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CalibratePhysics

11.7

Particle Nature of Light: The Photon

Dual Nature of Radiation and Matter · CBSE

Theory

What this section has to deliver

  • The list of photon properties — CBSE asks for "three basic properties of photons used to obtain Einstein's photoelectric equation" as a standing 2- or 3-mark question — plus the energy and momentum relations and the definition of intensity. Numericals here are one-liners.

From quanta to photons

  • The photoelectric effect showed that light, in interacting with matter, behaves as if made of packets of energy hνh\nu.
  • Einstein went further and showed that such a quantum also carries momentum hν/ch\nu/c. A definite energy together with a definite momentum is a strong sign of a particle, and the quantum was later named the photon.
  • A. H. Compton's 1924 experiment on the scattering of X-rays by electrons confirmed the particle-like behaviour of light directly.
  • Einstein received the 1921 Nobel Prize in Physics for his contribution to theoretical physics and the photoelectric effect; Millikan received the 1923 Prize for the elementary charge and for his work on the photoelectric effect.

The photon picture: the five statements

  • In the interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
  • Each photon has energy E=hνE = h\nu, momentum p=hν/cp = h\nu/c, and speed cc, the speed of light.
  • All photons of light of a particular frequency ν\nu, or wavelength λ\lambda, have the same energy E=hν=hc/λE = h\nu = hc/\lambda and the same momentum p=hν/c=h/λp = h\nu/c = h/\lambda, whatever the intensity of the radiation. Increasing the intensity of light of a given wavelength only increases the number of photons crossing a given area per second; each photon still carries the same energy. Photon energy is independent of intensity.
  • Photons are electrically neutral and are not deflected by electric and magnetic fields.
  • In a photon-particle collision, such as a photon-electron collision, the total energy and the total momentum are conserved. The number of photons, however, need not be conserved: a photon may be absorbed, or a new photon created.

The three properties used to establish Einstein's equation

  • In the interaction of a photon with a free electron, the entire energy of the photon is absorbed by that one electron.
  • The energy of a photon is directly proportional to its frequency, E=hνE = h\nu; and for a given frequency the intensity of the radiation is fixed by the number of photons falling per unit area per unit time.
  • In a photon-electron collision, total energy and total momentum are conserved.
  • Applying the third to the absorption event, with the escape costing at least ϕ0\phi_0, gives Kmax=hν−ϕ0K_{max} = h\nu - \phi_0.

Energy, momentum and intensity

E=hν=hcλp=Ec=hνc=hλλ=hcEE = h\nu = \dfrac{hc}{\lambda} \qquad p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda} \qquad \lambda = \dfrac{hc}{E}

  • A photon has zero rest mass; its momentum comes from E=pcE = pc, not from p=mvp = mv.

I=n hνA tI = \dfrac{n\,h\nu}{A\,t}

  • Intensity in the photon picture: the energy of the photons crossing unit area in unit time, where nn photons fall normally on area AA in time tt. SI unit W m−2^{-2}, equivalently J m−2^{-2} s−1^{-1}.
  • Contrast to state when asked: in the wave picture, intensity is determined by the square of the amplitude of the wave; in the photon picture, at a given frequency, it is determined by the number of photons crossing unit area per unit time.

N=Phν=PλhcN = \dfrac{P}{h\nu} = \dfrac{P\lambda}{hc}

  • Number of photons emitted per second by a source of power PP watt.

The KmaxK_{max} versus ν\nu plot

  • From Kmax=hν−ϕ0K_{max} = h\nu - \phi_0: a straight line of slope hh, cutting the ν\nu-axis at ν0=ϕ0/h\nu_0 = \phi_0/h and rising linearly for ν>ν0\nu > \nu_0. It is drawn only for ν≥ν0\nu \ge \nu_0; for ν<ν0\nu < \nu_0 there is no emission, so no KmaxK_{max} to plot. If extrapolated backwards the line would meet the vertical axis at −ϕ0-\phi_0.

Worked numericals

  • Wavelength of a photon of energy 3.3×10−193.3\times10^{-19} J: λ=hc/E=(6.63×10−34×3×108)/(3.3×10−19)≈6.03×10−7\lambda = hc/E = (6.63\times10^{-34}\times3\times10^{8})/(3.3\times10^{-19}) \approx 6.03\times10^{-7} m.
  • A laser of power 2.0×10−32.0\times10^{-3} W at 6.0×10146.0\times10^{14} Hz: each photon carries E=3.98×10−19E = 3.98\times10^{-19} J, so N=P/E=5.0×1015N = P/E = 5.0\times10^{15} photons per second.
  • Photon of wavelength 331.5331.5 nm: E=1240/331.5≈3.74E = 1240/331.5 \approx 3.74 eV, and p=E/c=(3.74×1.6×10−19)/(3×108)≈2×10−27p = E/c = (3.74\times1.6\times10^{-19})/(3\times10^{8}) \approx 2\times10^{-27} kg m s−1^{-1}.
  • X-rays from 3030 kV electrons (NCERT exercise): νmax=eV/h=7.24×1018\nu_{max} = eV/h = 7.24\times10^{18} Hz and λmin=hc/(eV)≈0.0414\lambda_{min} = hc/(eV) \approx 0.0414 nm.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

  1. 1

    What is the wavelength of a photon of energy 3.3×10−193.3\times10^{-19} J?

    1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 1 ↓
  2. 2

    Define intensity of radiation on the basis of the photon picture of light. Write its SI unit.

    1 markCBSE Class 12 Physics — Very Short Answer · AI 2014 · Section A, 1 markAnswer 2 ↓
  3. 3

    Write three basic properties of photons which are used to obtain Einstein's photoelectric equation. Use this equation to draw a plot of maximum kinetic energy of the electrons emitted versus the frequency of incident radiation.

    2 marksCBSE Class 12 Physics — SA · AI 2014C · Section B, 2 marksAnswer 3 ↓
  4. 4

    In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?

    3 marksCBSE Class 12 Physics — SA · AI 2016 · Section C, 3 marksAnswer 4 ↓
  5. 5

    (a) Write the important properties of photons which are used to establish Einstein's photoelectric equation. (b) Use this equation to explain the concept of (i) threshold frequency and (ii) stopping potential.

    3 marksCBSE Class 12 Physics — SA · AI 2015, U · Section C, 3 marksAnswer 5 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. 1.↑ back to the question

      λ=hcE=6.63×10−34×3×1083.3×10−19=19.89×10−263.3×10−19≈6.03×10−7 m\;\lambda=\dfrac{hc}E=\dfrac{6.63\times10^{-34}\times3\times10^8}{3.3\times10^{-19}}=\dfrac{19.89\times10^{-26}}{3.3\times10^{-19}}\approx6.03\times10^{-7}\text{ m}

  2. 2.↑ back to the question

    The amount of light energy (i.e. photon energy) incident per unit area per unit time is called intensity of radiation. SI unit: W m−2^{-2} or J m−2^{-2} s−1^{-1}.

  3. 3.↑ back to the question

    The main features of photons used to establish Einstein's equation: (i) in the interaction of photons with free electrons, the entire energy of the photon is absorbed; (ii) energy of a photon is directly proportional to its frequency, E=hνE=h\nu, and intensity of incident radiation depends on the number of photons falling per unit area per unit time for a given frequency; (iii) in photon-electron collision, total energy and total momentum remain conserved. Plot of KmaxK_{max} vs ν\nu (from Kmax=hν−ϕ0K_{max}=h\nu-\phi_0): a straight line of slope hh, cutting the ν\nu-axis at the threshold frequency ν0=ϕ0/h\nu_0=\phi_0/h, and rising linearly for ν>ν0\nu>\nu_0 (no emission, i.e. no meaningful KmaxK_{max}, for ν<ν0\nu<\nu_0).

  4. 4.↑ back to the question

    In the photon picture, for a given frequency, the intensity of light is determined by the energy of photons crossing unit area in unit time:   I=energy of photonsarea×time=n×hνA×t\;I=\dfrac{\text{energy of photons}}{\text{area}\times\text{time}}=\dfrac{n\times h\nu}{A\times t}, where nn is the number of photons incident normally on area AA in time tt.

  5. 5.↑ back to the question

    (a) (i) In the interaction of photons with free electrons, the entire energy of the photon is absorbed. (ii) Energy of a photon is directly proportional to frequency; intensity of incident radiation depends on the number of photons falling per unit area per unit time for a given frequency. (iii) In a photon-electron collision, total energy and total momentum remain conserved. Einstein's equation: Kmax=hν−ϕ0K_{max}=h\nu-\phi_0. (b) (i) Einstein's photoelectric equation: according to Einstein, when light is incident on a metal surface, incident photons are absorbed completely by valence electrons of atoms on its surface. Energy hνh\nu of each photon is partially utilized by an electron to become free of its "work function" W0W_0, and the rest provides the maximum kinetic energy to the photoelectron, i.e. hν=12mvmax2+W0h\nu=\dfrac12mv_{max}^2+W_0. The minimum value of the frequency of incident radiation below which photoelectric emission stops (i.e. kinetic energy of photoelectron is zero) is called threshold frequency, ν0=W0/h\nu_0=W_0/h. (ii) When the work done by the collecting-electrode potential on a photoelectron equals its maximum kinetic energy, that electrode potential is known as the stopping potential, V0=Kmax/eV_0=K_{max}/e.