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Homework Help: Photons emitted from a light bulb problem

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A 100 W incandescent light bulb converts approximately 2.5% of the electrical energy supplied to it into visible light. Assume that the average wavelength of the emitted light is λ = 530 nm, and that the light is radiated uniformily in all directions. How many photons per second, N, would enter an aperture of area A = 3 cm2 located a distance D = 6 m from the light bulb?

    https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys214/fall08/homework/03/photons/05.05.gif [Broken]

    2. Relevant equations

    Energy of photon = h x c / lambda (wavelength)
    Power = change in energy / change in time
    Surface area of sphere = 4πr

    3. The attempt at a solution

    I basically used the formula above to calculate the energy of the photons being emitted:

    E = 1240 eV nm / 530 nm = 2.3396 eVm

    The total power emitted by the light bulb should be 2.5% of what is put in - 2.5 watts. I know that if I have a sphere of radius 6m, the surface should "pick up" all of the power emitted. Therefore, I can set up a ratio of W/m2 : 2.5 / (4π x 6) W/m2 (not sure if this is a valid reasoning, because the aperture is not necessarily curved). Since the element(aperture) I am looking at is 3 cm2 :

    Power through element = 2.5 / (4π x 6) x 0.0003m2 = 9.947183943 x 10-6 W

    This power should be equal to the number of photons per second (n) multiplied by the energy of one photon, simply because both quantities represent the energy per second.

    9.947183943 x 10-6 W = n x (2.3396 eV x 1.602 x 10-19 J / eV)

    n = 2.653970093 x 1013 photons per second

    When I enter this answer (the homework is online), it is wrong. I'm pretty sure the problem lies with how I found the power through the aperture; I'm not sure what is wrong.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 13, 2008 #2


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    Homework Helper

    The fraction of light that enters the aperture is the area of the hole 3cm^2/the area of the sphere (2*pi*600*600) be careful to tuse the same units for radius and area.
  4. Sep 14, 2008 #3
    Surface area of a sphere is 4*Pi*r^2 not 4*Pi*r. I am in the same physics class at UIUC as you and i was able to get the correct solution with this method.
  5. Sep 15, 2008 #4
    Thanks, I got the right answer using 4 pi r^2. I guess I should review my geometry...
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