Question

One watt (W) is equal to 1 J/s. Assume that 5.0% of the energy output of a 75 watt light bulb is visible light and that the average wavelength of the light is 550 nm, how many photons per second are emitted by the light bulb? If 5.0% of the energy of a 75 watt light bulb is visible light, then the rate of energy of all the photons emitted is J/s (3 s.f.). A 550 nm wavelength is the same as meters (2 s.f.). One photon with a wavelength of 550 nm has J (2 s.f.) of energy. The rate of photons emitted by the light bulb is photons per second (2 s.f.). (Use E to express powers of 10, e.g 6.02x1023 entered as 6.02E23.)

Answer 1

Solution: Powes off bulb= 7swat = 75 J/s Visible 7 Too eneugy emilled -as =375 J/s audeugth of emited photon emiks photon Rыема = SSOnm 12400 Ewergy Cev) in. 16х10187 | SS00A 12400 : 22S4 ev a Tev=767 0OSS -19 = 2-254*16XIOT =360 E19 JT tolal energy as visile ligh Hence number of emitted photous per Sccondl a 3.15 2-254 xl-6 xiO9 =1-034X10 photons = 1039 E19 Ans