Assuming that 10.0% of a 100-W light bulb's energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.) (answer in km)

R = 1.81 10² km

Explanation:

Let's start by looking for the power in the visible range emitted this is 10W, the energy of that power is one second is

         P = E₁ / t

         E₁ = P t

         E₁ = 10 J

 Let's find the energy of a photon with Planck's equation

          E = h f

          c = λ f

we substitute

         E = h c /λ

         E = 6.63 10⁻³⁴ 3 10⁸/580 10⁻⁹

         E = 3.42 10⁻¹⁹ J

we can use a direct proportions rule to find the number of photons in the energy E₁

          #_photon = E₁ / E

          #_photon = 10 / 3.42 10⁻¹⁹

          #_photon = 2.92 10¹⁹ photons

This number of photons is distributed on the surface of a sphere. Let's find what the distance is so that there are 500 photons in 3 mm = 0.003 m.

the area of the sphere is

            A = 4π R²

area of the circle is

           A´ = π r²

as the intensity is constant over the entire sphere

         P = #_photon / A = 500 / A´

       # _photon / 4π R² = 500 / π r²

       R² = #_photon r² / 4 500

       r = d / 2 = 0.003 / 2 = 0.0015 m

       R² = 2.92 10¹⁹ 0.0015 2/2000

       R = √ (3,285 10¹⁰)  

        R = 1.81 10⁵ m

       R = 1.81 10² km

seeing the graph would

option number c is the right answer.