A bulb lamp emits light of mean wavelength of 4500 A. The lamp is rated at 150 W and 8% of the energy appears as emitted light. How many photons are emitted by the lamp per second ?

The energy of photon =hcλ=6.63×10−343×1084500×10−10According to the given problem, power of the lamp is 150 W. As 8% of the light energy is utilized for emission of light, hence the utilized energy for emission per second is given by Useful power=150×8100=12W=12 Js−1Therefore, number of photons emitted per second by the Iamp Useful power  Energy of photon =12×4500×10−106.63×10−343×109=27.17×1018