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IIT Medico Guide In collaboration with Ashoka Group of Schools

Physics · Class 12 · Chapter 11

Dual Nature of Radiation and Matter

Light behaves as a wave in interference and as a stream of particles in the photoelectric effect, and electrons, which we think of as particles, show wave behaviour too. The chapter is short and formula-driven; the photoelectric graphs are the part students most often misread.

In this chapter: electron emission and work function, the photoelectric effect, effect of intensity, potential and frequency on photocurrent, why the wave theory fails, Einstein's photoelectric equation, the photon, de Broglie waves and the wavelength of an accelerated electron.

Electron emission

Free electrons in a metal are held inside by the attraction of the positive ions. The minimum energy needed to pull an electron out of the surface is the work function φ0, usually given in electron volts (1 eV = 1.602 × 10−19 J). It depends on the metal and on the nature of its surface; platinum has one of the highest values (about 5.65 eV) and caesium one of the lowest (about 2.14 eV). The energy can be supplied in three ways:

  • Thermionic emission: heating the metal.
  • Field emission: a very strong electric field (about 108 V m−1) pulls electrons out, as in a spark plug.
  • Photoelectric emission: light of suitable frequency falls on the surface.

The photoelectric effect

Hertz noticed in 1887 that ultraviolet light helped sparks jump across a gap. Hallwachs and Lenard then showed that when UV falls on a metal plate (the emitter) in an evacuated tube, a current flows to a collector plate. No current flows, however bright the light, if its frequency is below a certain value, the threshold frequency ν0. Alkali metals such as lithium, sodium, potassium, caesium and rubidium respond even to visible light.

Experimental results

  • Intensity: for a fixed frequency (above ν0) and fixed potential, the photocurrent is directly proportional to the intensity of light. Intensity decides the number of electrons emitted per second.
  • Potential: as the collector is made more positive, the current rises to a saturation value, when all emitted electrons reach it. Making the collector negative reduces the current, which becomes zero at the stopping potential V0. Then Kmax = eV0.
  • Stopping potential and intensity: for a given frequency, V0 does not depend on intensity. Brighter light gives more electrons, not faster ones.
  • Frequency: V0 (and so Kmax) increases linearly with frequency. Below ν0 there is no emission at all.
  • No time lag: emission starts within about 10−9 s of switching on the light, even for very dim light.
Photoelectric effect graphswww.iitmedicoguide.comVII₃I₂I₁−V₀(a) Same ν, intensities I₃ > I₂ > I₁saturation current ∝ intensity;same stopping potential V₀νV₀metal Ametal Bν₀(A)ν₀(B)−φ₀(A)/e−φ₀(B)/e(b) Stopping potential vs frequencyparallel lines: slope = h/e for every metalintercept on ν axis = threshold frequency ν₀www.iitmedicoguide.com
Left: for one frequency, brighter light raises the saturation current but leaves the stopping potential unchanged. Right: stopping potential rises linearly with frequency, with the same slope h/e for every metal but a different threshold frequency.

Why the wave theory fails

On the wave picture, energy is spread uniformly over the wavefront and absorbed continuously. A brighter wave should then give faster electrons, any frequency should work if the light is bright enough, and for dim light electrons would need hours to collect enough energy. All three predictions disagree with experiment.

Einstein's photoelectric equation

Einstein proposed in 1905 that light energy comes in packets (quanta) of energy hν, each absorbed completely by a single electron. Part of the energy frees the electron and the rest becomes its kinetic energy:

Kmax = hν − φ0eV0 = hν − φ0,   so V0 = heν − φ0eThreshold: ν0 = φ0h,   λ0 = hcφ0h = 6.63 × 10−34 J s; a handy value is hc ≈ 1240 eV nm

The equation explains every observation. Emission needs hν ≥ φ0, hence the threshold. Kmax depends on ν and not on intensity. More intense light means more photons and so more electrons. And a single photon hands over its energy at once, so there is no delay. Millikan's careful measurements confirmed the straight-line graph of V0 against ν, and its slope gave a value of h in agreement with other methods.

Worked example: Light of wavelength 500 nm falls on a metal of work function 2.0 eV. Find the photon energy, the maximum kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength.
Solution: E = hc/λ = 1240/500 = 2.48 eV. Kmax = 2.48 − 2.0 = 0.48 eV, so V0 = 0.48 V. Threshold wavelength λ0 = 1240/2.0 = 620 nm; light of wavelength longer than 620 nm will not eject electrons.

The photon

  • In interaction with matter, radiation behaves as if made of particles called photons.
  • Each photon has energy E = hν = hc/λ and momentum p = hν/c = h/λ, and travels at c in vacuum.
  • All photons of a given frequency have the same energy and momentum, whatever the intensity. Raising the intensity only raises the number of photons crossing unit area per second.
  • Photons are electrically neutral and are not deflected by electric or magnetic fields.
  • In a photon-particle collision, total energy and total momentum are conserved, but photons may be absorbed or created, so their number need not be conserved.

Wave nature of matter

De Broglie argued in 1924 that if radiation has a dual nature, so should matter. A particle of momentum p has an associated wavelength

λ = hp = hmv = h√(2mK)Electron accelerated through V volts: λ = h√(2meV) = 1.227√V nm

For everyday objects the wavelength is so small that no wave effect can ever be seen. For a 0.12 kg ball at 20 m s−1, λ = 6.63 × 10−34/(0.12 × 20) ≈ 2.8 × 10−34 m. For an electron accelerated through 100 V, λ ≈ 0.123 nm, about the spacing between atoms in a crystal, which is why electron beams show diffraction by crystals and why electron microscopes can resolve far finer detail than light microscopes.

A particle and a photon with the same wavelength have the same momentum but not the same energy. For particles with the same kinetic energy, the heavier one has the shorter wavelength, since λ ∝ 1/√m.

Worked example: Find the de Broglie wavelength of an electron accelerated through 400 V, and that of an alpha particle with the same kinetic energy as a proton moving with λ = 2 pm.
Solution: Electron: λ = 1.227/√400 = 1.227/20 ≈ 0.061 nm. For the second part, λ = h/√(2mK); with the same K, λ ∝ 1/√m. The alpha particle has four times the mass of the proton, so λα = 2/√4 = 1 pm.
Common mistakes: (1) Thinking brighter light gives faster photoelectrons; intensity changes only the number. (2) Reading the slope of the V0 against ν graph as h; it is h/e (the slope of Kmax against ν is h). (3) Mixing units: convert eV to joules, or use hc = 1240 eV nm consistently. (4) Using λ = 1.227/√V nm for particles other than electrons. (5) Forgetting that Kmax is the maximum; most photoelectrons come out slower because they lose energy before leaving the surface.

JEE and NEET focus

  • Einstein's equation with threshold frequency, threshold wavelength and stopping potential.
  • Reading and sketching current-voltage graphs and V0 against ν graphs, and how they change with intensity, frequency and metal.
  • Photon energy, momentum and number of photons emitted per second by a source of given power.
  • de Broglie wavelength of electrons, protons and alpha particles; comparisons at the same speed, momentum, kinetic energy or accelerating voltage.

Practice questions

The stopping potential in the photoelectric effect depends on:

  1. the intensity of light
  2. the frequency of light
  3. the area of the emitter
  4. the time of exposure
Show answer
B. eV0 = hν − φ0.

The intensity of light of fixed frequency (above threshold) is doubled. Then:

  1. stopping potential doubles
  2. saturation current doubles, stopping potential unchanged
  3. both double
  4. neither changes
Show answer
B. Twice as many photons eject twice as many electrons with the same Kmax.

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

  1. h
  2. e/h
  3. h/e
  4. φ0/e
Show answer
C. V0 = (h/e)ν − φ0/e.

The work function of a metal is 4.2 eV. Its threshold wavelength is about:

  1. 195 nm
  2. 295 nm
  3. 420 nm
  4. 520 nm
Show answer
B. λ0 = 1240/4.2 ≈ 295 nm.

The energy of a photon of wavelength 620 nm is:

  1. 1 eV
  2. 2 eV
  3. 3 eV
  4. 6.2 eV
Show answer
B. E = 1240/620 = 2 eV.

An electron and a proton have the same kinetic energy. The one with the shorter de Broglie wavelength is:

  1. the electron
  2. the proton
  3. neither; they are equal
  4. it depends on the speed of light
Show answer
B. λ = h/√(2mK); the proton is heavier.

An electron is accelerated from rest through 100 V. Its de Broglie wavelength is about:

  1. 1.23 nm
  2. 0.123 nm
  3. 12.3 nm
  4. 0.0123 nm
Show answer
B. λ = 1.227/√100 nm.

Light of frequency 1.5ν0 falls on a metal of threshold frequency ν0. If the frequency is raised to 2ν0, the maximum kinetic energy of the photoelectrons becomes:

  1. the same
  2. 1.33 times
  3. twice
  4. four times
Show answer
C. K = h(ν − ν0): 0.5hν0 becomes hν0.
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