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A 25watt bulbemitsmonochromaticyellowlightofwavelengthof 0.57 mu m.Calculate therate ofemissionof quanta per second. |
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Answer» SOLUTION :ENERGYOF bulb=25watt=25 `js^(-1)` energy(E ) E = HV= `(HC)/(LAMBDA)` `=(6.626 xx 3.0)/( 0.57)xx 10^(20)J` perphoton Quantum velocity`=("Emitenergybulb")/( "Emitenergyof photon")` `=(25 js^(-1)xx 0.57)/(6.626 xx 3.0 xx 10^(20)J)` `7.169 xx 10^(19)s^(-1)` `=7.17 xx 10^(19) s^(-1)` |
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