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ISC 2027
All chaptersPhysics · Unit 7

Dual Nature of Radiation and Matter

5 articles22 formulas29 ways the board asks it
PHYPhotoelectric Effect

Photon Energy, Momentum & Number

Treating light as a stream of photons, each carries energy E=hf=hc/λE=hf=hc/\lambda and momentum p=E/c=h/λp=E/c=h/\lambda. From the source power you get the photon emission rate by dividing total power by the energy per photon.

ISC numericals here test unit conversion between joule and eV, between wavelength and frequency, and the idea that a modest macroscopic power corresponds to an enormous number of photons per second.

Photon energy
E=hf=hcλE = h f = \dfrac{h c}{\lambda}
EE = energy per photon (J), h=6.63×10−34h=6.63\times10^{-34} J s, ff = frequency (Hz), λ\lambda = wavelength (m), c=3×108c=3\times10^{8} m/s.
Energy in electron-volts
E(eV)=E(J)1.6×10−19=1240λ(nm)E(\text{eV}) = \dfrac{E(\text{J})}{1.6\times10^{-19}} = \dfrac{1240}{\lambda(\text{nm})}
Shortcut using hc≈1240hc\approx1240 eV nm; λ\lambda in nanometres gives EE directly in eV.
Photon momentum
p=Ec=hλp = \dfrac{E}{c} = \dfrac{h}{\lambda}
pp = momentum of one photon (kg m/s). For a photon E=pcE=pc, so p=h/λp=h/\lambda; it depends only on the wavelength.
Number of photons per second
n=PE=Pλhcn = \dfrac{P}{E} = \dfrac{P \lambda}{h c}
nn = photons emitted per second, PP = source power (W = J/s), EE = energy of one photon (J).
  • Photon energy scales inversely with wavelength: visible light (around 5000 \AA) has energy near 2.5 eV, i.e. ∼4×10−19\sim 4\times10^{-19} J per photon.
  • Momentum of a photon, p=h/λp=h/\lambda, is tiny (order 10−2710^{-27} kg m/s for visible light) yet non-zero even though the photon has zero rest mass.
  • The photon emission rate n=P/En=P/E for a few-watt source is astronomically large (∼1018\sim10^{18}–102010^{20} per second), which is why light looks continuous and the quantum graininess is hidden.
  • Use E=hfE=hf when frequency is given and E=hc/λE=hc/\lambda when wavelength is given — both yield the same energy, so pick whichever matches the data.
  • To go from a photon energy back to wavelength: λ=hc/E\lambda = hc/E, with EE in joule, or use λ(nm)=1240/E(eV)\lambda(\text{nm})=1240/E(\text{eV}).
  • Power is energy per second; nn already carries units of photons per second because watt = joule per second, so no extra time factor is needed.
  • Always keep the energy of a single photon in joule when dividing power by it, then convert the final photon-count to a pure number.
Where the marks go
  • Dividing power by the photon energy expressed in eV instead of joule, giving a wrong photon count by a factor of 1.6×10−191.6\times10^{-19}.
  • Forgetting to convert \aa ngstr\"om or nm to metres before using E=hc/λE=hc/\lambda — 1 A˚=10−101\,\text{\AA}=10^{-10} m, 1 nm=10−91\,\text{nm}=10^{-9} m.
  • Writing photon momentum as h/cλh/c\lambda or hc/λhc/\lambda instead of p=h/λp=h/\lambda; only the energy carries the extra factor of cc.
  • Treating a small wattage as a small number of photons — the count is huge; check the order of magnitude is ∼1018\sim10^{18} or more for a several-watt visible source.
How the board asks it
  • Numericalphoton energy E=hf=hc/λE=hf=hc/\lambda
    A monochromatic source emits light of wavelength 5000 A˚5000\,\text{\AA}. Calculate the energy of a single photon, expressing your answer in joule and in electron-volt. (h=6.6×10−34 J s, c=3×108 m/s)(h=6.6\times10^{-34}\,\text{J s},\ c=3\times10^{8}\,\text{m/s})
  • Numericalnumber of photons per second n=P/En=P/E
    A bulb radiates a power of 25 W25\,\text{W} at a wavelength of 6000 A˚6000\,\text{\AA}. Calculate the number of photons emitted by the source per second. (h=6.6×10−34 J s, c=3×108 m/s)(h=6.6\times10^{-34}\,\text{J s},\ c=3\times10^{8}\,\text{m/s})
  • Numericalenergy to wavelength λ=hc/E\lambda=hc/E, λ(nm)=1240/E(eV)\lambda(\text{nm})=1240/E(\text{eV})
    A photon has an energy of 3.1 eV3.1\,\text{eV}. Calculate the wavelength of the radiation in nanometre.
  • Numericalphoton momentum p=h/λp=h/\lambda
    Calculate the momentum of a photon of light of wavelength 4000 A˚4000\,\text{\AA}. (h=6.6×10−34 J s)(h=6.6\times10^{-34}\,\text{J s})
  • Give reasonsn=P/En=P/E is very large for a few-watt source
    A 10 W10\,\text{W} lamp emits about 101910^{19} photons per second. Explain why, despite this graininess, ordinary light appears perfectly continuous to the eye.
  • Define / statephoton carries energy E=hfE=hf and momentum p=h/λp=h/\lambda
    State two properties of a photon and write the expressions for its energy and momentum in terms of the wavelength λ\lambda, noting why its momentum is non-zero even though its rest mass is zero.

Practise this topic

PreviousStopping Potential & Maximum KENextde Broglie Wavelength

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.