The energy of a photon of wavelength (\( \lambda \)) is given by the expression: (Symbols have their usual meaning)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( h\lambda^{-1}c \)
By Planck's relation the energy of a photon is \( E = h\nu \), and the frequency and wavelength are related by \( c = \nu\lambda \).
\( \displaystyle E = h\nu = \frac{hc}{\lambda} = h\lambda^{-1}c \)
So the energy is inversely proportional to the wavelength and directly proportional to \( h \) and \( c \).
Note: A similar question was also asked in CUET (UG) 2025 on 16-05-2025 (Slot 1).
NCERT (2026-27):
Particle Nature of Light: The Photon – “… (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν = hc/λ) and momentum p (= hν/c = h/λ), whatever the intensity of radiation may be.”