- #1
ahk
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- TL;DR 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")?
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.
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.