Answer to Question #306807 in Physics for Precious

Question #306807

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.


1
Expert's answer
2022-03-06T15:16:13-0500

The amount of energy emitted as light per second is:


E=60J×5%=3JE = 60J\times 5\% = 3J

The energy in one photon is:


E1=hνE_1 = h\nu

where h=6.63×1034Jsh = 6.63\times 10^{-34}J\cdot s is the Planck's constant, and ν\nu is the frequency of the light. The frequecncy is given as follows:


ν=cλ\nu = \dfrac{c}{\lambda}

where c=3×108m/sc = 3\times 10^8m/s is the speed of light, and λ=5100nm=5100×109m\lambda = 5100nm = 5100\times 10^{-9}m is the wavelength.

Thus, the number of the photons:


N=EE1=EλhcN=3J5100×109m6.63×1034Js3×108m/s7.69×1019N = \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. 7.69×10197.69\times 10^{19}.


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