Calculating Visible Light Photons Emitted Per Second & Distance

In summary: Thanks for pointing that out.In summary, the conversation discusses finding the number of visible-light photons emitted per second from a 200 W light bulb and the distance at which it corresponds to 1x10^11 photons per square centimeter per second. The first part is solved by finding the energy and dividing it by the amount of energy per photon. The second part is solved by using the surface area formula for a sphere, 4pi*r^2, instead of pi*r^2 as initially attempted.
  • #1
EK86017
2
0
a.) If the average frequency emitted by a 200 W light bulb is 5x10^14, and 10.0% of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second?

I solved this part of the problem by finding the energy (E = hf) and dividing 10% of 200W by the amount of energy per photon to get 6.03 x 10^19 photons per second.

b. At what distance would this correspond to 1x10^11 visible-light photons per square centimeter per second if the light is emitted uniformly in all directions?

I am stuck on this part of the problem. I tried dividing 6.03 x 10^19 by 1x10^11 to get the how many square centimeters would correspond to 1x10^11 photons per square centimeter per second. I got 6.04x10^8 cm^2. Because the light is emitted uniformly in all directions, I did A = pi*r^2, using the value 6.04x10^8 as "A". I got r = 138.7 m, but this is incorrect.

The problem I think I am having is A = pi*r^2 does not correspond to visible-light in the 3rd dimension, but I am confused as to why the question then does not ask "per CUBIC centimeter?"

Any suggestions would be very much appreciated.
Thanks!
 
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  • #2
EK86017 said:
a.) If the average frequency emitted by a 200 W light bulb is 5x10^14, and 10.0% of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second?

I solved this part of the problem by finding the energy (E = hf) and dividing 10% of 200W by the amount of energy per photon to get 6.03 x 10^19 photons per second.

b. At what distance would this correspond to 1x10^11 visible-light photons per square centimeter per second if the light is emitted uniformly in all directions?

I am stuck on this part of the problem. I tried dividing 6.03 x 10^19 by 1x10^11 to get the how many square centimeters would correspond to 1x10^11 photons per square centimeter per second. I got 6.04x10^8 cm^2. Because the light is emitted uniformly in all directions, I did A = pi*r^2, using the value 6.04x10^8 as "A". I got r = 138.7 m, but this is incorrect.

The problem I think I am having is A = pi*r^2 does not correspond to visible-light in the 3rd dimension, but I am confused as to why the question then does not ask "per CUBIC centimeter?"

Any suggestions would be very much appreciated.
Thanks!

Welcome to the PF. The surface area of a sphere is not PI * R^2...

http://www.teacherschoice.com.au/Maths_Library/Area and SA/area_2.htm

.
 
  • #3
Thank you! It worked with 4pi*r^2
 

1. How do you calculate the number of visible light photons emitted per second?

The number of visible light photons emitted per second can be calculated using the formula: Number of photons = Power output (in watts) x Time (in seconds) x Efficiency (in percentage) / Energy per photon (in joules).

2. What is the relationship between distance and the number of visible light photons emitted per second?

The number of visible light photons emitted per second decreases as the distance from the source increases. This is because the photons spread out over a larger area, resulting in a lower concentration of photons.

3. How does the efficiency of a light source affect the number of visible light photons emitted per second?

The efficiency of a light source, which is the percentage of energy converted to visible light, directly affects the number of photons emitted per second. A more efficient light source will emit a higher number of photons per second compared to a less efficient one.

4. Can the number of visible light photons emitted per second be affected by external factors?

Yes, the number of visible light photons emitted per second can be affected by external factors such as the quality of the light source, atmospheric conditions, and obstacles in the path of the light. These factors can cause some photons to be absorbed or scattered, resulting in a lower number of visible light photons emitted per second.

5. How can the calculation of visible light photons emitted per second be useful in scientific research?

The calculation of visible light photons emitted per second can be useful in various scientific research fields, such as optics, astronomy, and biology. It can help in understanding the properties of light sources, measuring distances, and studying light interactions with different materials and environments.

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