A: Calculating Photon Count from Electrical Energy and Efficiency

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The discussion focuses on calculating the number of photons emitted by a 60-watt incandescent bulb operating at 2.2% efficiency, specifically for green light at a wavelength of 5500 Å. To determine the photons per second, one must first calculate the energy produced as visible light and then use the formula E=hf to find the energy per photon. The second part of the problem involves calculating how many photons enter the eye when standing 3.5 meters away from the bulb for 0.10 seconds, considering the pupil's diameter. Participants express uncertainty about the equations and approach needed for the calculations. Overall, the thread emphasizes the need for a structured approach to solve both parts of the problem effectively.
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Homework Statement


In converting electrical energy into light energy, a sixty-watt incandescent light bulb operates at
about 2.2% efficiency. Assuming that all of the light is green light (wavelength approximately
5500 Å), determine the number of photons per second given off by the bulb and B:if you were to look directly at the bulb, the light
from it would travel directly to your eye without reflecting from anything. If you stand 3.5
meters away and look at the bulb for only 0.10 seconds, how many photons will enter your eye?
(assume the pupil of your eye has a diameter of 2.1 mm)


Homework Equations


I don't really know what to use.
I think E= hf


The Attempt at a Solution


First I converted angstroms to wavelength then I found frequency. From frequency and planks constant I found E but then I got nothing . DOn't even know how to touch B.
 
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For (A), you need to answer two questions first:

1. How many Joules per second are produced in the form of visible light?

2. How many Joules of energy are in one photon? (Use the answer you got from using E=hf)
 
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