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 $=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

$=\frac{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}{4500\times {10}^{-10}}$

According 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

$=150\times \frac{8}{100}=12\mathrm{W}$

$=12 {\mathrm{Js}}^{-1}$

Therefore, number of photons emitted per second by the Iamp

$\frac{\text{ Useful power }}{\text{ Energy of photon }}=\frac{12\times 4500\times {10}^{-10}}{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^9\right)}$

$=27.17\times {10}^{18}$