Dual Nature of Radiation and Matter · TOPIC LESSON
Photoelectric Effect
Separate energy per photon, photon arrival rate and particle momentum to make duality problems predictable.
JEE Main · JEE Advanced · NEET UG75 min study guide9 indexed sections
BY THE END, YOU CAN
Explain photoelectric thresholds, stopping potential and intensity/frequency graph differences.
Calculate photon energy, momentum, count rate and radiation pressure with stated geometry.
Use Compton conservation results and identify their assumptions.
Calculate matter wavelengths and connect electron diffraction to experimental evidence.
BEFORE YOU START
Potential energy qV, kinetic energy and momentum.
Basic wave frequency/wavelength relations, conservation laws and trigonometry.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Main 2020. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
JEE Main · 2020 · 8 January · Shift 1 · Paper 1 · Physics · export Q17
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
Illumination of metals A and B uses photons of 4.0 eV and 4.5 eV respectively. Their fastest photoelectrons have kinetic energies T_A and T_B, with T_B=T_A−1.5 eV, and wavelengths λ_B=2λ_A. Find B’s work function.
Reveal answer & worked solution
Answer: 4.0 eV (independently derived; source option 2).
For nonrelativistic electrons, λ=h/√(2mT), so doubling wavelength reduces kinetic energy by four: T_B=T_A/4.
Combine T_A/4=T_A−1.5 eV to obtain T_A=2.0 eV and T_B=0.5 eV.
Source: JEE (Main) 2020, 8 January, first shift, Paper 1, Physics export Q17; independently derived solution; final-key status unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Light transfers energy in quanta
A photon of frequency ν has energy E = hν = hc/λ in vacuum and momentum magnitude p = E/c = h/λ. It has zero rest mass, yet carries energy and momentum. Do not substitute a photon into the nonrelativistic massive-particle formula K = p²/(2m); the energy–momentum relation for light is different.
Frequency determines energy per photon. Intensity describes energy arriving per area per time. At fixed frequency, increasing intensity increases the photon arrival rate rather than increasing each photon’s energy. A beam’s total energy may be large because it contains many low-energy photons, which does not make each photon capable of overcoming a larger single-photon threshold.
For monochromatic power P, photon rate is P/(hν). Over time t the number delivered is Pt/(hν) if all the stated power belongs to that wavelength. When a beam crosses between transparent media, frequency remains fixed at a stationary interface while phase speed and wavelength change. The simple vacuum photon momentum h/λ with λ interpreted as vacuum wavelength is the relation used here; momentum in material media involves additional electromagnetic–material distinctions.
The electron-volt is an energy unit: 1 eV is the energy associated with one elementary charge across one volt, equal to approximately 1.602 × 10⁻¹⁹ J. The convenient product hc ≈ 1240 eV nm converts between photon energy in eV and vacuum wavelength in nm. Keep the length units explicit.
Eγ = hν = hc/λ; pγ = Eγ/c = h/λ
Vacuum photon relations. λ is vacuum wavelength.
Photon rate = P/Eγ
Monochromatic beam of power P; this is a count rate, not an electric current.
ORIGINAL WORKED EXAMPLE
Count photons rather than confuse power and energy
A monochromatic 620 nm beam has power 2.0 mW. Estimate photon energy and photon arrival rate using hc = 1240 eV nm and 1 eV = 1.6 × 10⁻¹⁹ J.
Photon energy is 1240/620 = 2.0 eV = 3.2 × 10⁻¹⁹ J.
Power is 2.0 × 10⁻³ J s⁻¹, so rate is (2.0 × 10⁻³)/(3.2 × 10⁻¹⁹) = 6.25 × 10¹⁵ s⁻¹.
Doubling the power at the same wavelength would double this rate but leave photon energy at 2.0 eV.
Answer 2.0 eV per photon; 6.25 × 10¹⁵ photons s⁻¹.
02 / UNDERSTAND & APPLY
2. Photoemission separates intensity from frequency
In the photoelectric effect, incident light ejects electrons from a material if the photon energy can overcome the relevant work function φ. In the elementary single-photon model, an absorbed photon transfers its energy to one electron. Some energy is needed to escape the material; the most energetic emitted electrons retain the remainder as kinetic energy.
Einstein’s relation is Kmax = hν − φ. Emission requires hν ≥ φ in the ideal threshold model, so threshold frequency ν₀ = φ/h and longest threshold wavelength λ₀ = hc/φ. Below threshold, increasing ordinary light intensity does not produce one-photon emission because each photon still lacks the necessary energy. Strong-field multiphoton processes are outside this model.
Above threshold, increasing frequency increases maximum kinetic energy. Increasing intensity at fixed frequency generally increases the number of emitted electrons per second, and thus saturation current, if the surface response is unchanged. It does not raise Kmax. Real electrons emerge with a distribution of kinetic energies because their initial states and paths to the surface differ.
A retarding potential opposes the emitted electrons. The stopping potential magnitude Vs satisfies eVs = Kmax for the fastest electrons. At sufficiently large accelerating collection voltage, the current approaches a saturation value when essentially all emitted electrons are collected. Stopping potential measures the energy limit; saturation current reflects the emission rate.
Kmax = hν − φ = eVs; ν₀ = φ/h
Single-photon photoemission. Vs is the positive magnitude of the stopping potential.
λ₀ = hc/φ
Longest vacuum wavelength capable of threshold emission in the ideal work-function model.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Qualitative polylines only, not a model of the emission-energy distribution. Same material and frequency; only intensity differs.
ORIGINAL WORKED EXAMPLE
Threshold and stopping voltage
A metal has work function 2.5 eV and is illuminated by 310 nm light. Use hc = 1240 eV nm. Find maximum kinetic energy, stopping potential and threshold wavelength.
Photon energy is 1240/310 = 4.0 eV.
Maximum kinetic energy is 4.0 − 2.5 = 1.5 eV, so the stopping potential magnitude is 1.5 V.
Threshold wavelength is 1240/2.5 = 496 nm. Light of wavelength greater than this is below the single-photon threshold.
Answer Kmax = 1.5 eV; Vs = 1.5 V; λ₀ = 496 nm.
03 / UNDERSTAND & APPLY
3. Read the intercepts as physical parameters
A graph of Kmax against frequency is a straight line above threshold, with slope h and frequency-axis intercept ν₀. Extending the line mathematically to zero frequency gives energy intercept −φ; negative kinetic energy is not a physical emission result. Below threshold the statement is no one-photon emission, not electrons with negative kinetic energy.
A stopping-potential versus frequency graph has slope h/e and intercept −φ/e. For two metals with different work functions but the same fundamental h, the lines are parallel in the ideal model and have different thresholds. A change in light intensity moves current levels, not this energy–frequency slope.
A Kmax versus wavelength graph is not linear because photon energy is proportional to 1/λ. A Kmax versus inverse wavelength graph is linear with slope hc. Check the horizontal axis before extracting h or comparing intercepts. A graph’s visual straightness alone does not identify which physical constant its slope represents.
Photoelectric observations challenged a simple continuous-energy account: a threshold, prompt emission at ordinary accessible time scales and energy controlled by frequency are naturally explained by quantised energy exchange. This does not erase light’s interference behaviour. The experiment tests how light exchanges energy with matter, not whether every optical phenomenon can be replaced by classical particles.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Schematic scales. Physical lines stop at threshold; no negative kinetic-energy emission is implied.
ORIGINAL WORKED EXAMPLE
Recover Planck’s constant from a slope
Increasing light frequency by 5.00 × 10¹⁴ Hz raises stopping potential by 2.07 V for the same surface. Estimate h using e = 1.60 × 10⁻¹⁹ C.
Subtract the two photoelectric equations: eΔVs = hΔν, eliminating the work function.
Thus h = eΔVs/Δν = (1.60 × 10⁻¹⁹)(2.07)/(5.00 × 10¹⁴).
The result is 6.62 × 10⁻³⁴ J s to three significant figures.
Answer h ≈ 6.62 × 10⁻³⁴ J s.
04 / UNDERSTAND & APPLY
4. Photon momentum produces a measurable push
A surface that absorbs light receives the incident normal momentum. At normal incidence, a beam of power P carries momentum per time P/c, so perfect absorption gives force P/c and pressure I/c, where I is intensity. A perfectly reflecting stationary mirror reverses the normal momentum and receives twice the impulse, giving force 2P/c and pressure 2I/c.
The factor of two comes from momentum change, not from doubling photon energy. For an ideal stationary mirror, reflected photons retain essentially the same frequency while their momentum direction reverses. If the mirror moves appreciably, photon energy changes and a Doppler/energy analysis is needed; the stationary approximation is what makes the simple formula work.
At oblique incidence, resolve momentum normal to the surface and also account for the projected beam interception area. If I is intensity per area perpendicular to the incident beam and θ is measured from the surface normal, pressure on a perfect specular reflector is 2I cos²θ/c. For a perfectly absorbing surface the normal pressure is I cos²θ/c, though there can also be a tangential force component.
For partial reflection at normal incidence with reflected energy fraction R and transmitted fraction T, the absorbed fraction is A = 1 − R − T. Under a simple same-medium, specular stationary model, normal force per intercepted power is (A + 2R)/c = (1 + R − T)/c. Specify what happens to the outgoing light before deciding the force.
ORIGINAL WORKED EXAMPLE
Absorb or reflect
A 3.0 W beam strikes a surface normally. Compare force for complete absorption and perfect reflection using c = 3.0 × 10⁸ m s⁻¹.
Absorbed momentum per second is P/c = 3.0/(3.0 × 10⁸) = 1.0 × 10⁻⁸ N.
Reflection reverses the photon momentum, doubling momentum transfer to 2.0 × 10⁻⁸ N.
Both forces point along the incident beam’s direction.
5. Scattering tests photon energy and momentum together
When a photon scatters from an initially stationary approximately free electron, both energy and momentum must be conserved. The outgoing photon can have lower energy and longer wavelength because the recoiling electron carries away kinetic energy and momentum. Treating light as energy without momentum cannot correctly describe the angular dependence.
The Compton wavelength shift is Δλ = h(1 − cosθ)/(mec), where θ is the photon’s scattering angle. The shift is zero in forward scattering and maximum at 180°. It depends on angle and electron mass, not the incident wavelength, although the fractional energy change does depend on the initial photon energy.
The stationary free-electron model is an approximation. Strong binding and collective material effects can change the interpretation in other regimes. For simple calculations, use the electron Compton wavelength h/(mec) ≈ 2.43 pm, add the shift to the initial photon wavelength, and then compute outgoing photon energy with hc/λ. The electron’s kinetic energy is the lost photon energy.
Compton scattering is conceptually distinct from photoelectric absorption. In one, an outgoing photon remains with changed direction and energy; in the other, the incident photon is absorbed and an electron escapes a material. Both support quantised energy and momentum transfer, but they use different conservation equations and physical conditions.
ORIGINAL WORKED EXAMPLE
A right-angle Compton event
A photon with initial wavelength 5.00 pm scatters through 90° from a stationary free electron. Take h/(mec) = 2.43 pm. Find outgoing wavelength and the fraction of initial photon energy retained.
At 90°, 1 − cosθ = 1, so wavelength shift is 2.43 pm.
Outgoing wavelength is 5.00 + 2.43 = 7.43 pm.
Photon energy is inversely proportional to wavelength, so E′/E = 5.00/7.43 ≈ 0.673. The electron receives the remaining approximately 32.7% as kinetic energy.
Answer 7.43 pm; about 67.3% of the initial photon energy retained.
06 / UNDERSTAND & APPLY
6. Momentum sets the matter wavelength
De Broglie’s relation assigns wavelength λ = h/p to matter with momentum p. For a nonrelativistic particle of mass m and kinetic energy K, p = √(2mK), so λ = h/√(2mK). Equal kinetic energies do not give equal wavelengths for different masses. Equal momenta do give equal wavelengths, regardless of mass.
A particle of charge magnitude |q| accelerated from rest through a potential difference of magnitude V gains kinetic energy |q|V when losses are neglected. In the nonrelativistic regime, λ = h/√(2m|q|V). For electrons this becomes approximately 12.27/√V ångström when V is in volts. At high voltages, relativistic momentum must replace the nonrelativistic expression.
For a composite neutral object the de Broglie relation still concerns its centre-of-mass momentum in an appropriate quantum description, but macroscopic mass and momentum make the wavelength extraordinarily small and environmental interactions suppress observable interference. Matter-wave behaviour is not restricted to electrically charged particles; charge merely provides a convenient way to accelerate electrons in an experiment.
The wavelength is not the diameter of a particle, the radius of an orbit or the distance the particle travels before colliding. It describes phase variation of the quantum wave associated with momentum. A localized particle requires a superposition of momentum components rather than one infinitely extended monochromatic wave.
λ = h/p; nonrelativistically λ = h/√(2mK)
Use relativistic momentum when K is not small compared with mc².
λelectron ≈ 12.27/√V Å
Electron accelerated from rest through V volts, nonrelativistic approximation, negligible losses.
ORIGINAL WORKED EXAMPLE
An electron accelerated through a voltage
Find the de Broglie wavelength of an electron accelerated from rest through 150 V using λ(Å) = 12.27/√V.
The electron gains 150 eV, much smaller than its rest energy about 511 keV, so the nonrelativistic estimate is appropriate.
λ = 12.27/√150 ≈ 1.00 Å = 0.100 nm.
If the accelerating voltage were quadrupled, momentum would double and wavelength would halve in this regime.
Answer Approximately 1.00 Å.
07 / UNDERSTAND & APPLY
7. Diffraction patterns and discrete detections answer different questions
Electron diffraction from a crystal demonstrates a wavelength because scattered amplitudes reinforce only for suitable path differences. In a simple Bragg description, constructive interference from parallel crystal planes of spacing d occurs when 2d sinθ = nλ, where θ is the glancing angle to the planes. The angle convention is crucial; the angle between incident and diffracted beams is 2θ in the usual symmetric geometry.
The Davisson–Germer experiment observed electron scattering intensities with diffraction structure consistent with de Broglie wavelengths. Its significance is the agreement between an electron wavelength inferred from momentum and one inferred from interference geometry. One need not treat electrons as tiny classical waves of spread-out charge to acknowledge this evidence.
In a two-slit experiment with sufficiently coherent electrons or photons, individual detection events are localized, while many events build an interference distribution when which-path information is not available in the relevant experimental arrangement. The quantum description connects amplitudes to probabilities. Saying a particle “must have been a classical ball through one known slit” is not an experimentally justified explanation of the full interference pattern.
Wave–particle duality is therefore not an instruction to assign two incompatible classical pictures simultaneously. Use wave superposition to predict interference probabilities and quantum energy/momentum relations for discrete interactions. The experiment determines which observable is being tested. The elementary formulas here do not claim to replace the full quantum theory.
ORIGINAL WORKED EXAMPLE
Diffraction as a wavelength measurement
Electrons of wavelength 0.20 nm diffract from planes spaced 0.20 nm apart. Find the first-order Bragg glancing angle.
Set n = 1 in 2d sinθ = λ.
sinθ = 0.20/[2(0.20)] = 0.5, so θ = 30°.
The corresponding angle between incident and reflected beam directions in the standard symmetric geometry is 60°.
Answer Glancing angle 30°.
08 / UNDERSTAND & APPLY
8. Original practice: identify the interaction first
Use hc = 1240 eV nm where helpful. These are original exercises, not quoted examination questions.
ORIGINAL PRACTICE · 1
Compare photon energies at 400 nm and 800 nm.
Reveal answer & reasoning
The 400 nm photon has twice the energy.
E = hc/λ, so halving wavelength doubles energy.
ORIGINAL PRACTICE · 2
At fixed light frequency above threshold, intensity triples. In the elementary model, what changes: stopping potential or saturation current?
Reveal answer & reasoning
Saturation current approximately triples; stopping potential is unchanged.
Photon rate rises while energy per photon stays fixed, assuming unchanged surface yield and collection.
ORIGINAL PRACTICE · 3
A metal work function is 3 eV. Can 2 eV photons eject electrons by the ordinary single-photon process if intensity is made very large?
Reveal answer & reasoning
No within the stated single-photon model.
Each photon is below threshold; strong-field multiphoton effects are a different model.
ORIGINAL PRACTICE · 4
A 5 eV photon ejects an electron from a 2 eV work-function surface. Find Kmax and stopping potential.
Reveal answer & reasoning
3 eV and 3 V.
Subtract the work function; dividing 3 eV by one elementary charge gives 3 V.
ORIGINAL PRACTICE · 5
What is the threshold wavelength for φ = 4 eV?
Reveal answer & reasoning
310 nm.
λ₀ = 1240/4 nm.
ORIGINAL PRACTICE · 6
A photon’s momentum doubles. What happens to its vacuum frequency?
Reveal answer & reasoning
It doubles.
E = pc and E = hν, so frequency is proportional to momentum.
ORIGINAL PRACTICE · 7
A stationary perfect mirror receives a normally incident photon of momentum p. What impulse magnitude does it receive?
Reveal answer & reasoning
2p.
The photon changes momentum from +p to −p along the normal, so the mirror gains +2p.
ORIGINAL PRACTICE · 8
What is the maximum Compton wavelength shift for an electron with Compton wavelength 2.43 pm?
Reveal answer & reasoning
4.86 pm.
At 180°, 1 − cosθ = 2.
ORIGINAL PRACTICE · 9
A proton and electron have equal nonrelativistic kinetic energy. Which has longer de Broglie wavelength?
Reveal answer & reasoning
The electron.
λ ∝ 1/√m at fixed kinetic energy; the lower mass has lower momentum.
ORIGINAL PRACTICE · 10
Two different particles have the same momentum. Compare their de Broglie wavelengths.
Reveal answer & reasoning
They are equal.
λ = h/p depends directly on momentum, not separately on mass.
ORIGINAL PRACTICE · 11
An electron accelerating voltage rises from 100 V to 900 V. How does its nonrelativistic wavelength change?
Reveal answer & reasoning
It becomes one third.
λ ∝ 1/√V and voltage rises by a factor of nine.
ORIGINAL PRACTICE · 12
In Bragg diffraction, d = 0.30 nm and θ = 30° for first order. Find wavelength.
Reveal answer & reasoning
0.30 nm.
λ = 2d sinθ = 2(0.30)(0.5).
09 / UNDERSTAND & APPLY
9. Keep the photon, particle and beam quantities separate
Begin by identifying whether the question concerns a whole beam, one photon, an emitted electron or a matter-wave interference pattern. Beam power determines photon count rate only after frequency is known. Photoelectric kinetic energy subtracts a work function; Compton scattering balances outgoing photon and recoil; de Broglie wavelength uses particle momentum.
Check the regime and geometry last: vacuum or medium, nonrelativistic or relativistic, normal or oblique incidence, and Bragg glancing angle or full scattering angle. The equations are compact because these choices have already been made. Stating them is part of understanding, not an optional disclaimer.
Frequency sets photon energy; intensity sets energy flow.
Stopping voltage measures maximum electron kinetic energy.
A photon has momentum despite zero rest mass.
Radiation pressure follows momentum transfer and reflection geometry.
Matter wavelength is h/p and is not a particle size.
Diffraction evidence and localized detection are both essential to quantum behaviour.
KEEP THE BIG IDEAS
Quick revision
A photon’s energy and a beam’s intensity answer different questions.
Photoemission has a work-function threshold in the single-photon model.
Both photons and matter carry momentum relevant to their wavelength.
Use the experimental context to choose a quantum relation instead of forcing one classical picture.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.