A 100 watt bulb emits monochromatic light of wavelength 400 nm . Calculate the number of photons emitted per second by the bulb.

Explanation

Energy emitted by the bulb,E = 100 watt or `100 Js^(-1)`
Energy of one photon, `E=hv=(hc)/(lambda)`
`lambda=400 nm =400xx10^(-9)m=4xx10^(-7)m, c=3xx10^(8)ms^(-1), h=6.626xx10^(-34)Js`
Energy of photon, `E=(6.626xx10^(-34)Jsxx3xx10^(8)ms^(-1))/(4xx10^(-7)m)=4.969xx10^(-19)J`
 Number of photons emitted per sec `=(100Js^(-1))/(4.969xx10^(-19)J)=2.01xx10^(20)s^(-1)`