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Class 12 · Physics NCERT Class 12 Physics · Ch. 115 min read · 15 questions

Dual Nature of Radiation and Matter

Physics

Dual Nature of Radiation and Matter

One of the most revolutionary ideas in physics is that both radiation and matter possess a dual nature — behaving as waves in some experiments and as particles in others. This chapter explores how classical physics failed to explain certain phenomena, leading to the quantum revolution.

Photoelectric Effect

When electromagnetic radiation of sufficient frequency falls on a metal surface, electrons are emitted. This is the photoelectric effect, discovered by Hertz and explained by Einstein in 1905.

  • Key Observations:
  • Emission occurs only when frequency of light exceeds a threshold frequency nu0 (characteristic of the metal).
  • Kinetic energy of emitted photoelectrons is independent of intensity.
  • Number of photoelectrons emitted is proportional to intensity of light.
  • Emission is instantaneous (no time delay).

Classical wave theory fails to explain the threshold frequency and instantaneous emission.

Einstein's Photoelectric Equation:
K.E.max = h nu - phi = h nu - h nu0

  • where:
  • h = Planck's constant = 6.626 x 10-34 J s
  • nu = frequency of incident radiation
  • phi = h nu0 = work function of metal
  • nu0 = threshold frequency

Stopping Potential (V0): The minimum retarding potential that stops all photoelectrons.
e V0 = K.E.max = h nu - phi
So V0 = h nu / e - phi / e

Photoelectric current increases with intensity (more photons, more electrons) but is cut off at V = -V0 regardless of intensity.

Photon

  • Einstein proposed that light is composed of discrete packets of energy called photons.
  • Energy of photon: E = h nu = hc / lambda
  • Momentum of photon: p = h / lambda = h nu / c
  • Rest mass of photon: zero
  • Speed: c in vacuum

de Broglie Hypothesis

Louis de Broglie (1924) proposed that matter also has a wave nature. For a particle of mass m moving with velocity v:
de Broglie wavelength: lambda = h / (mv) = h / p

This was experimentally confirmed by Davisson and Germer (1927) through electron diffraction.

For electron accelerated through potential V:
lambda = h / √(2 m e V) = 1.227 / √(V) nm (approx, with V in volts)

Heisenberg Uncertainty Principle

deltax x deltap ≥ h / (4 pi)
deltaE x deltat ≥ h / (4 pi)

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Example 1

The work function of sodium is 2.3 eV. Find the threshold frequency.
nu0 = phi/h = (2.3 x 1.6 x 10-19) / (6.626 x 10-34) = 3.68 x 10-19 / 6.626 x 10-34 = 5.55 x 1014 Hz.

Example 2

Light of frequency 6 x 1014 Hz falls on sodium (phi = 2.3 eV). Find maximum kinetic energy.
K.E.max = h nu - phi = (6.626 x 10-34 x 6 x 1014) - (2.3 x 1.6 x 10-19) = 3.976 x 10-19 - 3.68 x 10-19 = 2.96 x 10-20 J = 0.185 eV.

Example 3

Find the de Broglie wavelength of an electron moving at 2 x 106 m/s. (m = 9.1 x 10-31 kg)
lambda = h/(mv) = 6.626 x 10-34 / (9.1 x 10-31 x 2 x 106) = 6.626 x 10-34 / 1.82 x 10-24 = 3.64 x 10-10 m = 0.364 nm.

Example 4

An electron is accelerated through 100 V. Find its de Broglie wavelength.
lambda = 1.227 / √(100) nm = 1.227 / 10 = 0.1227 nm.

Example 5

Find stopping potential for a metal with work function 1.8 eV when light of wavelength 300 nm is used. (h = 6.626 x 10-34 J s)
E = hc/lambda = (6.626 x 10-34 x 3 x 108) / (300 x 10-9) = 6.626 x 10-19 J = 4.14 eV.
V0 = (E - phi)/e = (4.14 - 1.8) = 2.34 V.

Example 6

Photons of energy 3.5 eV are used on a metal of work function 2.0 eV. Find maximum speed of emitted electrons. (m = 9.1 x 10-31 kg)
KE = (3.5 - 2.0) x 1.6 x 10-19 = 2.4 x 10-19 J. (1/2)mv2 = 2.4 x 10-19. v = √(4.8 x 10-19 / 9.1 x 10-31) = √(5.27 x 1011) = 7.26 x 105 m/s.

Example 7

Compare de Broglie wavelengths of a proton and an alpha particle accelerated through the same potential V.
lambdaproton / lambdaalpha = √(malpha qalpha) / √(mproton qproton). malpha = 4mp, qalpha = 2e, qp = e. Ratio = √(4 x 2) / √(1 x 1) = √(8) = 2 √(2). So lambdaproton = 2 √(2) x lambdaalpha.

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Common mistakes

  • Using frequency in threshold formula but wavelength in energy formula without converting.
  • Forgetting to convert eV to Joules (1 eV = 1.6 x 10-19 J).
  • Confusing stopping potential (in volts) with maximum KE (in joules or eV).

Summary

The photoelectric effect proves light has particle (photon) nature. Einstein's equation KEmax = h nu - phi explains all observations. de Broglie extended this duality to matter: lambda = h/p. Wave-particle duality is confirmed experimentally by electron diffraction.

Practice Problems

15 questions with instant feedback.

Question 1 of 15Score 0

The photoelectric effect was explained by: