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).

Explanation

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.