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CBSE Class 12 Physics 2014 Delhi Set 1 Paper

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Question : 18 of 30
Marks: +1, -0
(i) Monochromatic light of frequency 6.0×1014 Hz6.0 \times 10^{14} \text{ Hz} is produced by a laser. The power emitted is 2.0×2.0 \times 10−3 W10^{-3} \text{ W}. Estimate the number of photons emitted per second on an average by the source.
(ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface.
Solution:  
(i) An energy,
E  =hvE\;=h v
  =6.63×10−34×6.05×1014\;=6.63 \times 10^{-34} \times 6.05 \times 10^{14}
  =3.98×10−19 J\;=3.98 \times 10^{-19} \text{ J}
Now,number of photons emitted per second
n  =  pE=2.0×  10−33.98×10−19 Jn\;=\;\frac{p}{E}=2.0 \times \;\frac{10^{-3}}{3.98} \times 10^{-19} \text{ J}
  =5×1015   photons per second   \;=5 \times 10^{15} \;\text{ photons per second }\;
(ii)
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