Question

A watt is a unit of energy per unit time, and one watt (W) is equal to one joule per second (J*s−1) .
A 60.0 W incandescent lightbulb produces about 7.00% of its energy as visible light. Assuming that the light has an average wavelength of 510.0 nm, calculate how many such photons are emitted per second by a 60.0 W incandescent lightbulb?

Answer 1

Answer:

np =  1*10^{19} photons/s

Explanation:

To find the number of photons you divide the total light energy emitted by the lightbulb over the energy of one photon of 510.0nm:

$n_p=\frac{E_{lb}}{E_p}$

E: energy of the light bulb (7.00% of the total) = 0.07*60.0J = 4.2J

E: energy of one photon

Thus, you calculate the energy of the of the photon by using the following formula:

$E_p=h\nu=h\frac{c}{\lambda}=(6.62*10^{-34}Js)\frac{3*10^{8}m/s}{510.0*10^{-9}m}=3.89*10^{-19}J$

By replacing this value of Ep for the expression for np you obtain:

$n_p=\frac{4.2J}{3.89*10^{-19}J}=1.07*10^{19}\approx1*10^{19}\ photons/s$

hence, the number of photons emitted by the light bulb is 1*10^{19} per second