A 25 watt bulb emits monochromatic yellow light of wavelength of ` 0. 57 mu` m. Calculate the rate of emission of quanta per second .

Explanation

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)`.