Question

Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1). A 100-W incandescent light- bulb produces about 4% of its energy as visible light. Assuming that the light has an average wavelength of 510 nm, calculate how many such photons are emit- ted per second by a 100-W incandescent lightbulb.

Answer 1

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$, where h is the Planck constant and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$.

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$, so we have $E_N=\frac{N h c}{\lambda}$, where c is the speed of light.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$.

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$.

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$