Question about irradiance from a UV lamp

[piscosour00]

Greetings all,

I'm normally used to researching topics related to biology but I am currently tackling a related side project with UV irradiation, and I don't have much experience with radiometry. I have a compact UV lamp that I am using to irradiate a suspension of small particles (submicron, about 100 nm), and I would like to come up with an estimate of the UV exposure the particles will receive. The details of my setup are as follows:

I have an 8 watt UV lamp emitting 254 nm UV-C light. The company gives a reference irradiance of 1.8 mW/cm^2 by a UV dosimeter probe placed directly under the center of the lamp at a distance of 7.6 cm. The UV lamp itself contains two cylindrically shaped bulbs of lengths 227 mm, diameters of 15.9 mm, and spaced center-to-center 20 mm apart with curved reflectors behind the bulbs to help redirect all generated light to emit out of the lamp. The UV light is emitted through a rectangular area of 227 mm by 57 mm.

I fill a cylindrical polystyrene well of 1.74 cm high and 1.56 cm wide with my particle suspension to a height of 1 cm and placed it in the center under the lamp such that the distance away from the bottom of the well is 7.6 cm away from the lamp.

I ultimately want to know what is the UV dose these particles will receive, but I thought I would start off by just considering what is the dose the liquid volume will receive.

I am assuming the liquid in the well is perfectly cylindrical with no meniscus, and that the UV light can only enter through the opening of the well, but not through the polystyrene walls of the well.

I would like to be able to the answer the following questions:

Questions:
A) What is the average irradiance (mW/cm^2) on the surface of the liquid in the well?
B) What is the average irradiance (mW/cm^2) on the bottom of the well under the liquid?
C) What is the energy transfer rate in a small volume element anywhere in the liquid?

For A), I thought I may be able to come up with a reasonable estimate of the irradiance on the surface of liquid using the reference irradiance of 1.8 mW/cm^2 at 7.6 cm. Using an inverse square law, I get an estimate of the average irradiance of ~2.39 mW/cm^2 at (6.6 cm away from the lamp). Considering how the UV cannot pass through the plastic, should I reduce this value even further since the liquid surface being irradiated is surrounded by impermeable walls 0.74 cm high essentially.

I'm not quite sure how to think about the other two questions. Once the UV light reaches the air-liquid interface, would I need to remove a portion of the UV that is reflected away by the water? Pure water appears to have an absorption coefficient around 10^-3 cm^-1 at 254 nm, but I'm not sure how this fits into the picture given also that the are particles in suspension.

I have attached cartoon diagram of the setup for further reference. I'm looking for ways to simplify the problem. Any insight would be greatly appreciated.

Thank you,

 

[Tom.G]

Just from a general physical standpoint, things smaller than a wavelength are invisible/transparent. Given that, I would expect zero absorption of 254nm radiation by 100nm particles.

You might try a Google search of 'effect of UV on viruses'.

Others here may have more direct experience/knowledge of your particular situation.

 

[piscosour00]

Thank you for your input, Tom. I'll rephrase the question I think, because I would still care to know what the radiant energy is imparted on the liquid volume.

 

[piscosour00]

Thank you for your input, Tom. I'll rephrase the question I think, because I would still care to know what the radiant energy is imparted on the liquid volume.

Sorry, I didn't realize I couldn't edit the after a certain amount of time. But as I mentioned, what I would still be interested in is the radiant energy transfer on the top and bottom of the liquid volume, as well as throughout the liquid. The parts pertaining to the particles can disregarded for the moment then.

 

[hutchphd]

Given that, I would expect zero absorption of 254nm radiation by 100nm particles.

By that logic no atom (size 1 Angstrom) would ever absorb light (wavelength 5000 Angstrom). And even absent resonances, the cutoff is not at all sharp (see Mie Scattering)

 

[hutchphd]

And the reflectance at the water interface should be given by Fresnel's Law. R is less than 10 %. Plastic will block UV depending on details.

 

[piscosour00]

Thank you for all the responses so far. I am thinking that I can approach this problem by determining the radiant powder traveling from the lamp to the surface of liquid and taking that as a starting point or input power to estimated how much of it will continue through.

I am using the estimate of the inverse square law to say that the irradiance on the liquid surface is about 2.39 mW/cm^2 (at 6.6 cm from the lamp). If I assume 10% is reflected, could I say that only 2.15 mW/cm^2 pass through.

Do I continue to think that the energy is decreasing due to being further away from the source even if the medium has changed? In addition, some of the energy is absorbed traveling through the liquid (some sources cite an absorption coefficient for 254 nm in water of 0.04 - 0.1 m^-1). So do I consider both of these together if I wanted to estimate what the irradiance is on the bottom of the well 1 cm down?