A monochromatic light of frequency \( 5.0\times10^{14} \) Hz is emitted by a source of power \( 1.98\times10^{-3} \) W. The number of photons per second, on average, emitted by the source will be:
(Take \( h = 6.6\times10^{-34} \) J s)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 6.0\times10^{15} \)
The power of the source is the energy of one photon times the number of photons emitted per second: \( P = n\,h\nu \).
\( E = h\nu = (6.6\times10^{-34})(5.0\times10^{14}) = 3.3\times10^{-19}\,\mathrm{J} \)
\( \displaystyle n = \frac{P}{h\nu} = \frac{1.98\times10^{-3}}{3.3\times10^{-19}} = 6.0\times10^{15}\,\mathrm{s^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Particle Nature of Light: The Photon – …
