Dear Dr. Saeedeh Shahbazkhany,

for an accurate light intensity measurement, you should use a digital power intensity meter, following the appropriate indications provided with the instrumentation.

If you prefer calculate it and your source of UV is isotropic , you can divide the total power (in watts) by 4π (12.57) to get the radiant intensity in watts/steradian. Then, you can estimate the irradiance (in watts/square meter) by dividing the radiant intensity by the square of the distance (in meters). This assumes the source is small enough to be considered a "point source" and your distance is large enough not to deviate from the inverse-square law.

I suggest you to have also a look at the following note:

Calculation of Luminaires Using Radiance by Krzysztof Wandachowicz

Available at: https://www.radiance-online.org//community/workshops/2004-fribourg/presentations/Wandachowicz_paper.pdf

Best regards, Pierluigi Traverso.

you can used the UV sensor which direct read the light intensity in mW/cm2

or using chemical actinometry in enstine/s and than change the unite to mW/cm2

This is measurable. Why do you need this number?

Thank you Mr. Traverso and dear Luma and Mr. Geletii.

I need this number for reporting in article.

How this number helps to understand the process you are reporting in your paper? You might not to report this number.

Dear professor Geletii

Our article was revised by editor. Editor wants this number is reported in article.

for 200 W intensity of light and distance 20 Cm. How I can calculate this?

Also editor wants the calculation of quantum yield.

please help me.

Thank you.

Saeedeh

Dear Mr. Traverso

Please check my calculation following :

For 200 W intensity of light and 20 Cm distance between light and sample:

200W/12.57=15.91 (W/Staradian)

15.91/(0.2)^2= 397.75 W/m2

39.775 mW/Cm2

Is it correct?

Thank you a lot.

Saeedeh

Dear Saeedeh Shahbazkhany ,

I guess that the 200W are attributed to the electrical power of your UV lamp and not to the total optical UV power. In general there is a huge difference between electrical and optical power. Please have look at the data sheet for the UV power.

" Also editor wants the calculation of quantum yield. " This is the reason why you need to characterize your light source. The determination of the quantum yield is a pretty complicated procedure. You have to determine the number of absorbed photons by your catalyst and the number of formed (or consumed) molecules in the reaction. Commonly it's determined experimentally by a proper design of the experiment, especially by using the right light source. Personally, I would not even attempt to calculate the quantum yield using 200 W tungsten lamp as a source of light. The reviewer knew that you would be unable to answer his question. He probably used this question as an argument to reject your paper. If I'm a reviewer, I would do the same. Sorry, but your paper would be probably rejected by a respectable journal.

Dear professor Martens

tungsten lamp emits visible light. it is not UV lamp.

Dear professor Geletii

With respect, I disagree with your opinion. Because the reviewer could reject my manuscript at first. However, your opinion may be right. rejecting is not important for me. Finding the answer of my questions is important for me.

Best regards

Saeedeh

Thanks for feed back Saeedeh Shahbazkhany , but what is the optical output in the wave length range you need.

From that you can apply the calculation you have done above under the constraint hat the distance (here 20cm) is larger than the source dimension. See the anwser of Pierluigi Traverso .

You need to look at the datasheet of the lamp to calculate its radiant flux from wattage and (color temperature or efficacy). Then assume it's a point source and use the inverse square law.

Dear Dr. Saeedeh Shahbazkhany ,

considering the various approximations indicated in my previous e-mail, the calculation you made is correct.

Good luck and best regards, Pierluigi Traverso

Dear Saeedeh Shahbazkhany ,

just a remark:

apart from correct calculation, in the paper you should restrict the result to maximum 2 significant digits and mention that this result is a rough estimation of intensity due to crude point source approximation.

Do not use for example: 39.775 mW/cm² as calulated above.

Here for example go for: 'approximately 40mW/cm²' as statement in the paper.

Good luck

Thank you dear Dr. Rehn and Martens and Mr. Traverso.

Saeedeh Shahbazkhany

Yours: " Finding the answer of my questions is important for me"

The science is the best way to satisfy my curiosity for taxpayer money

Thank you dear Dr. Akif Zeb.

For calculation of quantum yield, I need a single wavelength, while my light source is visible lamp and it has a range of wavelength (380-760 nm). is it calculable?