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Dual Nature of Radiation & Matter revision notes

Every idea in the Dual Nature of Radiation & Matter chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 19 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Photons

Photon energy
12× asked, last 2026
E=hν=hcλ,E(eV)≈1240λ(nm)E = h\nu = \frac{hc}{\lambda},\qquad E(\text{eV}) \approx \frac{1240}{\lambda(\text{nm})}
h=6.63×10−34h=6.63\times10^{-34} J s =4.14×10−15=4.14\times10^{-15} eV s; 1 eV =1.6×10−19=1.6\times10^{-19} J. Some papers use hc=1242hc=1242 eV nm.
  • Visible light is about 400–700 nm, i.e. 3.1–1.8 eV. Shorter than 400 nm is UV (more energy per photon), longer than 700 nm is IR; X-rays are near 1 Å.
  • Not ν\nu = the number in sin⁡(ωt)\sin(\omega t). That number is ω\omega, so E=hω/2πE=h\omega/2\pi. Sum of sines: the highest ω\omega sets Kmax⁡K_{\max}; a product such as (1+cos⁡ω1t)cos⁡ω2t(1+\cos\omega_1t)\cos\omega_2t has highest ω1+ω2\omega_1+\omega_2.
Photon momentum and mass
4× asked, last 2025
p=hλ=Ec,E2=p2c2+m02c4p = \frac{h}{\lambda} = \frac{E}{c},\qquad E^2 = p^2c^2 + m_0^2c^4
Photon: rest mass m0=0m_0=0, so E=pcE=pc; speed cc in vacuum for every λ\lambda. Moving mass E/c2=h/cλE/c^2=h/c\lambda.
  • Not less momentum at shorter λ\lambda. p=h/λp=h/\lambda and E=hc/λE=hc/\lambda both rise as λ\lambda falls. Two photons with equal momentum have equal wavelength.
Photon flux
10× asked, last 2026
n=Phν=Pλhc,nA=IAλhcn = \frac{P}{h\nu} = \frac{P\lambda}{hc},\qquad n_A = \frac{I A\lambda}{hc}
nn = photons per second; II in W m⁻², area AA in m². At fixed power n∝λn\propto\lambda.
  • Point source of power PP: photons per second per m² at distance rr =Pλ/(4πr2hc)=P\lambda/(4\pi r^2hc). Pressure and force of light: EM Waves sheet.
  • Not n∝1/λn\propto1/\lambda. That is the energy per photon. At the same power, longer-wavelength photons are weaker, so more of them are emitted each second.
  • Quick method. Convert first: mW to W, nm to m, cm² to m² (×10−4\times10^{-4}). Or find EE in eV from 1240/λ1240/\lambda and multiply by 1.6×10−191.6\times10^{-19} for joules.

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16 more concepts in 5 topics

Photoelectric effect: observations, Einstein's photoelectric equation, Photoelectric graphs, de Broglie waves and Matter waves: links and evidence. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Photoelectric effect: observations

  • Hertz and Lenard observations10×
  • Why wave theory fails4×

Einstein's photoelectric equation

  • Einstein's equation23×
  • Threshold frequency and wavelength13×
  • Two lights on one metal17×
  • Speeds of photoelectrons6×
  • Photoelectrons in a magnetic field4×

Photoelectric graphs

  • Stopping potential vs frequency10×
  • Photocurrent vs anode potential4×

de Broglie waves

  • de Broglie relation16×
  • Charge accelerated through V14×
  • Comparing particles28×
  • Thermal particles5×
  • Charge moving in a field7×

Matter waves: links and evidence

  • Photon vs material particle7×
  • Photoelectron wavelength2×

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