A 337 nm nitrogen laser puts out 5.00 W of power with a beam 5.90 mm in diameter. The beam is pointed directly at a pinhole which has a diameter of 1.10 mm. How many photons of light will travel through the pinhole per second? Assume that the intensity of the light is equally distributed across the whole area of the beam.

#_foton2 = 2.96 10¹⁶ photon / s

Explanation:

We are going to solve this exercise in parts, first we will calculate how many photons are in the beam and then the amount that pass through the pinhole.

To find the energy of a photon, let's use the Planck relationship

          E = h f

          c = λ f

we substitute

          E₀ = h c /λ

          E₀ = 6.63 10⁻³⁴ 3 10⁸/337 10⁻⁹

          E₀ = 5.90 10⁻¹⁹ J

Now we can use a direct ratio rule to find out the number of photons in the beam. If 1 photon has an energy E₀, how many photons are in an energy 5.00W

         # _foton = 1  5/ E₀ = 1 5 / 5.90 10⁻¹⁹

         #_foton = 8.5 10¹⁸ photons / s

This number of photons is uniformly distributed in the area of the circle with diameter 5.90mm = 5.90 10⁻³m

      R= d/2=  2.95 10⁻³ m

     r = d/2 = 0.55 10⁻³ m

let's find the beam area

         A = π R²

         A = π (2.95 10⁻³)²

         A = 2.73 10⁻⁵ m²

the pinhole area

        a = π r²

        a = π (0.55 10⁻³)²

        a = 9.50 10⁻⁷ m²

Let's use another direct ratio (rule of three) if there are 8.5 1018 photons / s in an area A how many photons pass through the area at

         # _foton2 = 8.5 10¹⁸ a / A

         # _fotn2 = 8.5 10¹⁸ 9.50 10⁻⁷ /2.73 10⁻⁵

         #_foton2 = 2.96 10¹⁶ photon / s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west