Photons emitted from a light bulb problem

[fishh]

Homework Statement

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

Homework Equations

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

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.

 

[mgb_phys]

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.

 

[fullofentropy]

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.

 

[fishh]

Thanks, I got the right answer using 4 pi r^2. I guess I should review my geometry...