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Written by Administrator
Tuesday, 01 December 2009 11:26

Tomorrow's penetrants

January 2010

Many of our American friends read the papers that we wrote and published on our website. Quite often we even get support letters and some questions are asked.

We received the following e-mail from a well-known US manufacturer.

‘‘1) There are two ways to make up for lower efficiency of excitation of penetrants by blue light as compared to UV. One is, as you suggest, to make new penetrants that are optimized for the new blue wavelengths. Another is to specify a higher intensity level for blue lights used with existing penetrants. If an excitation wavelength is half as efficient as 365 but you use 2x the intensity you would have used with a 365 source you should get a similar intensity of fluorescence.

2) The situation with magnetic particles is more complicated than with penetrants. As you point out all manufacturers use a similar dye for penetrants, so the excitation spectrum is similar. While there are indeed some similarities, the excitation spectra for the pigments used for magnetic particles are more variable from manufacturer to manufacturer and from product to product within a single manufacturer. And while the excitation spectrum for penetrants is moderately sharply peaked at 365 nm, I have not found this to be so for particles. In many cases the excitation is just as efficient well into the 400's as it is in the UV, and in some cases even more efficient.’’

Our answer:

‘‘An irradiance at 450 nm twice that at 365 nm should make us think ... twice!! i.e., first, is any effect of saturation of the dyes, as seen with 5,000 µW/cm² and above at 365 nm, possible when using high 450 nm irradiance? Second: what is the effect of high irradiance at 450 nm for long exposures, as some times experienced in UV-A booths? Are dyes/brighteners able to withstand such a situation (kind of "UV-A fading" or "blue-light fading")?

These two points will need numerous tests on ALL the penetrants which are currently listed ... just to be sure. Sure if these two points are Okayed, better to keep current formulae!!

The other point is more ... administrative: no current standard, no specification allows for using blue light instead of UV-A.

Maybe not the easiest task to achieve!’’

His answer was:

I absolutely agree that the task of acceptance for primary inspections is difficult, and that great care must be taken to ensure that the same performance as with current methods can be achieved. This will require extensive testing and validation. And if the procedure is ever approved, it will indeed put additional requirements for internal procedure management, auditing, etc.

But now on to the interesting chemistry/physics questions relating to saturation and bleaching. The whole process of fluorescence first requires absorption of a photon to put the fluorescing substance into an excited state. Several things can happen to the excited photon, only one of which is fluorescence. Some can go into mechanical action (heat), and another is that the energy can go into chemical reactions. Is it correct that these chemical reactions are what ‘‘damages’’ the dye and causes bleaching?

A molecule does not ‘‘know how’’ it got into an excited state, whether from absorption of a UV photon or a blue photon. Once in an excited state the relative probability of fluorescence or other use of the excited state energy is the same (a little more on this below). So if blue is for example 50% as efficient at causing fluorescence, that would mean that it is only 50% as likely to be absorbed in the first place. Saturation and bleaching both result from excited state electrons, so if blue light produces only 50% as many excited state electrons for the same input energy, then by using twice as much blue light you would have the same saturation/bleaching issues as UV light - that is, no better, no worse. I do not say that this is true, as my understanding of the processes involved might not be completely correct. I know that the two of you know much more about the chemistry of penetrants than I do, so if I am making a significant error in this reasoning please tell me.

Is it possible that another factor is whether incoming photons are putting electrons into the first excited state or into a higher excited state? Fluorescence occurs from the first excited state. Could it be that the energy of a UV photon first puts the electron into the 2nd or higher excited state? And if so, could the photobleaching be related to this higher energy level? If so, and if blue photons put the electron only into the 1st excited state, would there then be less potential for bleaching? My own suspicion is that this would not be the case since the higher excited states decay to the lowest excited state very very fast, so the potential for much chemistry to occur would be low.

Is the above discussion correct according to your knowledge, or am I missing important points? I am not arguing here for blue or UV, only trying to fully understand the science behind the process.

Our answer was:

Fluorescence is clearly explained on the following Website:
http://www.ndt-ed.org/EducationResources/educationresource.htm
Then click on ‘‘resources’’, then on ‘‘penetrant testing (PT)’’, then on ‘‘color and fluorescent’’ and also on ‘‘Why Things Fluoresce’’.

Basically, in PT, fluorescence is achieved using 2 dyes: a yellow, yellow-green or green dye which emits a fluorescent colour accordingly and an optical brightener which emits a blue/white fluorescence under UV-A irradiation.

Both dyes are heterocyclic molecules with N (nitrogen) and C (carbon) atoms with both σ and π linkages. Fluorescence is due to the excitation of the π electrons of the π linkages.

Contrary to σ electrons, π electrons are free electrons moving on an electrons track.

π linkages are not so strong as σ linkages. If some π linkages are destroyed the fluorescence brightness under UV-A irradiance decays.

π linkages may be destroyed by:
• Heat.
• Photolysis.
• Curing (polymerisation) reactions.
• Chemical reactions: as by example, with acids, basic compounds, oxidisers such as chromic acid.

When UV-A irradiance is too high (from 5,000 µW/cm² and above) the yellow/green indication becomes more and more bluish/whitish. The indication seeability is reduced to an unacceptable brightness level.
We could spend quite a long time discussing about the physics of penetrants' fluorescence.
We suggest you contact a PT materials manufacturer, just to make tests comparing 450 and 365 nm responses, as per the American specification SAE-AMS 2644E, paragraphs 3.3.8.3.1 (color), 3.3.8.3.2 (brightness), 3.3.8.3.3 (ultraviolet stability, which would be used also for blue-light).

Should you want to perform the tests by yourself you would need the underneath mentioned equipments:
• A radiometer able to measure the irradiance at 450 nm, with bandwidth parameters close to those written in ISO 3059:2001 standard named ‘‘Viewing Conditions’’ for UV-A radiometers.
• A fluorometer, as described in ASTM E 1135, but using a blue light source instead of the
UV-A source currently used.

It is likely you may find help, at a lower cost, if working with a manufacturer.

Last Updated ( Wednesday, 25 May 2011 20:09 )