A monochromatic source emitting light of wavelength 300 nm has a power output of 33 W. Calculate the number of photons emitted by this source in 2 minutes. \( (h = 6.62\times10^{-34}\,\mathrm{J\,s}) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6\times10^{21} \)
The energy emitted in a given time is power × time, and it is carried in photons of energy \( E = hc/\lambda \) each. So the number of photons is the total energy divided by the energy of one photon.
\( \lambda = 300\,\mathrm{nm} = 3\times10^{-7}\,\mathrm{m},\qquad t = 2\,\mathrm{min} = 120\,\mathrm{s} \)
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.62\times10^{-34}\times3\times10^{8}}{3\times10^{-7}} = 6.62\times10^{-19}\,\mathrm{J} \)
\( E_{\text{total}} = Pt = 33\,\mathrm{W}\times120\,\mathrm{s} = 3960\,\mathrm{J} \)
\( \displaystyle N = \frac{E_{\text{total}}}{E} = \frac{3960}{6.62\times10^{-19}} \approx 6\times10^{21} \)