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If the efficiency of an electric bulb of 1 watt is 10% what is the number of photons emitted by it in one second ?(The wavelength of light emitted by it is 500 nm.)(h=6.625xx10^(-34)J.s). |
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Answer» Solution :As the bulb is of 1 W,if its efficienct is 100% ,it may emit 1 J radiant energy in 1s.But here the efficiency is 10% ,HENCE it emits `(1)/(10)J=10^(-1)J` adiatn energy in 1s. Note:The efficiency of bulb is 10% .It MEANS it emits 10% of energy consumed in form of light and remaining 90% is WASTED in form of heat energy (due to the resistance of filamet.) `therefore` Radiant energy otained from the bulb in 1s=`10^(-1)`J If it consists of n PHOTONS, `nhf=10^(-1)J` `therefore n=(10^(-1))/(hf)` `=(0.1)/(6.625xx10^(-34)xx(c)/(lambda))` `=(lambdaxx1-^(-1))/(6.625xx10^(-34)xx3xx10^(8))(because f=(c)/(lambda))` `therefore n=(0.1xx500xx10^(-9))/(6.625xx10^(-34)xx3xx10^(8))` (`because` velocity of light ,`c=3xx10^(8)ms^(-1))` `therefore n=2.53xx10^(17)` photons. |
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