Dual Nature of Matter and Radiation

A blazing red lamp shines on a metal plate. A dim ultraviolet torch shines on the same plate. Which one ejects electrons?

Most say the bright one. More light, more energy.

Only the ultraviolet torch — and it works the instant you switch it on. The red lamp can be made as bright as you like and will never eject a single electron.

The previous chapter established light as a wave. This one shows the description is incomplete. Two facts organise everything:

• Light delivers energy in indivisible packets, and packet size is set by frequency alone. Intensity controls only how many packets arrive per second, never how big each one is. Almost every error in this chapter is a failure to hold that distinction.
• The duality runs both ways. If waves carry momentum in packets, why should particles not have a wavelength? $\lambda =h/p$ applies to everything, and we miss it for everyday objects purely because of scale.

1. The Classical Picture and Why It Failed

Hertz found the effect by accident in 1887, while performing the experiments that confirmed Maxwell's waves: a spark jumped more readily when ultraviolet fell on the electrodes. Lenard mapped the details over the next decade, and every one contradicted wave theory.

Trap. The time-lag failure is the most damning of the four, because it is not a matter of degree. A wave spreads its energy over the whole wavefront, so one electron should wait while its tiny share builds up. That wait never happens.

Illustration 1

A 1 W lamp is 1 m from a metal of work function 2 eV. Taking an electron's collecting area as roughly one square angstrom, how long does classical physics say emission should take?

$I=\frac{P}{4\pi r^2}=\frac{1}{4\pi }=0.080{\text{W m}}^{-2}\Rightarrow P_{atom}=(0.080)(10^{-20})=8.0\times 10^{-22}\text{W}$

$t=\frac{\varphi }{P_{atom}}=\frac{3.2\times 10^{-19}}{8.0\times 10^{-22}}\approx 400\text{s}$

Nearly seven minutes of patient accumulation. The measured delay is under a nanosecond — a discrepancy of eleven orders of magnitude, not a correction. Something was wrong with the picture, not the numbers.

2. Einstein's Photoelectric Equation

Einstein's 1905 proposal: light itself is quantised into packets of energy $h\nu$, and one photon interacts with one electron in an all-or-nothing exchange.

$h\nu =\varphi +K_{max},\varphi =h{\nu }_0$

The work function $\varphi$ is the minimum energy needed to free an electron from the surface — a property of the metal.

All four failures are repaired at once:

Einstein received the Nobel Prize for this rather than for relativity, which says something about how the resolution was regarded.

Illustration 2

Metals A ($\varphi =2.0$ eV) and B ($\varphi =4.0$ eV) are lit by the same 300 nm source at the same intensity. Compare stopping potentials and saturation currents.

$E=\frac{1240}{300}=4.13\text{eV}$

$V_{0A}=4.13-2.0=2.13\text{V},V_{0B}=4.13-4.0=0.13\text{V}$

Same intensity means the same number of photons per second, so both metals emit the same number of electrons and the saturation currents are equal. Only the stopping potentials differ. Intensity fixed the count; the work function fixed the energy.

Illustration 3

Caesium has a work function of 2.14 eV. Will green light at 550 nm eject electrons from it? Will the same light work on zinc, where $\varphi =4.3$ eV?

$E=\frac{1240}{550}=2.25\text{eV}$

For caesium $2.25>2.14$, so emission occurs with $K_{max}=0.11$ eV. For zinc $2.25<4.3$, so nothing is emitted at any intensity whatsoever.

The same comparison runs in wavelength: ${\lambda }_0=1240/\varphi$ gives 579 nm for caesium and 288 nm for zinc, and emission needs $\lambda <{\lambda }_0$. Caesium's threshold falling inside the visible band is exactly why photocells are built from it.

3. Stopping Potential and the Graphs

The apparatus is a photosensitive plate and a collector sealed into an evacuated tube, with light admitted through a quartz window and a variable potential applied across the two electrodes. Making the collector negative pushes the emitted electrons back, and the current falls as that reverse voltage grows.

The stopping potential is the reverse voltage that just turns back the most energetic electrons.

$e{V}_0=K_{max}\Rightarrow V_0=\frac{h}{e}\nu -\frac{\varphi }{e}$

Those two panels are the whole content of the graph questions, and both follow from one line: intensity controls number, frequency controls energy.

Saturation current is proportional to intensity, because once the voltage collects every emitted electron the current is set purely by how many are emitted.

The current does not fall abruptly, because emitted electrons carry a spread of energies from zero up to $K_{max}$. Electrons freed from deeper in the metal lose some energy escaping, so only those from the surface reach $K_{max}$.

Illustration 4

Metal A has a threshold frequency $5\times 10^{14}$ Hz. Metal B has twice the work function. Sketch both $V_0$–$\nu$ lines and give B's threshold.

$\varphi =h{\nu }_0\Rightarrow {\nu }_{0B}=2{\nu }_{0A}=10\times 10^{14}\text{Hz}$

The two lines are parallel: the slope is $h/e$ for every metal alike, and only the intercept moves. Millikan spent a decade on precision measurements meaning to disprove the photon hypothesis, confirmed it instead, and got the best value of $h$ then available.

Illustration 5

A student measures stopping potentials at several frequencies, fits a straight line, and obtains a slope of $4.14\times 10^{-15}$ V s with an intercept on the frequency axis at $5.5\times 10^{14}$ Hz. Find $h$ and the work function.

The slope is $h/e$, so

$h=e(4.14\times 10^{-15})=(1.6\times 10^{-19})(4.14\times 10^{-15})=6.6\times 10^{-34}\text{J s}$

The frequency intercept is ${\nu }_0$ itself, since $V_0=0$ there:

$\varphi =h{\nu }_0=(4.14\times 10^{-15})(5.5\times 10^{14})=2.28\text{eV}$

That second line saves a conversion. With the slope in volt seconds and ${\nu }_0$ in hertz the product already comes out in volts, and a volt of stopping potential is an electronvolt of energy.

4. Photons

$E=h\nu =\frac{hc}{\lambda },p=\frac{h}{\lambda }=\frac{E}{c},n=\frac{P}{h\nu }$

Memorise $hc=1240$ eV nm. A 400 nm photon carries 3.1 eV, and that one line removes most of the arithmetic from this chapter.

Photons have zero rest mass, always travel at $c$, and are undeflected by fields. In the photoelectric effect the photon is absorbed entirely — which is why partial absorption never happens.

A 100 W lamp emits around $10^{20}$ photons per second, so the graininess averages out completely. That is why quantisation stayed invisible for so long.

Illustration 6

A 10 mW laser pointer at 650 nm shines straight onto a mirror. How many photons strike it each second, and what force do they exert?

$E=\frac{1240}{650}=1.91\text{eV}=3.05\times 10^{-19}\text{J}n=\frac{10^{-2}}{3.05\times 10^{-19}}=3.3\times 10^{16}{\text{s}}^{-1}$

Each photon arrives with momentum $E/c$ and leaves with that momentum reversed, so the mirror receives $2E/c$ from every one:

$F=\frac{2P}{c}=\frac{2(10^{-2})}{3\times 10^8}=6.7\times 10^{-11}\text{N}$

Thirty-three thousand million million photons a second, and the push is a hundred-billionth of a newton. On a black surface, which absorbs instead of reflecting, the momentum is not reversed but merely stopped, and the force halves to $P/c$.

Illustration 7

A sodium atom (23 u) at rest emits a 589 nm photon. Find its recoil speed.

Momentum is conserved, and the photon carries $h/\lambda$:

$p=\frac{6.63\times 10^{-34}}{589\times 10^{-9}}=1.13\times 10^{-27}{\text{kg m s}}^{-1}$

$v=\frac{p}{m}=\frac{1.13\times 10^{-27}}{23\times 1.66\times 10^{-27}}\Rightarrow v\approx 3{\text{cm s}}^{-1}$

Three centimetres per second per photon. Fire millions of photons a second at an atom moving toward the beam and you can bring it almost to rest — this is laser cooling, and it is how atoms are held at microkelvin temperatures.

The inverse process: X-rays

Run the effect backwards and a fast electron is stopped, its kinetic energy becoming a photon. The most energetic photon appears when an electron gives up everything in one stop:

${\lambda }_{min}=\frac{hc}{eV}=\frac{1240}{V}\text{nm}$

Trap. This cut-off depends only on the accelerating voltage, never on the target material — the exact mirror of the photoelectric threshold, which depends only on the metal and never on the light. The target sets the sharp characteristic lines superimposed on the spectrum, not where the continuous spectrum ends.

The continuum below the cut-off comes from electrons stopping in several stages, each surrendering part of their energy.

Illustration 8

An X-ray tube runs at 30 kV on a molybdenum target. Find the cut-off wavelength, then say what changes if the voltage is halved and the target replaced by tungsten.

${\lambda }_{min}=\frac{1240}{30000}\text{nm}=0.0413\text{nm}=0.413\text{A˚}$

Halving the voltage doubles the cut-off to 0.826 Å, since ${\lambda }_{min}\propto 1/V$. Swapping the target moves the characteristic lines, which sit at wavelengths fixed by tungsten's inner energy levels, and leaves the cut-off exactly where it was.

One caution on that second half. At 15 kV the tungsten $K$ lines may not appear at all, because an electron must first carry enough energy to knock a $K$-shell electron out. The continuum is always produced; a characteristic line appears only above its own threshold.

Photocells

The photoelectric effect packaged as a device, turning a light signal straight into a current. Automatic street lighting, burglar alarms, automatic doors, camera light meters and the sound track of old film projectors all rely on it. The emitter metal sets what the cell responds to: caesium, with $\varphi \approx 2$ eV, answers to visible light where a higher-work-function metal would ignore it.

5. de Broglie Waves

If radiation has particle properties, de Broglie argued in 1924, matter should have wave properties.

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$

For an electron accelerated through $V$ volts this collapses to a formula worth memorising:

$\lambda =\frac{12.27}{\sqrt{V}}\text{A˚}$

So 150 V gives about 1 Å — comparable to atomic spacing, which is precisely why electron diffraction from crystals is observable. Davisson and Germer saw exactly that from a nickel crystal in 1927, with the measured wavelength matching de Broglie's prediction.

The relation is universal. A 150 g cricket ball at 30 m s⁻¹ has $\lambda \approx 10^{-34}$ m, twenty orders of magnitude below a nucleus. No experiment could ever detect it — which is why the wave nature of matter went unnoticed for so long.

For a particle in thermal equilibrium, $\lambda =h/\sqrt{3m{k}_{B}T}$.

Illustration 9

Find the de Broglie wavelength of a nitrogen molecule (28 u) at 300 K, and compare it with the average spacing between molecules at STP.

$\lambda =\frac{h}{\sqrt{3m{k}_{B}T}}=\frac{6.63\times 10^{-34}}{\sqrt{3(4.65\times 10^{-26})(1.38\times 10^{-23})(300)}}\Rightarrow \lambda \approx 0.28\text{A˚}$

Spacing at STP is $(22.4\times 10^{-3}/N_A{)}^{1/3}\approx 33$ Å — over a hundred times larger.

The wave packets never overlap, so the molecules behave as independent particles and classical kinetic theory works. Cool the gas far enough, or squeeze it hard enough, and $\lambda$ grows until packets do overlap: that is where a Bose-Einstein condensate forms and the classical picture collapses.

Illustration 10

Light of 200 nm falls on a metal of work function 4.0 eV. Find the maximum kinetic energy of the emitted electrons and their de Broglie wavelength.

$E=\frac{1240}{200}=6.2\text{eV}\Rightarrow K_{max}=6.2-4.0=2.2\text{eV}$

Since $K_{max}=e{V}_0$ with $V_0=2.2$ V, the accelerated-electron form applies directly:

$\lambda =\frac{12.27}{\sqrt{2.2}}=8.3\text{A˚}=0.83\text{nm}$

The freed electron's wavelength is about 240 times shorter than that of the light which freed it. That one comparison is the whole case for electron microscopy: resolution is limited by wavelength, and a modest voltage buys a wavelength no lamp can match.

Illustration 11

A photon and an electron each carry 1 keV. Compare their wavelengths.

For the photon, $hc=1240$ eV nm is used directly:

${\lambda }_{photon}=\frac{1240}{1000}=1.24\text{nm}$

For the electron, 1 keV of kinetic energy is what 1000 V delivers:

${\lambda }_{electron}=\frac{12.27}{\sqrt{1000}}=0.388\text{A˚}=0.0388\text{nm}$

$\frac{{\lambda }_{photon}}{{\lambda }_{electron}}\approx 32$

Set this beside the reverse question — equal wavelength gives wildly unequal energy — and the reason is the same in both. A massless photon obeys $E=pc$; a slow massive particle obeys $E=p^2/2m$. Matching one quantity never matches the other, and knowing which relation belongs to which object is the entire trick.

6. Seeing the Waves: Davisson-Germer and Bohr's Orbits

A wavelength nobody can detect is a wavelength anybody can doubt. Two things settled the question — one an experiment, the other a piece of arithmetic that had been sitting unexplained for a decade.

Davisson and Germer fired electrons at a nickel crystal in 1927 and counted how many scattered in each direction. Classically the intensity should slide smoothly down with angle. Instead a distinct bump appeared at one particular angle, shifted as the accelerating voltage changed, and behaved in every respect like a diffraction maximum.

Illustration 12

At 54 V the scattered intensity peaks 50 degrees from the incident beam. The relevant nickel planes are spaced 0.91 Å apart and are inclined so that the Bragg angle is 65 degrees. Check that the two routes to the wavelength agree.

From de Broglie, through the accelerating voltage:

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

From Bragg's condition at first order:

$\lambda =2d\sin \theta =2(0.91)\sin 65^{\circ }=1.65\text{A˚}$

The two agree to about one part in a hundred, and they share no common assumption — one comes from a voltmeter, the other from a protractor and a known crystal spacing. That independence is what made the result decisive rather than suggestive.

The second confirmation was older than the experiment. Bohr had assumed, with no justification offered, that angular momentum arrives in multiples of $h/2\pi$. de Broglie's relation supplies the missing reason.

Illustration 13

Show that requiring the electron's wave to close on itself around a circular orbit reproduces Bohr's quantisation rule.

The wave must join up smoothly after one circuit, so the circumference has to hold a whole number of wavelengths:

$2\pi r=n\lambda =\frac{nh}{p}=\frac{nh}{mv}$

Rearranging gives

$mvr=\frac{nh}{2\pi }$

which is Bohr's postulate exactly. Leave a fraction of a wavelength over and the wave meets itself out of step on the next circuit, cancelling by destructive interference, so no such orbit survives. An assumption invented to fit the hydrogen spectrum turned out to be a statement about standing waves.

7. What Duality Actually Means

Trap. Duality is not a compromise, with light "partly a wave and partly a particle". Light is neither. Wave and particle are two models drawn from everyday experience, and there is no reason a photon should have to fit either.

The two descriptions are never needed simultaneously — Bohr called this complementarity. No single experiment forces both at once.

The double slit run with single electrons makes it concrete. Each electron arrives as one localised dot: particle. After many thousands the dots build an interference pattern: wave. Only one electron is in the apparatus at a time, so each interferes with itself. Try to detect which slit it took, however gently, and the pattern vanishes. The two descriptions are mutually exclusive, not merely alternative.

Summary

• Four classical predictions fail; one idea — light quantised into packets of $h\nu$ — repairs all four.
• Intensity controls the number of photons, never the energy of each. Almost every question tests this.
• $h\nu =\varphi +K_{max}$ with $\varphi =h{\nu }_0$: below threshold, nothing at any intensity.
• No time lag, because nothing accumulates — classically the wait would be minutes; the measured delay is under a nanosecond.
• $e{V}_0=K_{max}$; the $V_0$–$\nu$ line has slope $h/e$ for every metal, differing only in intercept.
• Varying intensity changes the saturation current but not $V_0$; varying frequency does the reverse.
• $E=hc/\lambda$, $p=h/\lambda$, $n=P/h\nu$. Zero rest mass, undeflected by fields, absorbed whole.
• Use $hc=1240$ eV nm.
• X-ray cut-off ${\lambda }_{min}=1240/V$ nm depends only on the voltage, never the target — the mirror of the photoelectric threshold.
• $\lambda =h/p$ is universal; $\lambda =12.27/\sqrt{V}$ Å for an accelerated electron, so 150 V gives about 1 Å.
• At fixed voltage heavier particles have shorter wavelengths, going as $1/\sqrt{m}$.
• A cricket ball's wavelength is $\sim 10^{-34}$ m, which is why matter waves went unnoticed.
• Equal wavelength means equal momentum but never equal energy; equal energy never means equal wavelength.
• Duality is not a compromise. The two descriptions are complementary and never needed at the same instant.