The question is incomplete the right solution for this question is:
E photon = 2.39 eV, b = 6.32 * 1048 fotones/s, c = 9.87 * 1048 fotones/s.
To compute the answer, we must take into account the Plack postulate, which states that each photon's energy is given by
Ephoton is equal to h*f, where f is the frequency of light and h is the Planck universal constant.
We also know that the relationship between wavelength and frequency for waves has the following form:
velocity equals wavelength times frequency (f) Assuming that velocity in our case is the same as the speed of light, we can derive
Ephoton= h*f= h*c/ and hc= 1240 eV/nm are related terms.
Consequently, Ephoton=1240/540 nm=2.39 eV
Since power is defined as energy/time, we may calculate the power emitted by considering a number per second (n) and multiplying it by its energy.
Power in this instance: n*h*f
On the other hand, we must take into account that the light is emitted isotropically in order to compute the number of photons per second at a distance of 5 m from a light bulb using an area of 2 cm2.
the percentage of the 2 cm2 at 5 meters is 425 of the total area radiated.
0.02*0.02/25=1,6*10^-5
To perform the calculations for photons at 810 nm, we must alter the photon energy to 1.53 eV.
As 810 nm photons have lower energy, 100 W light bulbs emit more of these photons than they do.
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