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InterviewSolution
| 1. |
A 25watt bulbemitsmonochromaticyellowlightofwavelengthof 0.57 mu m.Calculate therate ofemissionof quanta per second. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/energyof-2613064" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGYOF">ENERGYOF</a> bulb=25watt=25 `js^(-1)` energy(E ) <br/> E = <a href="https://interviewquestions.tuteehub.com/tag/hv-1033771" style="font-weight:bold;" target="_blank" title="Click to know more about HV">HV</a>= `(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>)` <br/>`=(6.626 xx 3.0)/( 0.57)xx 10^(20)J` perphoton <br/> Quantum velocity`=("Emitenergybulb")/( "Emitenergyof photon")` <br/>`=(25 js^(-1)xx 0.57)/(6.626 xx 3.0 xx 10^(20)J)` <br/> `7.169 xx 10^(19)s^(-1)` <br/> `=7.17 xx 10^(19) s^(-1)`</body></html> | |