JEE Main Physics · Chapter 20

Dual Nature of Matter and Radiation

Photoelectric effect, Einstein’s equation, photons and de Broglie matter waves.

NCERT Class 12JEE Main: 1 Q per shift 8 min read

1. Chapter at a Glance

Light behaves as a wave in interference, but as a stream of particles (photons) when it knocks electrons out of metals. Matter, in turn, shows wave behaviour. This chapter — Dual Nature of Matter and Radiation — covers the photoelectric effect, Einstein's equation, photon energy and momentum, de Broglie waves and the Davisson–Germer experiment. It belongs to "modern physics", the highest-yield area of JEE Main Physics.

What JEE asks most
  • Einstein's photoelectric equation: Kmax, stopping potential, threshold wavelength.
  • Graphs: photocurrent vs voltage and stopping potential vs frequency.
  • Effect of intensity vs frequency on photoelectrons.
  • de Broglie wavelength of electrons, protons and neutrons.

2. The Photoelectric Effect

  • Light of frequency above a threshold ν₀ ejects electrons from a metal surface instantly (no time lag).
  • The maximum kinetic energy of photoelectrons depends on frequency, not on intensity. The number of photoelectrons (current) is proportional to intensity.
  • These facts cannot be explained by the wave theory; Einstein explained them with photons.
hν = φ + Kmax    Kmax = eV₀    φ = hν₀ = hc/λ₀
  • Stopping potential V₀: the reverse voltage that just stops the fastest electrons.
  • V₀ versus ν is a straight line of slope h/e and intercept ν₀ on the frequency axis (the same slope for every metal).
  • Saturation current rises with intensity; the stopping potential stays the same for the same frequency.
ν₀(A)ν₀(B) metal Ametal B slope = h/e (same)V₀ν
Stopping potential vs frequency: parallel lines for different metals; the intercept gives the threshold frequency.

3. Photons

  • Energy E = hν = hc/λ; a handy form: E(eV) = 1240/λ(nm).
  • Momentum p = h/λ = E/c; photons have zero rest mass and always travel at c.
  • Number of photons emitted per second by a source of power P: n = Pλ/hc.
  • Photons transfer momentum when absorbed or reflected (radiation pressure).

4. Matter Waves

λ = h/p = h/mv = h/√(2mK)
  • Electron accelerated through V volts: λ = 1.227/√V nm.
  • Particle with thermal energy: λ = h/√(3mkT).
  • For equal kinetic energy, lighter particles have longer wavelengths (λ ∝ 1/√m). For equal momentum, all have the same λ.
  • Davisson–Germer experiment: electrons of 54 eV diffracted by a nickel crystal gave a wavelength matching de Broglie's prediction — the first proof of matter waves. Electron microscopes use the short wavelength of fast electrons.
  • Everyday objects have immeasurably small de Broglie wavelengths, so their wave nature is not noticed.

5. Formula Master Table

QuantityFormula
Photon energyE = hν = hc/λ = 1240/λ(nm) eV
Photon momentump = h/λ
Einstein's equationhν = φ + Kmax; Kmax = eV₀
Thresholdφ = hν₀ = hc/λ₀
Photon fluxn = Pλ/hc
de Broglieλ = h/√(2mK)
Electron through Vλ = 1.227/√V nm

6. Mind Map

Dual Nature
Photoelectric effect
  • Threshold frequency
  • Intensity vs frequency
Einstein
  • hν = φ + Kmax
  • V₀–ν graph
Photons
  • E = hc/λ
  • p = h/λ
Matter waves
  • λ = h/p
  • Davisson–Germer

7. Tips and Tricks

Shortcuts
  • Work in electron-volts with hc = 1240 eV nm — it saves a lot of arithmetic.
  • Doubling intensity doubles the current but leaves V₀ unchanged; raising frequency raises V₀.
  • For two frequencies on the same metal: h = e(V₂ − V₁)/(ν₂ − ν₁).
  • λp/λe for equal kinetic energy = √(me/mp) ≈ 1/43.
Mistakes students make
  • Saying more intense light gives faster photoelectrons.
  • Forgetting that no photoemission occurs below the threshold, however bright the light.
  • Using λ = h/mv for photons with a rest mass (photons have none — use p = E/c).

8. Important One-liners

  • Photoelectric emission is instantaneous.
  • Maximum kinetic energy of photoelectrons depends on frequency, not intensity.
  • The slope of the V₀–ν graph is h/e for all metals.
  • The Davisson–Germer experiment confirmed the wave nature of electrons.
  • Photons carry momentum even though they have no rest mass.
  • Electron microscopes resolve finer detail than optical microscopes because electron wavelengths are shorter.

9. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, e = 1.6 × 10⁻¹⁹ C, hc = 1240 eV nm.

Average level

N1Average

Find the energy of a photon of wavelength 500 nm in electron-volts.

Show answer & solution

Concept: E(eV) = 1240/λ(nm).

  1. E = 1240/500.
  2. = 2.48 eV (≈ 3.97 × 10⁻¹⁹ J).

Answer: 2.48 eV

N2Average

Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find Kmax and the stopping potential.

Show answer & solution

Concept: Kmax = hν − φ; V₀ = Kmax/e.

  1. Photon energy = 1240/400 = 3.1 eV.
  2. Kmax = 3.1 − 2.0 = 1.1 eV.
  3. V₀ = 1.1 V.

Answer: 1.1 eV; 1.1 V

N3Average

The work function of a metal is 2.5 eV. Find its threshold wavelength.

Show answer & solution

Concept: λ₀ = hc/φ.

  1. λ₀ = 1240/2.5.
  2. = 496 nm (blue-green); longer wavelengths will not eject electrons.

Answer: 496 nm

N4Average

Find the de Broglie wavelength of an electron accelerated from rest through 100 V.

Show answer & solution

Concept: λ = 1.227/√V nm.

  1. λ = 1.227/√100 = 1.227/10.
  2. = 0.1227 nm ≈ 1.23 Å (X-ray range).

Answer: about 0.123 nm

N5Average

A 10 W source emits light of wavelength 600 nm. How many photons does it emit per second?

Show answer & solution

Concept: n = P/(hc/λ).

  1. Energy per photon = 6.63 × 10⁻³⁴ × 3 × 10⁸/(6 × 10⁻⁷) ≈ 3.31 × 10⁻¹⁹ J.
  2. n = 10/(3.31 × 10⁻¹⁹) ≈ 3.0 × 10¹⁹ s⁻¹.

Answer: about 3 × 10¹⁹ photons per second

Advanced level

N6Advanced

For a metal, the stopping potential is 1 V at frequency ν₁ and 3 V at ν₂, where ν₂ − ν₁ = 4.84 × 10¹⁴ Hz. Estimate Planck's constant.

Show answer & solution

Concept: eV₀ = hν − φ; subtract the two equations to remove φ.

  1. e(V₂ − V₁) = h(ν₂ − ν₁).
  2. h = 1.6 × 10⁻¹⁹ × 2/4.84 × 10¹⁴.
  3. h ≈ 6.6 × 10⁻³⁴ J s.

Answer: about 6.6 × 10⁻³⁴ J s

N7Advanced

An electron and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths λe : λp (mp/me ≈ 1835).

Show answer & solution

Concept: λ = h/√(2mK) ∝ 1/√m for equal K.

  1. λe/λp = √(mp/me).
  2. = √1835 ≈ 42.8.

Answer: about 43 : 1

N8Advanced

Ultraviolet light of 300 nm falls on a metal of work function 2.2 eV. Find the maximum speed of the photoelectrons (me = 9.1 × 10⁻³¹ kg). What changes if the intensity is doubled?

Show answer & solution

Concept: Kmax = hc/λ − φ; v = √(2K/m).

  1. Photon energy = 1240/300 ≈ 4.13 eV; Kmax ≈ 1.93 eV = 3.09 × 10⁻¹⁹ J.
  2. v = √(2 × 3.09 × 10⁻¹⁹/9.1 × 10⁻³¹) ≈ 8.2 × 10⁵ m s⁻¹.
  3. Doubling intensity doubles the number of photoelectrons (current) but not their maximum speed.

Answer: about 8.2 × 10⁵ m s⁻¹; only the current doubles

N9Advanced

Find the de Broglie wavelength of a neutron at 300 K, using its average thermal kinetic energy (3/2)kT (mn = 1.675 × 10⁻²⁷ kg, k = 1.38 × 10⁻²³ J K⁻¹).

Show answer & solution

Concept: λ = h/√(2mK) = h/√(3mkT).

  1. 3mkT = 3 × 1.675 × 10⁻²⁷ × 1.38 × 10⁻²³ × 300 ≈ 2.08 × 10⁻⁴⁷.
  2. √(3mkT) ≈ 4.56 × 10⁻²⁴ kg m s⁻¹.
  3. λ = 6.63 × 10⁻³⁴/4.56 × 10⁻²⁴ ≈ 1.45 × 10⁻¹⁰ m — close to atomic spacing, so thermal neutrons diffract from crystals.

Answer: about 1.45 Å

N10Advanced

Find the momentum of a 600 nm photon. A 10 W laser beam of this light is completely absorbed by a surface; find the force on the surface.

Show answer & solution

Concept: p = h/λ; force = rate of momentum delivered = P/c.

  1. p = 6.63 × 10⁻³⁴/6 × 10⁻⁷ ≈ 1.1 × 10⁻²⁷ kg m s⁻¹.
  2. F = n × p = (P/E) × (E/c) = P/c.
  3. F = 10/(3 × 10⁸) ≈ 3.3 × 10⁻⁸ N.

Answer: 1.1 × 10⁻²⁷ kg m s⁻¹; 3.3 × 10⁻⁸ N

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The slope of the graph of stopping potential against frequency equals:

  1. h
  2. e/h
  3. h/e
  4. φ/e
Show solution

V₀ = (h/e)ν − φ/e, so the slope is h/e.

Answer: (C)

Q2Easy

If the intensity of incident light (above threshold) is increased, the maximum kinetic energy of photoelectrons:

  1. Increases
  2. Decreases
  3. Remains unchanged
  4. Becomes zero
Show solution

Kmax depends on frequency only; intensity changes the number of electrons.

Answer: (C)

Q3Numerical

A metal has threshold wavelength 620 nm. Find its work function in eV.

Show solution

φ = 1240/620 = 2 eV.

Answer: 2

Q4Medium

An electron is accelerated through V volts. If V is made four times, its de Broglie wavelength becomes:

  1. Four times
  2. Twice
  3. Half
  4. One-fourth
Show solution

λ ∝ 1/√V, so λ halves.

Answer: (C)

Q5Hard

The Davisson–Germer experiment established:

  1. The particle nature of light
  2. The wave nature of electrons
  3. The existence of the nucleus
  4. Quantisation of charge
Show solution

Electrons were diffracted by a crystal, just like waves, with λ matching h/p.

Answer: (B)

Q6Numerical

Photons of energy 5 eV fall on a metal with work function 3 eV. Find the stopping potential in volts.

Show solution

V₀ = (5 − 3) = 2 V.

Answer: 2

11. Quick Revision Summary

Remember before the exam
  • hν = φ + eV₀; φ = hc/λ₀; E(eV) = 1240/λ(nm).
  • Intensity → number of electrons; frequency → their energy.
  • V₀–ν slope = h/e, same for all metals.
  • λ = h/p = h/√(2mK); electron: 1.227/√V nm.
  • Photon momentum h/λ; force of absorbed beam P/c.
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