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

np = 1*10^{19} photons/s

Step-by-step 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=(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\\u=h(c)/(\lambda)=(6.62*10^(-34)Js)(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=(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