A lamp gives out \(4.0\times10^{15}\) photons per second of light of wavelength 663 nm. The power given out as light is (Take \(h=6.63\times10^{-34}\ \mathrm{J\,s}\) and \(c=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.2\times10^{-3}\ \mathrm{W}\)
Power is the energy given out per second. For a beam of light, it is the number of photons sent out each second times the energy of one photon. Each photon carries the energy \( h\nu \), which is \( hc/\lambda \) in terms of wavelength.
Given:
\( N=4.0\times10^{15}\ \mathrm{s^{-1}} \)
\( \lambda=663\ \mathrm{nm}=663\times10^{-9}\ \mathrm{m} \)
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{6.63\times10^{-34}\times3\times10^{8}}{663\times10^{-9}} \\ &= 3.0\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} P &= NE \\ &= 4.0\times10^{15}\times3.0\times10^{-19} \\ &= 1.2\times10^{-3}\ \mathrm{W} \end{aligned} \)
Hence, the lamp gives out 1.2 mW as light.
NCERT (2026-27):
Particle Nature of Light: the Photon – “(ii) Each photon has energy E (=hν) and momentum p (= h ν/c), and speed c, the speed of light.”
Example 11.1 – “(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. …”