Question

800 W bulb emits monochromatic light of wavelength 1400 angstrom and only 10% of the energy is emitted as light the number of photons emitted by the bulb per second will be

Answer 1

Each photon has energy of  hν  h = planks constant ν = frequency of light If there are N photons emitted then the total energy radiated is  Nhν. Average power is given by  P=Energy/time=E/t=Nhν/t But  N/t=n is the number of photons emitted per second. Therefore,  n=P/hν .ν=c/λ So,  n=Pλ/hc If all the power is not radiated then the effective power = rated power x efficiency. So the number of photons emitted per second  n=Peffλ/c = P×ηλ/hc where η is the efficiency. In the problem power P = 150W, wavelength λ = 4500A and efficiency is 8% = 0.08 So number of photons emitted per second, n=(150×0.08*4500*10-10)/(6.6×10−34) *(3*×108) = 27.2*1018 27.2*1018 number of photons will be emitted per second by the bulb..hope it will help you..