Question

The energy efficiency of an incandescent light bulb (= the percentage of consumed power that is actually converted into radiated light) is typically around 5%. Suppose a certain brand of 75 W incandescent bulb emits light primarily at wavelength ???? = 570 nm.

Calculate the total number of photons emitted by one of those bulbs if it has been shining for a 6-hour period.

Answer 1

Answer:

4*10^22 photons

Explanation:

To find the number of photons is necessary to calculate the total energy of the light emitted in one hour = 3600s:

$E'=0.05E=0.05Pt=0.05(75J/s)(3600s)=13.500J$

Furthermore, is necessary to find the  associated energy to the photon 0f 570nm with following formula:

$E'=h\nu=h\frac{c}{\lambda}$

h: Planck's constant = 6.62*10^-34 Js

c: speed of light = 3*10^8 m/s

wavelength = 570*10^-9 m

$E_p=(6.62*10^{-34}Js)\frac{3*10^{8}m/s}{570*10^{-9}m}=3.484*20^{-29}J$

Finally you divide E' between Ep to find the number of photons:

$n=\frac{E'}{E_p}=\frac{13.500J}{3.484*10^{-19}}\approx4*10^{22}photons$

the number of emitted photons is 4*10^22