11. Dual Nature Of Radiation And Matter

Energy can arrive continuously on average while being exchanged in separate events. This chapter connects light emission, photon energy and the wave behaviour of moving particles. Constants such as hc and the elementary charge are rounded classroom values in numerical examples.

Planck’s energy quantisation opened a route beyond classical radiation theory. Einstein applied a light-quantum description to photoelectric energy transfer. De Broglie later extended wavelength–momentum reasoning to matter. Experiments on electron diffraction tested that proposal.

Light can spread across two openings like a wave, yet transfer energy to a surface in individual events.

A scientific model answers a particular kind of question. Wave optics predicts interference and diffraction. A particle description predicts the energy and momentum transferred in a single light–matter interaction. We need both descriptions because neither a classical wave alone nor tiny classical pellets explain every observation.

The photoelectric effect is the emission of electrons when suitable electromagnetic radiation reaches a material. The emitted particles are electrons already present in the material. Light does not manufacture the metal electrons. Their release provides a direct way to compare the predictions of classical and quantum models.

“Dual nature” does not mean that a detector photographs a light wave turning into a miniature ball. Quantum theory gives probabilities for measurement outcomes. At school level, use wavelength and frequency for propagation, and photon energy and momentum for individual transfers.

A beam makes an interference pattern, while a weak version of the beam produces separated detector clicks. Which description is useful for each observation?

The position-dependent pattern needs wave interference. The individual transfer events need quantised detection. Repeating many events builds the distribution predicted by the wave description.

Result: Use wave propagation and discrete energy-transfer events together.

1. Does the photoelectric effect eject newly created electrons?
2. Which observation requires adding amplitudes rather than just particle counts?

Answers

1. No. It releases electrons initially in the material.
2. Interference, where phase relationships affect the detected intensity.

Before an electron can leave a surface, it must overcome the surface energy barrier.

In a metal, some electrons can move through the solid. Attraction to the remaining positive structure nevertheless makes escape into the space outside energetically costly. The work function is the minimum energy needed to take an electron from the relevant occupied states at the surface to just outside with zero kinetic energy. It depends on the material and surface condition.

The work function is written $\phi$. Energy may be measured in joules or electronvolts. One electronvolt is the energy associated with moving one elementary charge through one volt: $1\,\mathrm{eV}=1.602\times10^{-19}\,\mathrm J$. The symbol eV is an energy unit, not a voltage.

Heating, illumination and a strong electric field are different ways of causing emission. Merely charging a metal does not automatically imply that electrons have escaped. Real surfaces can be oxidised, contaminated or rough, so tabulated work functions should be treated as stated model data when solving a problem.

A surface has work function $2.3\,\mathrm{eV}$. Express this in joules.

Multiply the value in electronvolts by the energy in joules per electronvolt.

$\phi=2.3\times1.602\times10^{-19}\,\mathrm J$.

Result: $3.6846\times10^{-19}\,\mathrm J$.

1. Why does a conducting electron not necessarily escape from the metal?
2. Name the emission process driven mainly by heating.

Answers

1. Motion inside the solid and escape through its surface barrier are different requirements.
2. Thermionic emission.

The key question is whether light can remove negative charge from an illuminated surface.

When sufficiently energetic light strikes a suitable clean surface, electrons can leave it. An initially negatively charged isolated plate may lose negative charge. A neutral isolated emitter becomes positive as electrons leave, until the resulting electric potential inhibits further escape. An external circuit can replace emitted electrons and maintain a photocurrent.

During investigations of electromagnetic waves, Hertz found that ultraviolet illumination could help a spark occur across a nearby detector gap. The observation connected radiation with easier electrical discharge. It was an experimental clue; by itself it did not provide the later photon energy equation.

Experiments on illuminated metal surfaces showed that light could release negative charge, later identified as electrons. Hallwachs used the discharge of a negatively charged surface as evidence. Lenard studied the emitted particles and how their collection changed with electrical conditions. These studies made the effect measurable rather than just visible as a spark.

An isolated neutral metal plate emits electrons under illumination. What is the sign of the charge it tends to acquire, and why can emission eventually be inhibited?

Each escaping electron removes negative charge. The plate therefore tends to become positive. A positive plate attracts an outgoing electron back and raises the energy needed for complete escape.

Result: Positive; its growing electric potential can prevent further escape unless electrons are replaced.

1. What particles leave an ordinary photoelectric metal emitter?
2. Why is an evacuated enclosure useful in a photoelectric experiment?

Answers

1. Electrons.
2. It reduces collisions with gas particles between emitter and collector.

Change one control at a time: intensity, collector potential or light frequency.

For a fixed frequency above threshold and unchanged collection conditions, stronger illumination generally releases more electrons per second. Once the collector attracts essentially all available emitted electrons, the current reaches saturation. Increasing the collecting voltage further does not create additional photoelectrons.

A positive collector attracts electrons. A negative collector retards them. Emitted electrons have a distribution of kinetic energies, so current falls gradually as the retarding potential grows. At the stopping potential the fastest emitted electrons are just unable to reach the collector in the ideal model. Use the magnitude of that retarding potential.

$$K_{\max}=q_{\rm e}V_0$$

Changing intensity at fixed frequency changes saturation current but does not change the ideal stopping potential. The maximum kinetic energy is controlled by photon energy and work function, not by how many photons arrive per second.

Increasing frequency raises the maximum emitted-electron energy and therefore raises the stopping-potential magnitude. Below the threshold frequency, the ordinary one-photon effect is absent even if the beam is made brighter. At this level we assume ordinary intensities, not multiphoton laser processes.

A photoelectric experiment has stopping potential $1.7\,\mathrm V$. What is the maximum electron kinetic energy?

The retarding electric field removes energy $q_{\rm e}V_0$ from the fastest electron. In electronvolts, multiplying an elementary charge by a voltage gives the same numerical value.

$K_{\max}=1.7\,\mathrm{eV}$.

Result: $2.7234\times10^{-19}\,\mathrm J$.

1. At fixed frequency, what ideally happens to saturation current when incident photon rate doubles?
2. Does increasing intensity ideally move the stopping-potential intercept?

Answers

1. It doubles if quantum efficiency and collection remain unchanged.
2. No, at fixed frequency and surface in the one-photon regime.

Classical waves predict energy delivery, but their simple continuous-absorption picture misses the photoelectric rules.

A classical electromagnetic wave carries more energy per unit time when its intensity is larger. If an electron could simply collect that energy continuously for as long as necessary, a sufficiently long exposure might eject it at almost any frequency. At very low intensity the collection time should become noticeable. Increasing intensity might also be expected to increase the energy of an individual emitted electron.

The measured ordinary photoelectric effect instead has a threshold frequency, emission without the large accumulation delay expected from that picture, and a maximum kinetic energy that follows frequency rather than intensity. These three facts must be explained together. Merely saying “light carries energy” is insufficient.

The failure is not that waves cannot carry energy. Interference, diffraction and electromagnetic propagation remain successful wave ideas. The missing ingredient is how radiation exchanges energy with matter in individual events.

A bright low-frequency beam gives no emission, but a faint higher-frequency beam does. Explain why intensity alone cannot be the threshold control.

Brightness controls the rate of incident energy. The observation shows that the energy available in one interaction matters. In the one-photon description, that energy is set by frequency.

Result: Each absorbed photon must individually supply at least the work function; a larger rate of insufficient-energy photons does not satisfy that condition.

1. What is the basic difficulty with slow classical energy accumulation?
2. Which quantity sets the energy of a photon?

Answers

1. It predicts emission after sufficient waiting and a large low-intensity delay, rather than the observed ordinary threshold behaviour.
2. Its frequency.

A single energy balance explains the threshold and the fastest emitted electrons.

A photon of frequency $\nu$ supplies energy $h\nu$. The escaping electron uses at least $\phi$ to overcome the surface barrier. For an electron with the greatest possible kinetic energy, the balance is $h\nu=\phi+K_{\max}$. Electrons that lose additional energy inside the surface emerge more slowly.

$$K_{\max}=h\nu-\phi=q_{\rm e}V_0$$

At threshold, the maximum kinetic energy tends to zero. Thus $h\nu_0=\phi$. Since $c=\nu\lambda$, the largest wavelength that can cause ordinary photoemission is $\lambda_0=hc/\phi$. Frequency must be at least the threshold, while wavelength must be no greater than the threshold wavelength.

A plot of $V_0$ against $\nu$ has slope $h/q_{\rm e}$ and vertical intercept $-\phi/q_{\rm e}$ when the above-threshold line is extrapolated. The extrapolation below threshold is a way to read the graph; it does not imply a physical negative maximum kinetic energy.

Light of wavelength $310\,\mathrm{nm}$ strikes a surface with $\phi=2.3\,\mathrm{eV}$. Use $hc=1240\,\mathrm{eV\,nm}$. Find photon energy, maximum electron energy and threshold wavelength.

$E=hc/\lambda=1240/310\,\mathrm{eV}$.

Subtract the work function once for one emitted electron.

For the threshold wavelength, set photon energy equal to the work function.

Result: $E=4\,\mathrm{eV}$, $K_{\max}=1.7\,\mathrm{eV}$, $\lambda_0=539.1304\,\mathrm{nm}$; $V_0=1.7\,\mathrm V$.

For the same beam, $\nu=967741935483871\,\mathrm{Hz}$. Either joules throughout or electronvolts throughout gives the same physical result. The measured slope is approximately $4.136\times10^{-15}\,\mathrm{V\,s}$ when frequency is in hertz.

1. A $3.1\,\mathrm{eV}$ photon reaches a $3.4\,\mathrm{eV}$ surface. What is the outcome?
2. Why do emitted electrons not all have the maximum energy?

Answers

1. No ordinary one-photon photoemission.
2. Their initial states and energy losses on the route to the surface differ.

A photon has energy and momentum even though its rest mass is zero.

For light in vacuum, $E=h\nu=hc/\lambda$ and $p=E/c=h/\lambda$. A photon travels at the vacuum speed of light. The massive-particle expression $p=mv$ with a fixed rest mass must not be used to assign it a nonzero rest mass.

Intensity can rise because more photons of the same energy cross a given area per second. Increasing frequency changes the energy per photon. At fixed total beam power, increasing photon energy reduces the photon arrival rate. Absorption transfers both energy and momentum to matter; reflection can also transfer momentum.

For complete absorption of a monochromatic beam, photon arrival rate is $\dot N=P/(h\nu)$. If only a fraction $\eta$ of photons produces collected electrons, $I=\eta q_{\rm e}\dot N$. The quantum efficiency $\eta$ includes the relevant emission and collection losses, and need not equal one.

A monochromatic beam has power $2.4\,\mathrm{mW}$ and photon energy $3.0\,\mathrm{eV}$. Find photon rate using $1\,\mathrm{eV}=1.602\times10^{-19}\,\mathrm J$.

Convert the power to joules per second. Divide by the energy in one photon.

$\dot N=(2.4\times10^{-3})/(3.0\times1.602\times10^{-19})$.

Result: Approximately $0.4994\times10^{16}\,\mathrm{s^{-1}}$.

1. If photon wavelength doubles, how do energy and momentum change?
2. Can a zero-rest-mass photon transfer momentum?

Answers

1. Both are halved.
2. Yes. Its momentum is E/c in vacuum.

Diffraction is not restricted to light: particles with momentum also have a wavelength.

De Broglie proposed $\lambda=h/p$ for a particle. For a nonrelativistic particle, $p=mv$ and $K=p^2/(2m)$, so $\lambda=h/\sqrt{2mK}$. An electron accelerated from negligible initial kinetic energy through potential difference $V$ has $K=q_{\rm e}V$ and $\lambda=h/\sqrt{2mq_{\rm e}V}$. Use the relativistic momentum if the speed is no longer small compared with c.

A matter wave is not a mechanical ripple of the electron’s substance. Its wavefunction is used to predict measurement probabilities. A large everyday object has an extremely small wavelength at ordinary momentum, which makes its diffraction difficult to observe in ordinary situations. The relation itself is not limited to electrons.

Electron diffraction by crystals supports the wave description. Crystal planes provide separations comparable to electron wavelengths. The positions of intensity maxima depend on wavelength and geometry. In a simple Bragg description, constructive scattering requires an integer number of wavelengths in the path difference: twice the plane spacing times the sine of the glancing angle.

$$2d\sin\theta=n\lambda$$

An electron starts nearly at rest and is accelerated through $81\,\mathrm V$. Use the nonrelativistic relation $\lambda\text{ (nm)}=1.227/\sqrt{V\text{ (V)}}$.

The square root of the accelerating voltage gives the denominator. Doubling voltage would reduce wavelength by a factor $\sqrt 2$, not by two.

Result: $\lambda=0.13633\,\mathrm{nm}$.

1. At fixed kinetic energy, which has the shorter wavelength: an electron or a heavier particle?
2. If accelerating voltage is multiplied by nine in the nonrelativistic regime, what happens to electron wavelength?

Answers

1. The heavier particle, because momentum is the square root of twice mass times kinetic energy.
2. It becomes one third as large.

• Photon energy is $h\nu$ and photon momentum in vacuum is $h/\lambda$.
• For ordinary photoemission, $K_{\max}=h\nu-\phi$ when $h\nu\geq\phi$.
• Stopping-potential magnitude satisfies $q_{\rm e}V_0=K_{\max}$.
• Matter waves satisfy $\lambda=h/p$; nonrelativistic electrons accelerated through V have wavelength proportional to $V^{-1/2}$.

• A graph extrapolation can have a negative intercept without representing a physical negative emitted-electron energy.
• The electron wavefunction is not an oscillating piece of classical charged material.
• Intensity and frequency control different features of a monochromatic photon beam.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

11.1 State one experiment that needs wave propagation and one that needs quantised energy transfer.

11.2 Distinguish thermionic, field and photoelectric emission.

11.3 Convert $1.8\,\mathrm{eV}$ to joules using $q_{\rm e}=1.602\times10^{-19}\,\mathrm C$.

11.4 A $2.6\,\mathrm{eV}$ work function is illuminated by $3.9\,\mathrm{eV}$ photons. Find maximum kinetic energy and stopping potential.

11.5 Find threshold wavelength for $\phi=3.1\,\mathrm{eV}$ with $hc=1240\,\mathrm{eV\,nm}$.

11.6 At fixed above-threshold frequency, compare doubling intensity with doubling collector voltage beyond saturation.

11.7 Why can electrons be emitted with less than the maximum kinetic energy?

11.8 A beam has wavelength $400\,\mathrm{nm}$ and strikes a $3.5\,\mathrm{eV}$ surface. Use $hc=1240\,\mathrm{eV\,nm}$. Is emission expected?

11.9 Why does a negative collector voltage reduce current?

11.10 What graph gives the work function and Planck constant from stopping-potential measurements?

11.11 Maximum kinetic energy changes from $0.6$ to $1.8\,\mathrm{eV}$. Find the change in stopping potential.

11.12 Two surfaces have different work functions. Compare the slopes of their ideal stopping-potential versus frequency plots.

11.13 A photon has energy $4.0\,\mathrm{eV}$. Find its wavelength using $hc=1240\,\mathrm{eV\,nm}$.

11.14 At the same beam power, compare photon rates for wavelengths λ and 2λ.

11.15 A photon has energy $6.0\times10^{-19}\,\mathrm J$. Find its momentum using $c=3.0\times10^8\,\mathrm{m\,s^{-1}}$.

11.16 An electron and a photon have equal momentum. Compare their de Broglie wavelengths.

11.17 An electron is accelerated through $36\,\mathrm V$. Find its wavelength from $1.227/\sqrt V$ in nm.

11.18 How does nonrelativistic electron wavelength change if accelerating voltage increases by a factor of four?

11.19 Why is electron diffraction easier to observe using crystals than using an ordinary wide doorway?

11.20 In first-order Bragg scattering, $d=0.20\,\mathrm{nm}$ and $\theta=30^\circ$. Find wavelength.

11.21 State the limitation of λ=h/√(2mK).

11.22 A neutral isolated plate loses electrons. What charge develops, and what is its effect on subsequent escape?

11.23 Does a higher photon frequency make the photon move faster in vacuum?

11.24 Explain why “the brighter beam must eject faster electrons” is incorrect at fixed frequency.

11.25 What does a finite photocurrent measure that a stopping-potential reading does not?

Answers

11.1 Interference needs wave propagation; photoelectric energy transfer needs photon energy.

11.2 Heating supplies thermal energy; a strong field alters the escape barrier; radiation supplies photon energy.

11.3 $2.8836\times10^{-19}\,\mathrm J$.

11.4 $K_{\max}=1.3\,\mathrm{eV}$ and $V_0=1.3\,\mathrm V$.

11.5 $\lambda_0=400\,\mathrm{nm}$.

11.6 Doubling intensity ideally doubles saturation current; extra collector voltage beyond saturation adds no emitted electrons.

11.7 Their initial states and losses within the surface differ.

11.8 Photon energy is $3.1\,\mathrm{eV}$, below the work function; no ordinary one-photon emission.

11.9 Its electric field opposes the motion of emitted electrons towards the collector.

11.10 Plot stopping-potential magnitude against frequency above threshold. The slope is h divided by elementary charge; the extrapolated frequency intercept gives φ/h.

11.11 $1.2\,\mathrm V$.

11.12 The slopes are equal to h divided by elementary charge; the intercepts differ.

11.13 $310\,\mathrm{nm}$.

11.14 The longer-wavelength photons each have half the energy, so their rate is twice as large.

11.15 $2\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}$.

11.16 They have equal wavelength h/p, although their energy–momentum relations differ.

11.17 $0.2045\,\mathrm{nm}$.

11.18 It is halved.

11.19 Crystal spacings can be comparable to electron wavelength; a doorway is enormously larger.

11.20 $\lambda=2d\sin\theta=0.2\,\mathrm{nm}$.

11.21 It uses nonrelativistic kinetic energy and must be replaced when speeds approach c.

11.22 Positive charge develops; its attraction opposes subsequent electron escape.

11.23 No. All photons propagate at c in vacuum.

11.24 Brightness raises photon arrival rate; each photon retains the same energy, so the ideal maximum electron energy stays fixed.

11.25 Current measures collected charge per time; stopping potential measures the upper kinetic-energy limit.