Light intensity at a distance -- UV light to kill bacteria

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SUMMARY

The discussion focuses on using UV light at approximately 254nm to decontaminate seeds of a rare rhododendron species. The user has a 25W Ag bulb that produces about 5W at 254nm and is attempting to calculate the intensity of UV light at a distance of 0.254m. The correct calculation shows that the intensity is 6.17 W/m² when the energy flow is spread over the surface area. Additionally, the UV dose required to kill or inactivate various organisms varies significantly, emphasizing the need for precise measurements.

PREREQUISITES
  • Understanding of UV light properties and wavelengths, specifically 254nm
  • Basic knowledge of intensity calculations (Intensity = Power/Area)
  • Familiarity with the concept of UV dose expressed in watts⋅sec/cm²
  • Experience with agricultural lighting systems, particularly Ag bulbs
NEXT STEPS
  • Research UV light intensity calculations for different distances
  • Explore UV dose requirements for various bacterial and fungal species
  • Investigate the effects of UV light on seed viability and germination
  • Learn about different UV light sources and their applications in agriculture
USEFUL FOR

Agricultural scientists, horticulturists, and anyone involved in seed decontamination or UV sterilization techniques will benefit from this discussion.

ahk
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TL;DR
If I've got a bulb producing about 5J/s of useful energy at 253.7nm then how much useful energy do I have at .254m (10")?
I have obtained seeds of a rare rhododendron species.

These seeds are contaminated (experimentally determined). While I have tried several common methods, none decontaminated the seeds.

Lately, I have read that so many joules of UV light at approximately 254nm will prevent replication and inactivate or kill bacteria, fungus, mold... even in the spore stage.

I don't need a perfect number as it seems that too much UV (within reason) won't damage seeds.

My light bulb is 25W. I have read that an average Ag bulb of this wattage produces about 5W at 254nm. Using 5W at 254nm and .254m as my distance, when I try to use (what I think is) the appropriate equation (Intensity=W/m^2), I get a ridiculous answer of 77.5W/m^2.

Can anybody help?

Thanks.
 
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ahk said:
Summary: If I've got a bulb producing about 5J/s of useful energy at 253.7nm then how much useful energy do I have at .254m (10")?

My light bulb is 25W. I have read that an average Ag bulb of this wattage produces about 5W at 254nm. Using 5W at 254nm and .254m as my distance, when I try to use (what I think is) the appropriate equation (Intensity=W/m^2), I get a ridiculous answer of 77.5W/m^2.

Your source is generating 5w of power which represents the flow of energy. At .254 m this flow is spread out over the surface area of 4π(.254)2 m2 = 0.811 m2 so the flow of energy can be expressed in terms of watts/cm2. So 5W/0.811 m2 = 6.17 W/m2 = 6.17x10-4W/cm2

ahk said:
Lately, I have read that so many joules of UV light at approximately 254nm will prevent replication and inactivate or kill bacteria, fungus, mold... even in the spore stage.

I believe UV dose is usually expressed in watts⋅sec/cm2.
 
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UV dose to kill (or inactivate) different species various among the different species, sometimes by a large amount.
Here is a pdf (from a different thread) that has a information on a lot of different organisms UV sensitivity. There are many other similar lists around.
 
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