Chapter 01 · Modern Physics · Full book

Dual Nature of Radiation and Matter

The previous chapter closed a long argument. Interference, diffraction, the resolving limit of a telescope, the polarisation of a reflected glare—every one of them demanded that light be a wave, a continuous disturbance spread across a wavefront, and every one of them was confirmed to exquisite precision. By the close of the nineteenth century the wave theory of light was not a hypothesis but a settled fact, sealed by Maxwell's discovery that light simply is an electromagnetic wave travelling at \(c\). The matter was closed.

This chapter reopens it.

It reopens it because a different set of experiments—experiments about what happens when light strikes a metal, and what happens when a beam of electrons strikes a crystal—refused to fit the wave picture at all. The trouble was not that the wave theory was slightly off; it was that the wave theory made confident predictions and nature did the opposite. Shine a faint blue light on a clean metal and electrons fly off the instant the light arrives; shine an intense red light on the same metal and nothing comes off, however long you wait, however bright the lamp. A wave cannot explain that. Something about light, when it is absorbed, behaves as though it arrives not as a smooth swell but in hard, indivisible lumps.

And then, in one of the boldest acts of imagination in the history of physics, a young French aristocrat asked the reverse question. If light, which everyone knew was a wave, can act like a particle, might matter—an electron, which everyone knew was a particle—act like a wave? He wrote down a wavelength for the electron, a single equation with no experiment behind it, purely from a sense that nature ought to be symmetric. Three years later electrons were diffracted by a crystal exactly as the equation foretold.

So this chapter is about a double surprise: radiation has a particle side, and matter has a wave side. Neither replaces what came before. Light is still a wave when it interferes; it is a stream of particles when it is absorbed one lump at a time. The electron is still a particle when it leaves a track in a chamber; it is a wave when it threads a crystal. The deep lesson—one we will only be able to state, not yet resolve—is that "wave" and "particle" are two faces of a single thing, and which face you see is decided by the question you ask of it.

We build the particle side of light first (the photoelectric effect and the photon), then the wave side of matter (de Broglie's hypothesis and its confirmation). The picture is for everyone (NEET, AP Physics 2); the full calculus of photon momentum, radiation pressure, and confinement estimates is for those going further (JEE Advanced, the Olympiads). Both describe the same strange, beautiful fact.

Core intuition: Chapter Roadmap

Electron emission and the work function — An electron sits in a metal inside an energy well. The minimum energy needed to lift it out through the surface is the work function \(\phi_0\)—a property of the metal, not of any atom. Four ways to supply that energy: heat, light, strong fields, and impact.
The photoelectric effect — Light ejects electrons. Four experimental laws emerge, and three of them flatly contradict the wave theory: a threshold frequency, a stopping potential set by colour not brightness, and instantaneous emission.
Einstein's photon — Light is absorbed in indivisible quanta of energy \(h\nu\). One photon, one electron. This single idea turns every contradiction into a prediction: \(KE_{\max}=h\nu-\phi_0\), and the stopping-potential line \(V_0=(h/e)\nu-\phi_0/e\) whose slope is the universal constant \(h/e\).
The photon as a particle — It carries energy \(E=h\nu\) and momentum \(p=h/\lambda\), despite having no mass. Counting photons explains the photocurrent; their momentum explains radiation pressure and the solar sail.
Wave–particle duality of light — The synthesis: the same light is wave and particle, and the two never appear in the same measurement.
de Broglie matter waves — The reverse symmetry: a particle of momentum \(p\) has a wavelength \(\lambda=h/p\). An electron accelerated through \(V\) volts has \(\lambda=1.227/\sqrt{V}\) nm—and a cricket ball has a wavelength so absurdly small it can never be seen.
Davisson–Germer — The experiment that caught the matter wave: slow electrons scattered off a nickel crystal peak at exactly the angle de Broglie's wavelength demands.

Perplexing Questions

Perplexing questions

Before any formula, sit with these six puzzles. Each isolates one crack in the wave theory of light, or one consequence of the strange symmetry between wave and particle. You will meet the explicit resolution at the point in the chapter where the relevant idea is built—look for the [Resolution: PQ N] callout.

Colour decides, brightness does not. Shine a dim violet lamp on a clean metal plate and electrons leap off the instant the light arrives. Now shine a blazing red searchlight on the same plate—a thousand times brighter—and wait an hour. Not one electron emerges. A wave carries energy in proportion to its intensity, so the searchlight pours far more energy onto the plate than the dim violet lamp. Why, then, does the feeble violet light succeed where the brilliant red light utterly fails?
The same light, two natures. In the previous chapter we proved beyond doubt that light is a wave: it bends round corners, it cancels itself in dark fringes, it spreads across a wavefront. In this chapter we will find that the very same light, when it strikes a metal, arrives in countable lumps and delivers a sharp kick of momentum like a tiny bullet. How can one thing be both a wave spread over space and a particle localised at a point?
Two knobs that do not interfere. Turn up the brightness of the light and more electrons come out—but each one emerges no faster than before. Turn up the frequency (make it bluer) and each electron comes out faster—but no more of them appear. Brightness controls how many; frequency controls how fast; and the two knobs never trade places. In a wave, energy and intensity are the same thing—so why do these two controls act on entirely separate features of the emission?
Energy with no time to gather. A clean wave theory says a single electron, soaking up energy from a faint beam spread over the whole plate, would need many seconds—sometimes minutes—to accumulate enough to escape. Yet experiment finds that emission begins within a nanosecond of the light arriving, even for the faintest beam that works at all. Where does a single electron find its escape energy all at once, with no time to gather it?
If a wave can be a particle, can a particle be a wave? Light, the textbook wave, turns out to have a particle side. Symmetry whispers the reverse: perhaps the electron, the textbook particle, has a wave side. But what would the "wavelength" of an electron even mean—and if electrons really are waves, why has no one ever seen a cricket ball diffract through a doorway?
Why slow electrons, and why a crystal? When the wave nature of matter was finally caught in the laboratory, it was caught using slow electrons—a mere \(54\) volts of accelerating push—scattered off the face of a nickel crystal. Fast electrons would not do, and a smooth metal surface would not do; it had to be slow electrons and a crystal lattice. Why should the speed of the electron and the orderliness of the target matter so much?

Every one of these is answered explicitly before the chapter ends.
Look for the [Resolution: PQ N] callout as you read.

Electron Emission and the Work Function

A piece of metal is a sea of electrons. The outermost electron of each atom is only loosely held; in a metal these loose electrons detach from their parent atoms and roam freely through the whole lattice—they are the same "free electrons" that carry current in Chapter 4. Free to move inside the metal, yes. But free to leave it—no. At ordinary temperatures the electrons stay confined within the boundary of the metal, held back at the surface by an attractive pull. If an electron tries to escape, the positive lattice it leaves behind tugs it straight back.

To picture this, think of the whole interior of the metal as the floor of a shallow pit and the world outside as ground level. An electron rolling around on the floor of the pit cannot get out unless something gives it enough energy to climb the wall. The height of that wall, expressed as an energy, is the central quantity of this section.

Core intuition: The Work Function Is the Height of the Wall

The work function \(\phi_0\) (also written \(W_0\)) of a metal is the minimum energy that must be supplied to free the least tightly bound electron from the metal surface—to lift it from the top of the electron sea just barely out into the vacuum, with no kinetic energy to spare. It is measured in joules, but almost always quoted in electron-volts: recall \(1\,\text{eV}=1.602\times10^{-19}\,\text{J}\), the energy an electron gains falling through one volt. Typical metals have \(\phi_0\) between about \(2\) and \(6\,\text{eV}\): caesium is low (\(2.14\,\text{eV}\)), platinum is high (\(5.65\,\text{eV}\)). The work function is the single most important number about a metal's surface for everything that follows.

Misconception alert: The Work Function Is Not an Atom's Ionisation Energy

It is tempting to read \(\phi_0\) as "the energy to ionise an atom of the metal." It is not. Ionisation energy is the energy to tear an electron off an isolated atom (for sodium, \(5.1\,\text{eV}\)). The work function is the energy to free an electron from the bulk metal through its surface (for sodium, only \(2.75\,\text{eV}\)). The two differ because in the solid the electron is already half-free—it belongs to the whole lattice, not to one atom—and because the surface itself, with its layer of escaped charge, sets up the barrier. \(\phi_0\) is a property of a surface, so it even changes if you coat or contaminate the metal. Never substitute an ionisation energy for a work function.

There are four standard ways to hand an electron the energy it needs to climb the wall, and the whole of this chapter is really about the second:

Thermionic emission. Heat the metal. Thermal agitation gives some electrons enough energy to spill over the wall. This is how the cathode of an old vacuum tube or an X-ray machine boils off electrons.
Photoelectric emission. Shine light on the metal. The light delivers the energy. This is our subject.
Field emission. Apply a fierce electric field (\(\sim10^{8}\,\text{V m}^{-1}\)) at the surface. The field effectively thins the wall until electrons leak through.
Secondary emission. Bombard the surface with fast particles, which knock electrons out by impact.

In every case the wall—the work function—is the same; only the source of the climbing energy differs. We now watch what happens when that source is light.

✓ Checkpoint · Electron Emission and the Work Function

Name the emission process that “boils off” electrons from a heated filament in a vacuum tube.
Thermionic emission.
A metal has \(\phi_0=3.2\,\mathrm{eV}\). Express the work function in joules.
\(3.2\times1.60\times10^{-19}=5.1\times10^{-19}\,\mathrm{J}\).
Two metals have \(\phi_0=2.1\,\mathrm{eV}\) and \(4.5\,\mathrm{eV}\). Which holds its electrons more tightly, and which responds to longer-wavelength light?
The \(4.5\,\mathrm{eV}\) metal holds electrons more tightly; the \(2.1\,\mathrm{eV}\) metal (lower \(\phi_0\), longer \(\lambda_0\)) responds to longer wavelengths.

The Photoelectric Effect: What the Experiments Show

◈ Photocell & V–I curve

live bench

Vary frequency and intensity and watch the photocurrent–voltage curve: brightness lifts the saturation current, but the stopping potential is set by colour alone.

◈ loading the interactive bench…

↗ open in the Simulations page

In 1887 Heinrich Hertz—ironically, while confirming Maxwell's wave theory of light—noticed that a spark jumped more readily between two metal balls when ultraviolet light fell on them. Hallwachs and Lenard pinned the effect down over the next two decades: ultraviolet light falling on a clean metal causes electrons to be emitted. The emitted electrons are called photoelectrons; the current they constitute is the photocurrent.

The apparatus

The experiment is done in an evacuated tube. Light of a chosen frequency and intensity falls on a metal plate, the emitter (cathode). A second electrode, the collector (anode), faces it. A battery with a sliding contact lets us put any potential difference—of either sign—between collector and emitter, and a sensitive microammeter reads the photocurrent. By turning two knobs (the light's intensity and frequency) and watching two readings (the current and the voltage needed to stop it), the whole effect is mapped.

◇ Figure 1

Fig 1. Schematic of the photoelectric apparatus. Light of frequency \(\nu\) ejects photoelectrons from the emitter; a variable potential difference (of either sign) between the plates controls their collection, and a microammeter reads the photocurrent.

Two measurements matter. First, with the collector made positive, it sweeps up every emitted electron; as we raise the positive voltage the current climbs and then levels off at a saturation current—every electron emitted is now being collected, and there are no more to gather. Second, with the collector made negative, it repels the electrons; only those launched with enough kinetic energy reach it. Make the collector negative enough and even the fastest photoelectron is turned back just short of the collector. That critical reverse voltage is the stopping potential \(V_0\), and it is the experimental handle on the maximum kinetic energy of the emitted electrons: \[ eV_0 = KE_{\max} = \tfrac{1}{2}mv_{\max}^2 . \] Everything the photoelectric effect has to teach is contained in how the saturation current and the stopping potential depend on the two knobs. There are four findings.

The four experimental laws

Saturation current \(\propto\) intensity. For a fixed frequency, doubling the intensity of the light exactly doubles the saturation current—twice as many electrons per second. Intensity controls how many.
Stopping potential is independent of intensity. For a fixed frequency, changing the intensity does not change \(V_0\). Bright or dim, the fastest electrons come off with the same maximum energy. So intensity does not control how fast.
Stopping potential rises with frequency; a threshold frequency exists. Raise the frequency of the light and \(V_0\) rises in a straight line. Lower the frequency and \(V_0\) falls—until, at a certain threshold frequency \(\nu_0\), it reaches zero. Below \(\nu_0\), no electrons are emitted at all, no matter how intense the light or how long you wait. Frequency controls how fast, and there is a frequency floor below which nothing happens.
Emission is instantaneous. Electrons appear within about \(10^{-9}\,\text{s}\) of the light striking the surface, even for the faintest light that works at all. There is no measurable time lag.

◇ Figure 2

Fig 2. Photocurrent versus collector voltage at three intensities (same frequency). Brighter light raises the saturation current in proportion to intensity, but the stopping potential \(-V_0\) is shared—intensity sets how many electrons, not how fast.

◇ Figure 3

Fig 3. Photocurrent versus collector voltage at three frequencies (same intensity). The saturation current is identical, but a higher frequency demands a larger stopping potential (\(-V_{03}\) for the bluest light)—frequency sets how fast, not how many.

These four facts are simple to state and were beautifully established by 1902. The drama is that, taken together, they demolish the wave theory of light.

Misconception alert: Brighter Light Does Not Make Faster Electrons

The most common and most costly error in this whole topic is to think that more intense light gives the photoelectrons more energy. It does not. For a fixed frequency, increasing the intensity increases only the number of electrons emitted per second (the saturation current); the maximum kinetic energy, and hence the stopping potential, stays exactly the same. To make the electrons faster you must raise the frequency, not the brightness. Burn this asymmetry into memory: intensity \(\to\) how many; frequency \(\to\) how fast. An examiner will offer you a "brighter light, so larger \(V_0\)" option in nearly every photoelectric question; it is always wrong.

Why the Wave Picture Fails

A wave carries energy spread continuously over its wavefront, and the energy it delivers per second per unit area is its intensity. From this one idea the wave theory makes three confident predictions—and experiment contradicts all three.

Wave theory says: brighter light should eject faster electrons. If the energy arrives as a continuous wave, a more intense wave deposits energy faster, so an electron should be shaken loose with more kinetic energy. Therefore \(KE_{\max}\) ought to rise with intensity. Experiment (Law 2): \(KE_{\max}\) does not depend on intensity at all.
Wave theory says: there should be no threshold frequency. Light of any frequency, if intense enough or shone long enough, should eventually pour enough energy into an electron to free it. There should be no frequency below which emission is forbidden. Experiment (Law 3): below \(\nu_0\) nothing is emitted, however intense the light.
Wave theory says: faint light should show a long time lag. A very weak wave deposits energy slowly. A simple calculation (an electron presenting a tiny target area, soaking up a faint, spread-out beam) gives a build-up time of seconds to minutes before any single electron could accumulate \(\phi_0\). Experiment (Law 4): emission is immediate, within a nanosecond.

Core intuition: [Resolution: PQ 1, PQ 3, PQ 4] One Idea Fixes All Three

The three failures share a single root: the wave theory insists that the energy an electron receives depends on how much light there is (intensity) and how long it shines, never on the light's colour. But experiment says the energy of each ejected electron depends only on colour (frequency), while intensity merely sets the number ejected. PQ 1 (colour decides, not brightness), PQ 3 (two knobs that do not interfere), and PQ 4 (energy with no time to gather) are the same puzzle seen from three angles. All three dissolve the moment we suppose that light is absorbed not as a continuous trickle but in indivisible packets, each carrying an energy fixed by the frequency—which is exactly what Einstein proposed.

✓ Checkpoint · The Photoelectric Experimental Laws

At fixed frequency the intensity is doubled. What happens to (a) the saturation current and (b) the stopping potential?
(a) doubles; (b) unchanged.
Which experimental law most directly refutes the wave-theory expectation of a build-up time?
Instantaneous emission—photoelectrons appear within \(\sim10^{-9}\,\mathrm{s}\), with no time lag.
Light below the threshold frequency is made \(1000\) times more intense. What photocurrent flows?
Zero—below \(\nu_0\) no electrons are emitted at any intensity.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Energy of a photon of green light

Find the energy of a photon of green light, wavelength \(\lambda=550\,\mathrm{nm}\), in both electron-volts and joules.

Solution. Given: \(\lambda=550\,\mathrm{nm}\).
Find: photon energy \(E\).
Setup: A photon carries \(E=hc/\lambda\); in working units \(E(\mathrm{eV})=1240/\lambda(\mathrm{nm})\).
Solve: \[ E=\frac{1240}{550}=2.25\,\mathrm{eV}=2.25\times(1.60\times10^{-19}\,\mathrm{J})=3.6\times10^{-19}\,\mathrm{J}. \] Answer: \(\boxed{E=2.25\,\mathrm{eV}=3.6\times10^{-19}\,\mathrm{J}}\)
Check: By the joule route, \(hc/\lambda=(6.63\times10^{-34})(3.0\times10^{8})/(550\times10^{-9})=3.6\times10^{-19}\,\mathrm{J}=2.25\,\mathrm{eV}\)—both routes agree, and \(2.25\,\mathrm{eV}\) sits squarely in the visible band (\(1.8\)–\(3.1\,\mathrm{eV}\)), as green light should ✓.

Example 2 · How many photons a laser pointer emits each second

A laser pointer emits \(P=5.0\,\mathrm{mW}\) of red light at \(\lambda=650\,\mathrm{nm}\). How many photons does it send out per second?

Solution. Given: \(P=5.0\times10^{-3}\,\mathrm{W}\), \(\lambda=650\,\mathrm{nm}\).
Find: photon rate \(n\).
Setup: Each photon carries \(hc/\lambda\); the number per second is \(n=P/(hc/\lambda)\).
Solve: \(E_{\text{ph}}=1240/650=1.91\,\mathrm{eV}=3.06\times10^{-19}\,\mathrm{J}\), so \[ n=\frac{P}{E_{\text{ph}}}=\frac{5.0\times10^{-3}}{3.06\times10^{-19}}=1.6\times10^{16}\ \text{photons/s}. \] Answer: \(\boxed{n\approx1.6\times10^{16}\ \text{photons/s}}\)
Check: Multiplying back, \(n\,E_{\text{ph}}=(1.6\times10^{16})(3.06\times10^{-19})\approx5\,\mathrm{mW}\) recovers the rated power. And \(\sim\!10^{16}\) photons every second is why laser light looks perfectly smooth—the graininess is far too fine to notice ✓.

Example 3 · Threshold wavelength from the work function

A metal has work function \(\phi_0=2.5\,\mathrm{eV}\). What is the longest wavelength of light that can eject electrons from it?

Solution. Given: \(\phi_0=2.5\,\mathrm{eV}\).
Find: threshold wavelength \(\lambda_0\).
Setup: Emission just stops when \(hc/\lambda_0=\phi_0\), i.e. \(\lambda_0=1240/\phi_0(\mathrm{eV})\,\mathrm{nm}\). Longer \(\lambda\) (lower frequency) cannot eject.
Solve: \[ \lambda_0=\frac{1240}{2.5}=496\,\mathrm{nm}. \] Answer: \(\boxed{\lambda_0=496\,\mathrm{nm}}\) (green); any light longer than this fails, however intense.
Check: Reading it back, \(1240/496=2.5\,\mathrm{eV}=\phi_0\), so a \(496\,\mathrm{nm}\) photon arrives with exactly zero energy to spare—the borderline. Shorter (bluer) light ejects; longer (redder) never does, whatever its intensity ✓.

Example 4 · Maximum kinetic energy and stopping potential

Light of wavelength \(\lambda=400\,\mathrm{nm}\) falls on a metal of work function \(\phi_0=2.0\,\mathrm{eV}\). Find the maximum kinetic energy of the photoelectrons, the stopping potential, and the maximum electron speed.

Solution. Given: \(\lambda=400\,\mathrm{nm}\), \(\phi_0=2.0\,\mathrm{eV}\).
Find: \(KE_{\max}\), \(V_0\), \(v_{\max}\).
Setup: Einstein's equation \(KE_{\max}=hc/\lambda-\phi_0\); \(eV_0=KE_{\max}\); \(KE_{\max}=\tfrac12 m v_{\max}^2\).
Solve: \(E_{\text{ph}}=1240/400=3.10\,\mathrm{eV}\), so \[ KE_{\max}=3.10-2.0=1.10\,\mathrm{eV},\qquad V_0=1.10\,\mathrm{V}. \] \[ v_{\max}=\sqrt{\frac{2KE_{\max}}{m_e}}=\sqrt{\frac{2(1.10\times1.60\times10^{-19})}{9.11\times10^{-31}}}=6.2\times10^{5}\,\mathrm{m/s}. \] Answer: \(\boxed{KE_{\max}=1.10\,\mathrm{eV},\ V_0=1.10\,\mathrm{V},\ v_{\max}=6.2\times10^{5}\,\mathrm{m/s}}\)
Check: \(eV_0=KE_{\max}\) gives \(V_0=1.10\,\mathrm{V}\) directly, and \(v_{\max}/c=2.1\times10^{-3}\ll1\), so the non-relativistic \(\tfrac12 mv^2\) was justified. Had \(\phi_0\) been larger than \(3.10\,\mathrm{eV}\), no electron would emerge—consistent with the threshold logic ✓.

Example 5 · de Broglie wavelength of a moving electron

An electron moves at \(v=2.0\times10^{6}\,\mathrm{m/s}\). Find its de Broglie wavelength.

Solution. Given: \(v=2.0\times10^{6}\,\mathrm{m/s}\).
Find: \(\lambda\).
Setup: \(\lambda=h/p=h/(m_e v)\) (non-relativistic, since \(v\ll c\)).
Solve: \(p=m_e v=(9.11\times10^{-31})(2.0\times10^{6})=1.82\times10^{-24}\,\mathrm{kg\,m/s}\), so \[ \lambda=\frac{h}{p}=\frac{6.63\times10^{-34}}{1.82\times10^{-24}}=3.6\times10^{-10}\,\mathrm{m}=0.36\,\mathrm{nm}. \] Answer: \(\boxed{\lambda=0.36\,\mathrm{nm}}\)
Check: Second route via energy: \(KE=\tfrac12 m_e v^2=11.4\,\mathrm{eV}\), an electron of \(11.4\,\mathrm{V}\)-equivalent, so \(\lambda=1.227/\sqrt{11.4}=0.364\,\mathrm{nm}\)—the same wavelength by the accelerating-voltage shortcut ✓.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Energy of a UV photon:
Find the energy of a \(200\,\mathrm{nm}\) ultraviolet photon in eV and in joules.
\(E=1240/200=6.20\,\mathrm{eV}=6.20\times1.60\times10^{-19}=9.9\times10^{-19}\,\mathrm{J}\).
Photon momentum:
Find the momentum of a \(600\,\mathrm{nm}\) photon.
\(p=h/\lambda=6.63\times10^{-34}/600\times10^{-9}=1.1\times10^{-27}\,\mathrm{kg\,m/s}\).
Counting photons:
A monochromatic source radiates \(25\,\mathrm{W}\) at \(500\,\mathrm{nm}\). How many photons does it emit per second?
\(E_{\text{ph}}=1240/500=2.48\,\mathrm{eV}=3.97\times10^{-19}\,\mathrm{J}\); \(n=P/E_{\text{ph}}=25/3.97\times10^{-19}=6.3\times10^{19}\,\mathrm{s^{-1}}\).
Work function from a stopping potential, then a prediction:
A metal gives a stopping potential of \(2.0\,\mathrm{V}\) under \(200\,\mathrm{nm}\) light. Find its work function, then predict \(V_0\) under \(250\,\mathrm{nm}\) light.
\(\phi_0=1240/200-2.0=6.20-2.0=4.2\,\mathrm{eV}\). At \(250\,\mathrm{nm}\): \(KE_{\max}=1240/250-4.2=4.96-4.2=0.76\,\mathrm{eV}\Rightarrow V_0=0.76\,\mathrm{V}\).
de Broglie's symmetry argument:
Explain the reasoning by which de Broglie assigned a wavelength to matter. Why is the wave a probability wave, not an electromagnetic one?
For light, \(p=h/\lambda\). de Broglie postulated this relation is universal, so any particle of momentum \(p\) has \(\lambda=h/p\). It is not an EM wave (no \(\vec E,\vec B\), not at speed \(c\)); its amplitude squared gives the probability of finding the particle—confirmed by Davisson–Germer's interference of single-type particles.

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

Two metals on one graph:
Sketch stopping potential \(V_0\) versus frequency \(\nu\) for metal A (\(\phi_0=2.0\,\mathrm{eV}\)) and metal B (\(\phi_0=3.0\,\mathrm{eV}\)). State the slope, and give each threshold frequency and wavelength.
Two parallel straight lines, common slope \(h/e=4.14\times10^{-15}\,\mathrm{V\,s}\). Thresholds: A at \(\nu_0=\phi_0/h=4.8\times10^{14}\,\mathrm{Hz}\) (\(\lambda_0=620\,\mathrm{nm}\)); B at \(7.3\times10^{14}\,\mathrm{Hz}\) (\(\lambda_0=413\,\mathrm{nm}\)). B is shifted right (larger \(\phi_0\)).
Why the current saturates:
Explain why the photocurrent versus collector-voltage curve levels off at high positive voltage, and why brighter light raises the saturation level but leaves the stopping potential unchanged.
At high positive voltage every emitted electron is collected, so the current cannot grow—saturation. Brighter light (same frequency) ejects more electrons per second, raising the saturation current, but each electron's maximum energy is still \(h\nu-\phi_0\), so \(V_0\) is unchanged.
The classical time-lag that never happens:
Estimate the time a faint beam of intensity \(10^{-6}\,\mathrm{W/m^2}\) would classically need to deliver \(\phi_0=2.3\,\mathrm{eV}\) to one atom of cross-section \(\sim3\times10^{-20}\,\mathrm{m^2}\). Compare with the observed delay.
Power intercepted \(=IA=10^{-6}\times3\times10^{-20}=3\times10^{-26}\,\mathrm{W}\); time \(=\phi_0/\text{power}=3.7\times10^{-19}/3\times10^{-26}\approx1.2\times10^{7}\,\mathrm{s}\) (\(\sim\)140 days). Observed: \(\lt 10^{-9}\,\mathrm{s}\). The wild discrepancy is fatal to the wave picture; a single photon delivers \(\phi_0\) at once.
Derive Einstein's equation:
Starting from the photon idea, derive \(KE_{\max}=h\nu-\phi_0\) and state which experimental law each feature explains.
One electron absorbs one photon of energy \(h\nu\); energy conservation \(h\nu=\phi_0+KE\), with the least-bound electron paying only \(\phi_0\) giving \(KE_{\max}=h\nu-\phi_0\). Intensity sets photon number (saturation current, Law 1) not \(h\nu\) (so \(V_0\) independent of intensity, Law 2); \(h\nu\) rises with frequency (Law 3, threshold at \(h\nu=\phi_0\)); one-photon delivery is instantaneous (Law 4).