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A 25 watt bulb emits monochromatic tellow light of wavelength of 0.57 mum. Calculate the rate of emission of quanta per second. |
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Answer» Solution :ENERGY emitted by the bulb `=25` watt `=25" J s"^(-1)""( :' 1" watt"=1" J s"^(-1))` Energy of one photon `(E)=hv=h c/lambda` Here, `lambda=0.57 MU m=0.57xx10^(-6) m""(1 mu m=10^(-6) m)` PUTTING `c=3xx10^(8)" m s"^(-1), h=6.62xx10^(-34)J` s, we GET `E=((6.62xx10^(-34) Js)(3xx10^(8) ms^(-1)))/(0.57xx10^(-6) m)=3.48xx10^(-19) J` `:.` No. of photons emitted per sec `= (25"J s"^(-1))/(3.48xx10^(-19) J)=7.18xx10^(19)`. |
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