Photons per second hitting an aperture

[SammC]

Homework Statement

>An aperture is 3m away from a light bulb, with an area of .09m^2.
>The light bulb is 100 Watts, and 2.5% of the energy is converted to visible light
> From the previous part: The average wavelength of the light is 520 nm, meaning that each photon has an energy of 2.386 eV (this answer is correct in the system)

How many photons per second will hit the aperture.

[The hint for this is "Remember to convert eV to Joules, because the intensity is given in W/m^2"
I'm not using intensity in my answer, nor am i sure how to, but the answer I'm getting seems to make sense anyway.]

Homework Equations

The Attempt at a Solution

First, i convert 2.386 eV to Joules, getting 3.82279*10^-19 Joules/Photon.
Now, given that its a 100 Watt bulb, converting 2.5% to light, i assume that means that it emits 2.5 Joules/Second of light.

So, 2.5 (J/S) / 3.82279*10^-19 (J/P) will give an answer in Photons/Second, which is
6.5397 * 10^18 Photons/Second. Emitted from the bulb (I assume spherically)

So, then I find what percentage of those photons will hit the aperture by dividing Aperture area by the surface area of a 3m radius sphere: .09m^2 / (4*pi*3^2) = 7.9577*10^-4 m^2/m^2

Multiplying that percentage by the number of photons per second: 5.204*10^15 photons/second, which is incorrect.

 

[turin]

Hmm... I haven't read the numerical details of your calculation, but the way you describe it sounds correct.

OK, I just did the calculation, and I agree with your result except for the last decimal place. Are you entering this into one of those online homework systems that is picky about sig figs? If so, post the numbers EXACTLY how they are given, and I will check for sig figs.

 

[thomate1]

After reviewing your approach, I can see that you have correctly converted the energy of each photon to Joules and have accounted for the percentage of energy converted to light from the 100 Watt bulb. However, it seems that you have made a mistake in your calculation of the percentage of photons hitting the aperture. Instead of dividing the aperture area by the surface area of a 3m radius sphere, you should divide it by the surface area of a 3m radius circle (since the aperture is a flat surface, not a sphere). This would give you a percentage of 2.6534*10^-5 m^2/m^2. Multiplying this by the number of photons per second emitted from the bulb (6.5397 * 10^18 photons/second) would give you a final answer of approximately 1.734*10^14 photons/second hitting the aperture. This seems like a more reasonable and accurate answer. Additionally, I would recommend using the equation for intensity (I = P/A, where P is power and A is area) to check your answer. I hope this helps!