Answer to Question #306807 in Physics for Precious

Question

A watt is a unit of energy per unit time, and one watt (W) is equal to one joule per second (Jasa 1) A 60.0 W incandescent lightbulb produces about 5.00% of its energy as has an average wavelength of 5100 nm. calculate how many such photons are emitted per second by a 60.0W incandescent lightbulb.

Answer 1

The amount of energy emitted as light per second is:

$$E = 60J\times 5\% = 3J$$

The energy in one photon is:

$$E_1 = h\nu$$

where $h = 6.63\times 10^{-34}J\cdot s$ is the Planck's constant, and $\nu$ is the frequency of the light. The frequecncy is given as follows:

$$\nu = \dfrac{c}{\lambda}$$

where $c = 3\times 10^8m/s$ is the speed of light, and $\lambda = 5100nm = 5100\times 10^{-9}m$ is the wavelength.

Thus, the number of the photons:

$$N = \dfrac{E}{E_1} = \dfrac{E\lambda}{hc}\\ N = \dfrac{3J\cdot 5100\times 10^{-9}m}{6.63\times 10^{-34}J\cdot s\cdot 3\times 10^8m/s} \approx 7.69\times 10^{19}$$

Answer. $7.69\times 10^{19}$.