Dual Nature of Radiation and Matter | ISC Class 12 Physics Notes

This note covers electron emission, the photoelectric effect, Hertz’s and Lenard’s observations, experimental graphs, Einstein’s photoelectric equation, photons, the determination of Planck’s constant, de Broglie matter waves and the qualitative interpretation of the Davisson-Germer experiment.

What allows an electron to escape from a metal?

What is the work function?

Metals contain free electrons, negatively charged particles that can move within the metal and contribute to electrical conduction. They cannot normally escape from its surface. Attractive forces hold them inside, so an electron needs additional energy to leave.

Definition: The work function of a metal is the minimum energy required by an electron to escape from its surface. It depends on the metal and the nature of its surface.

We use W₀ for the work function. The subscript zero identifies this minimum escape energy. Work function is an energy, not a force or a voltage. Different electrons need different additional energies to escape, because they do not all initially have the same energy.

The electron volt, symbol eV, is the energy gained by an electron accelerated through a potential difference of one volt. A potential difference measures energy transferred per unit charge. The symbol J denotes the joule, the SI unit of energy; SI means International System of Units.

1 eV = 1.602 × 10⁻¹⁹ J

How can the required energy be supplied?

Frequency is the number of oscillations per second. Its unit is the hertz, symbol Hz. Electrons emitted because light falls on a surface are called photoelectrons. The emission of these electrons is the photoelectric effect.

The word “free” describes motion inside a metal; it does not mean that escape needs no energy. Similarly, illumination alone does not guarantee emission. The frequency must be suitable for the particular surface, a condition explored through the emission experiments below.

What did Hertz and Lenard observe about photoelectric emission?

How did light affect electrical discharge?

Hertz observed that high-voltage sparks across a detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp. Ultraviolet radiation is electromagnetic radiation at frequencies higher than visible violet light.

Light falling on a metal could help charged particles escape. In the electron description, some electrons near the surface absorb enough energy from the radiation to overcome the attraction holding them within the material.

Lenard illuminated an emitter plate inside an evacuated glass tube containing two metal electrodes. An electrode is a conducting part through which charge enters or leaves a device. “Evacuated” means that gas has been removed from the tube.

When ultraviolet radiation reached the emitter, current flowed in the circuit. When the radiation stopped, the current stopped. The electrons released by the emitter were attracted towards the positive collector, giving a current through the apparatus.

Why does the material matter?

Hallwachs and Lenard found that emission failed below a certain minimum frequency. This threshold frequency depends on the emitter material. It is the lower frequency boundary for photoelectric emission from a given surface.

Zinc, cadmium and magnesium respond to ultraviolet light. Some alkali metals, including lithium, sodium, potassium, caesium and rubidium, are sensitive even to visible light. Thus “light causes emission” needs the qualification that the radiation frequency is suitable for the surface.

The experiments raised distinct questions: how many electrons are emitted, how much kinetic energy they have, and which frequencies can release them. Kinetic energy is energy associated with motion. Measuring current and measuring electron energy therefore investigate different features of photoemission.

A larger current indicates a greater rate of charge flow. It does not by itself establish that each emitted electron has greater energy. The experimental arrangement must allow these two effects to be separated.

How does the photoelectric apparatus measure emission?

What are the parts of the apparatus?

A thin photosensitive plate, meaning a plate capable of releasing electrons under suitable illumination, acts as the emitter. A second metal plate collects the electrons. Both are enclosed in an evacuated tube, with a quartz window admitting ultraviolet radiation.

Monochromatic light is light of a single frequency. Light from a source passes through the window and falls on the emitter. A battery maintains an adjustable potential difference between the plates. A commutator, a switch that reverses connections, changes the polarity.

Here S labels the source, C the emitter and A the collector. The voltmeter measures potential difference. The microammeter measures the small current due to collected photoelectrons, called the photocurrent. These letters label apparatus components, not numerical quantities.

Light from source S passes through a quartz window towards the photosensitive plate C inside an evacuated glass tube. Electron arrows run towards collector A. The external circuit shows a battery, commutator, voltmeter and microammeter.

See Fig. 11.1 in your NCERT textbook

Which variables are controlled?

1. Select an emitter material and illuminate it with light whose frequency is above its threshold frequency.
2. Keep frequency and plate potential fixed while changing the intensity, the radiation energy incident per unit area per unit time.
3. Keep frequency and intensity fixed while changing the collector potential, including its polarity relative to the emitter.
4. Measure the stopping potential at different frequencies to examine how the maximum electron energy changes.

A positive collector potential attracts electrons towards the collector. A negative collector potential opposes their motion and is called a retarding potential. Reversing polarity changes the collection of emitted electrons; it does not redefine the work function of the emitter.

The intensity can be varied by changing the distance between the source and emitter. Appropriate coloured filters select different frequencies. Keeping the other variables fixed is essential when interpreting a change in photocurrent.

How do intensity and collector potential affect photocurrent?

What changes when intensity increases?

For a fixed photosensitive material and frequency above threshold, photocurrent is directly proportional to incident intensity when the collecting conditions remain the same. More intense illumination releases more photoelectrons per second. The observation concerns the number emitted, rather than their maximum kinetic energy.

The horizontal axis shows intensity of light and the vertical axis shows photoelectric current. A straight line rises from the origin, showing direct proportionality under fixed experimental conditions.

See Fig. 11.2 in your NCERT textbook

At fixed frequency and intensity, raising the positive collector potential increases the current until all emitted photoelectrons are collected. The maximum current then reached is the saturation current. Increasing the positive potential further does not increase this current.

What does stopping potential measure?

The stopping potential is the minimum magnitude of negative collector potential that reduces photocurrent to zero. We write its positive magnitude as V₀. The corresponding signed collector potential is −V₀ relative to the emitter. The unit V denotes the volt.

Let Kmax denote maximum photoelectron kinetic energy and e the positive magnitude of electron charge. The electron’s charge is −e. Using e = 1.602 × 10⁻¹⁹ C, where C here denotes the coulomb, the charge unit, gives:

Kmax = eV₀

Electrons have a range of kinetic energies. A retarding potential first prevents less energetic electrons from arriving. At the stopping potential, even the most energetic electrons cannot reach the collector. Thus the measurement gives the maximum energy, not the average energy.

Three curves, labelled by intensities I₁, I₂ and I₃ with I₃ > I₂ > I₁, reach successively higher saturation currents. All meet the potential axis at the same negative stopping value.

See Fig. 11.3 in your NCERT textbook

Here I₁, I₂ and I₃ denote light intensities. Their common stopping potential means that changing intensity at fixed frequency changes the electron emission rate without changing maximum kinetic energy.

Worked example 1. A photoelectric experiment has a cut-off voltage of 1.5 V. Find the maximum kinetic energy, using e = 1.6 × 10⁻¹⁹ C.

Formula: Kmax = eV₀. Substitute: Kmax = (1.6 × 10⁻¹⁹) × 1.5. Answer: Kmax = 2.4 × 10⁻¹⁹ J. The retarding potential of 1.5 V stops even the most energetic photoelectrons.

How does frequency reveal the limitations of classical wave theory?

What are the experimental laws?

For a given emitter, increasing frequency above threshold increases maximum photoelectron kinetic energy and stopping-potential magnitude linearly. The threshold itself is a material property. Below threshold, increasing intensity does not produce photoelectric emission.

Above threshold, emission starts without any apparent time lag, even when the radiation is very dim. The observed time is of the order of 10⁻⁹ s or less, where s denotes the second. “Without apparent time lag” does not assert a measured duration of exactly zero.

Why is continuous energy absorption inadequate?

The classical wave picture treats light as electric and magnetic fields with energy continuously distributed over the wave. If electrons continuously absorbed this energy, greater intensity should supply each electron with more energy and increase maximum kinetic energy. That prediction contradicts the stopping-potential observations.

It also suggests that sufficiently intense radiation acting for sufficient time could release electrons at any frequency. This does not explain a threshold frequency. At low intensities, continuous accumulation would require a delay before an electron acquired enough energy to escape.

The failure concerns the explanation of photoelectric emission. Wave descriptions still explain interference, diffraction and polarisation. Interference is the redistribution of intensity when waves superpose; diffraction is wave spreading around obstacles or apertures; polarisation describes restriction of transverse oscillations to a particular direction.

The evidence therefore requires a description that preserves the successful wave account of propagation phenomena while explaining discrete energy transfer between radiation and electrons. The photon picture supplies that account of the photoelectric effect.

How does Einstein’s equation explain the photoelectric effect?

What is a light quantum?

Quantisation of radiation means that radiation exchanges energy in discrete packets called quanta. A quantum of light is a photon. Einstein extended Planck’s quantum idea to explain the interaction between light and electrons.

Let E denote the energy of one photon, h Planck’s constant, and ν the light frequency. Use h = 6.626 × 10⁻³⁴ J s unless a numerical problem supplies a rounded value. Then:

E = hν

Derivation: Einstein’s photoelectric equation

1. A single electron absorbs a single quantum of incident radiation, receiving energy hν.
2. The minimum energy required to escape from the metal surface is the work function W₀.
3. For an electron emerging with maximum kinetic energy, conservation of energy gives hν = W₀ + Kmax.
4. Subtract the escape energy from the photon energy to obtain the maximum possible kinetic energy.

Kmax = hν − W₀

More tightly bound electrons emerge with less than this maximum energy. The equation therefore does not assign the same kinetic energy to every emitted electron. It identifies the upper limit for a given radiation frequency and emitter surface.

Let ν₀ denote the threshold frequency. At the threshold boundary, photon energy equals the work function, so W₀ = hν₀. Substitution gives two useful forms:

ν₀ = W₀/h

Kmax = h(ν − ν₀)

How are the observations explained?

At fixed frequency, greater intensity means more incident photons per second over a given area. This allows more electrons to absorb photons, increasing photocurrent. It does not change the energy of each photon, so it does not change Kmax.

Below threshold, a photon has insufficient energy to supply the work function. Above threshold, absorption of one photon supplies energy in one elementary interaction, explaining emission without any apparent time lag even at low intensity.

Note: A negative calculated value of hν − W₀ means that photoelectric emission is not possible under the stated conditions. It is not the kinetic energy of an emitted electron. At threshold, the limiting maximum kinetic energy is zero.

How can a stopping-potential graph determine Planck’s constant?

Derivation: The straight-line graph

1. For a given emitter above threshold, Einstein’s equation gives Kmax = hν − W₀.
2. The stopping measurement gives Kmax = eV₀, so eV₀ = hν − W₀.
3. Divide by e to express stopping potential as a linear function of frequency: V₀ = (h/e)ν − W₀/e.

V₀ = (h/e)ν − W₀/e

Let a be the gradient, meaning change in vertical coordinate divided by change in horizontal coordinate. With V₀ vertically and ν horizontally, a = h/e. Therefore:

h = ea

The horizontal axis shows incident frequency and the vertical axis shows stopping potential. Two rising straight lines represent metals A and B. They have different threshold-frequency intercepts on the horizontal axis.

See Fig. 11.5 in your NCERT textbook

The horizontal intercept is ν₀, the frequency at which the line reaches zero stopping potential. The gradient is independent of the emitter material. Extending the line algebraically to zero frequency gives vertical intercept −W₀/e; this extension does not describe emission below threshold.

If maximum kinetic energy in joules is plotted vertically instead, the gradient is h. Distinguish the two graphs before interpreting a measured slope. Their horizontal intercept is the same threshold frequency for the same surface.

Worked example 2. The gradient of cut-off voltage against frequency is 4.12 × 10⁻¹⁵ V s. Calculate Planck’s constant using e = 1.6 × 10⁻¹⁹ C.

Formula: h = ea. Substitute: h = (1.6 × 10⁻¹⁹) × (4.12 × 10⁻¹⁵). Answer: h = 6.592 × 10⁻³⁴ J s, or approximately 6.59 × 10⁻³⁴ J s. For the unit conversion, 1 V C = 1 J, so charge multiplied by the graph gradient gives joule seconds.

Which units must accompany the quantities?

The symbol λ denotes wavelength. Points in the same phase are at corresponding stages of oscillation, such as successive crests. In numerical work, convert electron volts to joules before combining an energy with h expressed in joule seconds.

How are photon energy, momentum and number calculated?

What properties does a photon have?

A photon carries energy and momentum, the quantity associated with motion and conserved in interactions. Let p denote momentum and c the speed of light in vacuum. Using c = 3.0 × 10⁸ m/s, with m/s meaning metres per second, the relations are:

E = hc/λ

p = E/c = hν/c = h/λ

Photons are electrically neutral and are not deflected by electric or magnetic fields. Photons of a particular frequency have the same energy and momentum, whatever the intensity. Increasing intensity at that frequency increases the number passing through a given area per second.

In a photon-particle collision, total energy and momentum are conserved. Photon number need not be conserved: a photon may be absorbed or a new one created. The photon description concerns discrete interaction with matter and does not remove the evidence for wave behaviour.

How is beam power related to photon rate?

Let P denote power, the energy emitted per second, and N the number of photons emitted per second. The power unit is the watt, symbol W. For a monochromatic beam:

P = NE

N = P/E

Worked example 3. A laser emits monochromatic light of frequency 6.0 × 10¹⁴ Hz with power 2.0 × 10⁻³ W. Find photon energy and photon emission rate. Use h = 6.63 × 10⁻³⁴ J s.

Formula: E = hν; N = P/E. Substitute: E = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴); N = (2.0 × 10⁻³)/(3.98 × 10⁻¹⁹). Answer: E = 3.98 × 10⁻¹⁹ J and N = 5.0 × 10¹⁵ photons per second. The 0.002 W beam supplies this many equal-energy photons each second.

Worked example 4. Caesium has work function 2.14 eV. Find its threshold frequency and the incident wavelength when stopping potential is 0.60 V. Use h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s and e = 1.6 × 10⁻¹⁹ C; 1 eV = 1.6 × 10⁻¹⁹ J for this calculation.

Formula: ν₀ = W₀/h; Kmax = eV₀; E = Kmax + W₀; λ = hc/E. Substitute: ν₀ = (2.14 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴). The required photon energy is 0.60 + 2.14 = 2.74 eV.

Answer: ν₀ = 5.16 × 10¹⁴ Hz and λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(2.74 × 1.6 × 10⁻¹⁹) = 454 nm. Here nm denotes the nanometre, equal to 10⁻⁹ m. At the stopping potential of 0.60 V, the most energetic photoelectrons have kinetic energy 0.60 eV.

The calculation adds the escape energy to maximum kinetic energy to recover the incident photon energy. Subtracting the work function again when reconstructing the photon energy would reverse the energy balance.

What does de Broglie’s relation say about matter waves?

How can a particle have a wavelength?

De Broglie proposed that moving particles should display wave-like properties under suitable conditions. The waves associated with moving material particles are called matter waves. Their wavelength is connected to the particle’s momentum, linking a wave property with a particle property.

λ = h/p

Let m now denote particle mass and v particle speed. In the non-relativistic description, meaning speeds sufficiently below the speed of light for ordinary Newtonian mechanics to apply, momentum is p = mv. Thus:

λ = h/(mv)

The symbol m in this equation is a variable for mass; m after a numerical length is the metre unit. Mass is measured in kilograms, symbol kg. Momentum is measured in kilogram metres per second, written kg m/s.

For equal speeds, a greater mass gives a shorter wavelength. For the same particle mass, a greater speed gives a shorter wavelength within this non-relativistic treatment. The wavelength relation depends on momentum, not on whether the particle is electrically charged.

Why is wave behaviour easier to observe for electrons?

Planck’s constant is extremely small. Ordinary moving objects have momenta that give wavelengths far below experimental measurement in the examples considered here. For subatomic particles, the wavelengths can be significant and measurable. The difference is a matter of scale, not a restriction of the hypothesis to electrons.

Worked example 5. Find the de Broglie wavelength of an electron moving at 5.4 × 10⁶ m/s. Use electron mass m = 9.11 × 10⁻³¹ kg and h = 6.63 × 10⁻³⁴ J s.

Formula: p = mv; λ = h/p. Substitute: p = (9.11 × 10⁻³¹)(5.4 × 10⁶) = 4.92 × 10⁻²⁴ kg m/s. Answer: λ = (6.63 × 10⁻³⁴)/(4.92 × 10⁻²⁴) = 1.35 × 10⁻¹⁰ m, or 0.135 nm. The conversion uses 1 m = 10⁹ nm.

Worked example 6. Find the de Broglie wavelength of a ball of mass 150 g travelling at 30.0 m/s. Use h = 6.63 × 10⁻³⁴ J s and 1000 g = 1 kg, where g denotes the gram.

Formula: p = mv; λ = h/p. Substitute: m = 0.150 kg and p = 0.150 × 30.0 = 4.50 kg m/s. Answer: λ = (6.63 × 10⁻³⁴)/4.50 = 1.47 × 10⁻³⁴ m. Its momentum of 4.50 kg m/s is far greater than the electron momentum in the preceding example, giving a far shorter wavelength.

A photon also satisfies λ = h/p, because its momentum is hν/c and its radiation wavelength is c/ν. For material particles, however, the proposed wavelength needed experimental testing. Electron diffraction provided such a test.

How does the Davisson-Germer experiment support wave-particle duality?

What is the qualitative arrangement?

The Davisson-Germer experiment investigates the scattering of electrons by a nickel crystal. A crystal has an ordered arrangement of atoms. Scattering means a change in the direction of motion after interaction with the target.

An electron gun, a device that produces and directs a beam of electrons, sends the beam towards the crystal in an evacuated chamber. An adjustable accelerating potential controls the electron energy. A movable detector measures the scattered electron intensity at different directions.

The detector signal measures how strongly electrons are scattered in a selected direction. Changing its direction allows the angular distribution to be examined. Changing the accelerating potential changes electron momentum and therefore the wavelength predicted by the de Broglie relation.

What observation supports matter waves?

1. Direct a beam of electrons towards the nickel crystal under controlled accelerating conditions.
2. Measure the scattered electron intensity as the detector direction changes.
3. Identify maxima, meaning directions with enhanced scattered intensity, in the angular distribution.
4. Interpret the directional maxima as electron diffraction and compare the inferred wavelength with the de Broglie prediction.

The agreement supports the wave nature of electrons. Electron diffraction is significant because diffraction is characteristic of waves, whereas the electrons also carry momentum and kinetic energy. The qualitative conclusion concerns both descriptions of the same physical entity.

No detailed diffraction calculation is needed to express this conclusion. The essential chain is an electron beam, scattering by an ordered crystal, directional diffraction maxima and agreement with the matter-wave wavelength.

Does duality mean choosing one permanent description?

Radiation exhibits wave behaviour in interference, diffraction and polarisation, and particle behaviour in photoelectric energy transfer. Matter possesses particle properties such as momentum and kinetic energy while displaying wave properties under suitable conditions.

Wave-particle duality means that both descriptions are needed. The nature of an experiment determines which description best explains its results. Treating light as discrete energy packets during photoemission does not invalidate its wave behaviour in interference experiments.

Glossary

• Work function — Minimum additional energy an electron needs to escape from a particular metal surface.
• Photoelectric effect — Emission of electrons from a material illuminated by light of suitable frequency.
• Photoelectron — An electron emitted from a surface after absorbing energy from incident light.
• Threshold frequency — Minimum radiation frequency forming the boundary for photoelectric emission from a given surface.
• Photocurrent — Electric current produced when emitted photoelectrons are collected in the experimental circuit.
• Saturation current — Maximum photocurrent reached when the collector receives all photoelectrons emitted under fixed illumination.
• Stopping potential — Minimum magnitude of negative collector potential needed to reduce the photocurrent to zero.
• Photon — A quantum of electromagnetic radiation carrying energy and momentum determined by its frequency.
• Electron volt — Energy gained by an electron accelerated through a potential difference of one volt.
• Monochromatic light — Radiation of a single frequency, whose photons therefore have the same energy.
• Intensity — Radiation energy incident per unit area per unit time at the receiving surface.
• De Broglie wavelength — Wavelength associated with a moving particle, equal to Planck’s constant divided by momentum.
• Wave-particle duality — The need for both wave and particle descriptions to explain radiation and matter.

Common errors and misconceptions

• Misconception: Free electrons need no energy to escape a metal. Correct: They can move inside it, but escaping its surface requires at least the work function.
• Misconception: Brighter light gives faster photoelectrons at fixed frequency. Correct: It increases the emission rate and saturation current, while maximum kinetic energy remains unchanged.
• Misconception: Enough intensity can overcome any threshold frequency. Correct: Below threshold, individual photons cannot supply the required work function in the photoelectric process described here.
• Misconception: Stopping potential measures average electron energy. Correct: Its magnitude multiplied by the elementary charge gives the maximum kinetic energy of emitted electrons.
• Misconception: A negative Einstein-equation result is an emitted electron’s energy. Correct: It indicates insufficient photon energy for emission; an emitted electron cannot have negative kinetic energy.
• Misconception: The stopping-potential graph has gradient h. Correct: Its gradient is h/e; the maximum-kinetic-energy graph in joules against frequency has gradient h.
• Misconception: The photon picture eliminates the wave picture. Correct: Different experiments require different descriptions, with interference and diffraction demonstrating wave behaviour.
• Misconception: Macroscopic objects have no de Broglie wavelength. Correct: The relation applies, but their wavelengths in ordinary examples are extremely small and beyond measurement.

Exam-style questions with model answers

Q1. Define work function and threshold frequency for a metal surface. [2 marks]

1. The work function is the minimum additional energy needed by an electron to escape from the metal surface.
2. The threshold frequency is the minimum incident-light frequency forming the boundary for photoelectric emission from that surface.

Q2. For a fixed emitter and light frequency above threshold, explain how increasing intensity affects saturation current, stopping potential and maximum photoelectron kinetic energy. [3 marks]

1. Saturation current increases in direct proportion to intensity because more photons arrive per second and more photoelectrons are emitted and collected per second.
2. The stopping-potential magnitude remains unchanged, since the incident frequency and emitter work function remain fixed during this comparison.
3. The maximum kinetic energy remains unchanged: each photon still has the same energy, and subtracting the unchanged work function gives the same maximum electron energy.

Q3. Describe the four principal parts of an arrangement for measuring photocurrent and stopping potential, explaining the role of each. [4 marks]

1. An evacuated tube contains a photosensitive emitter and a collector, allowing emitted electrons to travel from the illuminated surface to the collecting electrode.
2. A quartz window admits suitable monochromatic light to the emitter; quartz allows ultraviolet radiation to pass into the tube.
3. A variable battery arrangement and commutator control the magnitude and polarity of the collector potential, providing attracting or retarding conditions.
4. A microammeter measures photocurrent and a voltmeter measures plate potential difference; the stopping potential is found when the retarding setting reduces current to zero.

Q4. Explain Einstein’s photon account of the photoelectric effect. Include its energy equation and explanations of threshold frequency, the intensity effect and emission timing. Define your symbols. [5 marks]

1. A photon is a discrete quantum of radiation. Its energy is E = hν, where E is photon energy, h is Planck’s constant and ν is incident frequency; one electron absorbs one photon in the elementary process.
2. Conservation of energy gives Kmax = hν − W₀, where Kmax is maximum emitted-electron kinetic energy and W₀ is the surface work function.
3. The threshold frequency ν₀ satisfies hν₀ = W₀. Below it, photon energy cannot supply the escape energy, so greater intensity does not produce emission.
4. At fixed frequency above threshold, increased intensity supplies more photons per second, increasing photocurrent but leaving maximum kinetic energy unchanged.
5. Absorption supplies energy in an elementary interaction, explaining emission without any apparent time lag even when illumination is very dim.

Q5. A cut-off-voltage-versus-frequency graph has gradient 4.12 × 10⁻¹⁵ V s. State its relation to Planck’s constant and calculate that constant. Use electron-charge magnitude e = 1.6 × 10⁻¹⁹ C. [3 marks]

1. The stopping-potential equation is V₀ = (h/e)ν − W₀/e, where V₀ is cut-off voltage, h is Planck’s constant, ν is frequency and W₀ is work function. Its gradient a is h/e.
2. Rearranging gives h = ea. Substitution gives h = (1.6 × 10⁻¹⁹)(4.12 × 10⁻¹⁵) in joule seconds.
3. Thus h = 6.592 × 10⁻³⁴ J s, approximately 6.59 × 10⁻³⁴ J s. The measured gradient must be multiplied by charge, not used directly as h.

Q6. Caesium has work function 2.14 eV. Find its threshold frequency and the incident wavelength when stopping-potential magnitude is 0.60 V. Use Planck’s constant h = 6.63 × 10⁻³⁴ J s, speed of light c = 3.0 × 10⁸ m/s, electron-charge magnitude e = 1.6 × 10⁻¹⁹ C and 1 eV = 1.6 × 10⁻¹⁹ J. [4 marks]

1. Convert the work function W₀ to joules: W₀ = 2.14 × 1.6 × 10⁻¹⁹ = 3.424 × 10⁻¹⁹ J.
2. The threshold frequency is ν₀ = W₀/h = (3.424 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 5.16 × 10¹⁴ Hz.
3. Stopping-potential magnitude V₀ gives maximum kinetic energy eV₀ = 0.60 eV. Incident photon energy E = eV₀ + W₀ = 2.74 eV.
4. Using wavelength λ = hc/E gives λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(2.74 × 1.6 × 10⁻¹⁹) = 4.54 × 10⁻⁷ m.

Q7. Calculate the de Broglie wavelength of an electron moving at 5.4 × 10⁶ m/s and explain its relation to momentum. Use electron mass m = 9.11 × 10⁻³¹ kg and Planck’s constant h = 6.63 × 10⁻³⁴ J s. [3 marks]

1. Using speed v = 5.4 × 10⁶ m/s, the electron momentum is p = mv = (9.11 × 10⁻³¹)(5.4 × 10⁶) = 4.92 × 10⁻²⁴ kg m/s.
2. The de Broglie wavelength is λ = h/p = (6.63 × 10⁻³⁴)/(4.92 × 10⁻²⁴) = 1.35 × 10⁻¹⁰ m.
3. Wavelength is inversely proportional to momentum. A larger momentum therefore produces a shorter associated wavelength; this relation connects a wave property with a particle property.

Q8. Describe the Davisson-Germer experiment qualitatively and explain its conclusion about electrons. Include the beam, target, measurement and interpretation. [4 marks]

1. An electron gun directs a beam of electrons towards a nickel crystal inside an evacuated chamber, with accelerating potential controlling electron energy.
2. A movable detector measures scattered electron intensity at different directions relative to the incident beam and crystal.
3. The scattered intensity has directional maxima characteristic of diffraction. The wavelength inferred from the diffraction agrees with the de Broglie prediction.
4. The experiment supports the wave nature of electrons, while their momentum and kinetic energy remain particle properties, demonstrating the need for both descriptions.

Key takeaways

• The work function is the minimum escape energy of a metal surface, while threshold frequency identifies the corresponding radiation boundary.
• At fixed frequency above threshold, greater intensity increases photocurrent and saturation current without increasing maximum photoelectron kinetic energy.
• Stopping potential measures maximum kinetic energy through multiplication by the electron-charge magnitude; it does not measure average energy.
• Einstein’s equation balances one photon’s energy against the work function and the maximum kinetic energy of the emitted electron.
• The gradient of stopping potential against frequency equals Planck’s constant divided by elementary charge, independently of the emitter material.
• Photons at a fixed frequency have identical energy and momentum; greater intensity increases their number passing per second.
• De Broglie wavelength equals Planck’s constant divided by momentum, linking particle motion with a measurable wave property under suitable conditions.
• Electron diffraction supports matter waves, while interference and photoelectric emission show why radiation requires both wave and particle descriptions.

Test yourself

Why can electrons move inside a metal without freely escaping it?

Attractive forces hold electrons within the surface. Escaping requires additional energy, even though the electrons can move inside the metal.

Can longer exposure to light below threshold produce ordinary photoelectric emission?

No. Below threshold, a photon has insufficient energy to supply the work function, regardless of the intensity or exposure duration.

Why can photocurrent exist with a negative collector potential before the stopping value is reached?

Some emitted electrons have sufficient kinetic energy to overcome the retarding effect and reach the collector. Less energetic electrons are turned back.

What does a larger work function imply about threshold frequency?

It implies a higher threshold frequency because the minimum photon energy must equal the greater escape energy.

What does the horizontal intercept of a stopping-potential-versus-frequency graph represent?

It represents the threshold frequency, where the limiting maximum photoelectron kinetic energy and the stopping potential are zero.

What changes in a monochromatic beam when intensity rises but frequency stays fixed?

More photons pass through a given area per second. The energy and momentum of each photon remain unchanged.

How does de Broglie wavelength vary when momentum increases?

It decreases because wavelength is inversely proportional to momentum. Planck’s constant is the constant of proportionality in this relation.

What is the central conclusion of the Davisson-Germer experiment?

Electrons display diffraction and therefore wave behaviour. This supports the de Broglie hypothesis while retaining their particle properties.