Question

Infrared lamps are used in restaurants to keep the food warm. The infrared radiation is strongly absorbed by water, raising its temperature and that of the food. If the wavelength of infrared radiation is assumed to be $1500nm$, then the number of photons per second of infrared radiation produced by an infrared lamp, that consumes energy at the rate of $100W$ and is $12\%$ efficient only, is $(x\times {10}^{19})$. The value of $x$ is:
[Given: $h=6.625\times {10}^{-34}J-s$]

Answer 1

use this equation to calculate energy per photon

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

where

E = energy per photon

h = planks constant $=6.626\times {10}^{-34}$ J . sec

c = speed of light $=3.00\times {10}^8$ m/s

$\lambda$ = wavelength $=1500$ nm $=1500nm\times \left(\frac{1m}{{10}^{9}nm}\right)=1.500\times {10}^{-}6$ m

so that

E / photon $=\frac{(6.626\times {10}^{-34}J.sec)\times \left(3.00m/sec\right)}{(1.500\times {10}^{-}6m)}=1.325x{10}^{-19}$ J / photon

Since 1 W = 1 J / sec, we just convert to photons per sec using that E/photon

$\left(\frac{1photon}{1.325\times {10}^{-19}J}\right)\times \left(\frac{1J/sec}{1W}\right)\times (0.12\times 100W)=9.03\times {10}^{50}$ photons / sec

Hence $x$ = 9 (approx)