg.

$p=\sqrt{3mkT}=\sqrt{3\left(1.67\times 10^{-27}\right)\left(1.38\times 10^{-23}\right)(300)}$

$=\sqrt{2.07\times 10^{-47}}=4.55\times 10^{-24}$ kg m s${}^{-1}$

$\lambda =\frac{6.63\times 10^{-34}}{4.55\times 10^{-24}}=1.46$ Å

Which is why thermal neutrons diffract from crystals. Neutron diffraction is a standard structural technique precisely because room temperature happens to give a wavelength matched to atomic spacing.

5. Davisson and Germer: the wave made visible

Electrons of $54$ eV were fired at a nickel crystal, and the scattered intensity peaked sharply at $50^{\circ }$ — a diffraction maximum, not the smooth spread a particle beam would give.

Bragg's law with the nickel spacing $d=0.91$ Å gives $\lambda =1.65$ Å. The de Broglie prediction is

$\lambda =\frac{12.27}{\sqrt{54}}=\frac{12.27}{7.35}=1.67\text{A˚}$

The agreement to within about one per cent settled the matter. Electrons diffract, and the wavelength is exactly the one de Broglie's relation predicts.

The practical payoff is the electron microscope. At $100$ kV the electron wavelength is around $0.04$ Å, tens of thousands of times shorter than visible light, and since the resolution limit is proportional to wavelength, the gain in detail is of the same order.

Illustration 9

Find the accelerating voltage needed to give electrons a de Broglie wavelength of $0.5$ Å.

$\lambda =\frac{12.27}{\sqrt{V}}\Rightarrow \sqrt{V}=\frac{12.27}{0.5}=24.5$

$V=602$ V

Well within an ordinary laboratory supply, which is why electron diffraction became a routine technique almost immediately after it was discovered.

6. X-rays: the photoelectric effect run backwards

Fire fast electrons at a heavy target and the process reverses: kinetic energy becomes photons. An electron decelerating in the target can give up any fraction of its energy, so the spectrum is continuous — but it stops abruptly at a shortest wavelength, because no photon can carry more energy than the electron brought:

$h{\nu }_{max}=eV\Rightarrow {\lambda }_{min}=\frac{hc}{eV}=\frac{12400}{V\left(\text{volts}\right)}\text{A˚}$

This Duane-Hunt limit depends only on the accelerating voltage and not at all on the target material, which is precisely what a quantised picture demands and a classical one cannot explain.

Superimposed on that continuous background are sharp characteristic lines, produced when an inner-shell vacancy is filled. These do depend on the target, through Moseley's law $\sqrt{\nu }=a\left(Z-b\right)$, and they are how the atomic number of an element is measured directly.

Illustration 10

An X-ray tube operates at $30$ kV. Find the shortest wavelength emitted, and state what changes if the tungsten target is replaced by molybdenum.

${\lambda }_{min}=\frac{12400}{30000}=0.413$ Å

Replacing the target leaves ${\lambda }_{min}$ exactly unchanged, since it depends only on $V$.

What does change is the set of characteristic lines, which shift to longer wavelengths for the lighter element in accordance with Moseley's law.

Illustration 11

The cut-off wavelength of an X-ray tube is $0.62$ Å. Find the operating voltage and the maximum photon energy in keV.

$V=\frac{12400}{0.62}=20000$ V $=20$ kV

$E_{max}=eV=20$ keV

The cut-off is a direct voltmeter. Measuring the shortest wavelength in the spectrum gives the tube voltage without any electrical measurement at all, which is one of the cleanest confirmations that $E=h\nu$.

7. When does something behave as a wave?

The single criterion is whether the de Broglie wavelength is comparable with the aperture or spacing the particle encounters. Far smaller, and the particle travels in straight lines; comparable, and it diffracts.

This is not a property of the object but of the experiment. The same electron behaves as a particle in a cathode-ray tube, where every relevant dimension is enormous compared with its wavelength, and as a wave at a crystal, where the spacing matches.

The deepest statement of the duality comes from dimming a double-slit source until photons arrive one at a time. Each arrival is a single point-like flash on the detector — unmistakably a particle. Yet let the flashes accumulate over hours and they build the full interference pattern, fringe for fringe.

Each photon therefore interferes with itself, and the wave does not describe where the photon is but how likely it is to arrive there. The same experiment has since been performed with electrons, neutrons and whole molecules, always with the same result.

Illustration 12

A beam of electrons of wavelength $1$ Å passes through a slit of width $1\mu$m. Estimate the angular spread, and comment.

$\theta \approx \frac{\lambda }{a}=\frac{10^{-10}}{10^{-6}}=10^{-4}$ rad

Over a screen $1$ m away this is a spread of only $0.1$ mm.

Which is why the beam looks like a straight line of particles. Shrink the slit to $10$ Å and the spread becomes $0.1$ rad, unmistakably wavelike — the same electrons, a different experiment.

Illustration 13

If an electron and a photon have the same energy of $100$ eV, which has the longer wavelength?

Photon: $\lambda =\frac{1240}{100}=12.4$ nm

Electron: $\lambda =\frac{12.27}{\sqrt{100}}=1.23$ Å $=0.123$ nm

The photon's wavelength is a hundred times longer. This is the whole basis of electron microscopy: at the same energy, matter waves are far shorter than light waves and therefore resolve far finer detail.

Summary

• Classical waves predict a lag of years for faint light; the measured photoelectric delay is under $10^{-9}$ s.
• Four failures: a threshold frequency, instantaneous emission, intensity setting current not energy, and $K_{max}$ linear in $\nu$.
• $h\nu =\varphi +K_{max}$, $e{V}_0=h\nu -\varphi$, ${\lambda }_0=\frac{hc}{\varphi }$.
• Use $hc=1240$ eV nm: photon energy in eV is $1240$ divided by wavelength in nm.
• $V_0$ against $\nu$ is a straight line of slope $h/e$ — the same slope for every metal — with intercept $-\varphi /e$.
• More intensity raises the saturation current only; the stopping potential is untouched.
• Photon flux $N=\frac{P\lambda }{hc}$; momentum $p=\frac{h}{\lambda }=\frac{E}{c}$.
• Same wavelength does not mean same energy: $E\propto 1/\lambda$ for a photon but $1/{\lambda }^2$ for a slow particle.
• $\lambda =\frac{h}{p}$; accelerated through $V$, $\lambda =\frac{h}{\sqrt{2mqV}}$, and for electrons $\lambda \left(\text{A˚}\right)=\frac{12.27}{\sqrt{V}}$.
• Thermal particle: $\lambda =\frac{h}{\sqrt{3mkT}}$ — which puts room-temperature neutrons at about $1.5$ Å.
• At fixed $V$, $\lambda \propto 1/\sqrt{m}$: an electron's wavelength is about $43$ times a proton's.
• Davisson-Germer: $54$ eV electrons peaked at $50^{\circ }$, giving $1.65$ Å by Bragg against $1.67$ Å by de Broglie.
• X-rays reverse the effect: ${\lambda }_{min}=\frac{12400}{V}$ Å, the Duane-Hunt limit, independent of the target material.
• Characteristic lines do depend on the target, via Moseley's law $\sqrt{\nu }=a\left(Z-b\right)$.
• One photon at a time still builds the interference pattern: each interferes with itself, and the wave gives probability, not position.
• Wave behaviour appears only when $\lambda$ is comparable with the aperture — a property of the experiment, not of the object.