What this chapter is about
This chapter introduces one of the most revolutionary ideas in modern physics: that light and matter both exhibit a dual nature. Light, which was understood as an electromagnetic wave, also behaves as a stream of particles called photons. Similarly, matter particles like electrons, which we normally think of as tiny balls, can also behave as waves. This wave-particle duality forms the foundation of quantum mechanics.
You will study the photoelectric effect, where light falling on certain metal surfaces ejects electrons. This phenomenon could not be explained by classical wave theory and required Einstein's photon hypothesis. The chapter also covers de Broglie's bold proposal that all matter has an associated wavelength, which was later confirmed by electron diffraction experiments.
After studying this chapter, you should be able to explain the photoelectric effect using photon theory, calculate the kinetic energy of photoelectrons, find the de Broglie wavelength of moving particles, and understand why wave nature of matter is observable only for microscopic particles.
Key ideas
- Light exhibits particle nature in the photoelectric effect: it consists of discrete packets of energy called photons, each with energy E = hν, where h is Planck's constant and ν is the frequency.
- The photoelectric effect shows that electrons are emitted instantly when light of sufficient frequency falls on a metal, and the kinetic energy of emitted electrons depends on frequency, not intensity.
- Each metal has a threshold frequency ν₀ below which no electrons are emitted, regardless of how intense the light is. The corresponding minimum energy hν₀ is called the work function φ of the metal.
- Einstein's photoelectric equation, K_max = hν − φ, relates the maximum kinetic energy of photoelectrons to the incident photon energy and the work function.
- The stopping potential V₀ is the minimum retarding potential that stops the most energetic photoelectrons: eV₀ = K_max.
- de Broglie proposed that all matter has wave properties, with wavelength λ = h/p, where p is the momentum of the particle.
- The wave nature of electrons was confirmed experimentally by Davisson and Germer through electron diffraction patterns from nickel crystals.
- The de Broglie wavelength of everyday objects is negligibly small, which is why wave behaviour of matter is only observable for microscopic particles like electrons.
Formulas and facts to remember
- Energy of a photon: E = hν = hc/λ, where h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s.
- Einstein's photoelectric equation: K_max = hν − φ, or equivalently, eV₀ = hν − φ.
- Threshold frequency: ν₀ = φ/h; threshold wavelength: λ₀ = hc/φ.
- de Broglie wavelength: λ = h/p = h/(mv) for a particle of mass m and velocity v.
- For an electron accelerated through potential V: λ = h/√(2meV), which simplifies to λ ≈ 1.227/√V nm when V is in volts.
- Work function φ is the minimum energy needed to remove an electron from the metal surface; it is characteristic of the metal.
- One electron volt (eV) = 1.6 × 10⁻¹⁹ J.
- Photon momentum: p = h/λ = hν/c = E/c.
Worked examples
Example 1: Calculating maximum kinetic energy of photoelectrons
Light of wavelength 400 nm falls on a metal surface with work function 2.0 eV. Find the maximum kinetic energy of the emitted photoelectrons.
Solution: First, find the energy of the incident photon. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) J E = 4.97 × 10⁻¹⁹ J
Convert to eV: E = 4.97 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3.1 eV
Using Einstein's equation: K_max = hν − φ = 3.1 − 2.0 = 1.1 eV
The maximum kinetic energy of photoelectrons is 1.1 eV.
Example 2: Finding de Broglie wavelength of an electron
An electron is accelerated from rest through a potential difference of 150 V. Calculate its de Broglie wavelength.
Solution: When an electron is accelerated through potential V, it gains kinetic energy equal to eV. K = eV = 1.6 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J
The momentum is found from K = p²/(2m): p = √(2mK) = √(2 × 9.1 × 10⁻³¹ × 2.4 × 10⁻¹⁷) p = √(4.37 × 10⁻⁴⁷) = 6.6 × 10⁻²⁴ kg·m/s
de Broglie wavelength: λ = h/p = 6.63 × 10⁻³⁴ / 6.6 × 10⁻²⁴ = 1.0 × 10⁻¹⁰ m = 0.1 nm
This is comparable to atomic spacings, explaining why electron diffraction is observable.
Example 3: Threshold wavelength calculation
The work function of sodium is 2.3 eV. What is the maximum wavelength of light that can cause photoemission from sodium?
Solution: At the threshold, the photon energy exactly equals the work function. hν₀ = φ, so hc/λ₀ = φ
λ₀ = hc/φ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(2.3 × 1.6 × 10⁻¹⁹) λ₀ = 1.99 × 10⁻²⁵ / 3.68 × 10⁻¹⁹ = 5.4 × 10⁻⁷ m = 540 nm
Light with wavelength greater than 540 nm will not cause photoemission from sodium.
Common mistakes
- Thinking brighter light gives faster electrons → intensity affects the number of photoelectrons, not their energy; only frequency determines maximum kinetic energy.
- Using classical momentum p = mv for photons → photons have zero rest mass; use p = h/λ or p = E/c instead.
- Forgetting unit conversions between joules and electron volts → always check whether energy is in J or eV before substituting in equations.
- Expecting wave behaviour from macroscopic objects → the de Broglie wavelength of large objects is so tiny that wave effects cannot be detected.
- Confusing threshold frequency with any frequency → emission occurs only when incident frequency exceeds the threshold frequency of the metal.
Quick revision
- Photon energy is E = hν; photon momentum is p = h/λ.
- Photoelectric equation: K_max = hν − φ; below threshold frequency, no emission occurs.
- Stopping potential V₀ measures maximum kinetic energy: eV₀ = K_max.
- de Broglie wavelength λ = h/(mv); smaller for heavier or faster particles.
- Wave nature of matter is significant only for microscopic particles like electrons.
- Davisson-Germer experiment confirmed electron waves through diffraction from crystals.