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Monochromatic light of requency 6.0xx10^(14) Hz is produced by a laser. The power emitted is 2.0xx10^(-3)W. Calculate the (i) energy of a photon in the light beam, and (ii) number of photons emitted on an average by the source. |
Answer» <html><body><p></p>Solution :Here <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> `v=6.0xx10^(<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>)Hz` and power of laser `P=2.0xx10^(-3)W` <br/> (i) Energy of a photon `E=hv=6.63xx10^(-34)xx6.0xx10^(14)=3.978xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)J=3.98xx10^(-19)J` <br/> (ii) Number of photons emitted <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> second by laser source <br/> `n=(P)/(E)=(2.0xx10^(-3))/(3.98xx10^(-19))=5.03xx10^(15)s^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`.</body></html> | |