Specific Absorption Rate (SAR) is the rate at which electromagnetic energy is absorbed by water, human tissue or any other material.
After performing an electromagnetic simulation i got the below value
SAR = 0.00151 W/kg
Excitation duration = 2.63299907 nanoseconds (used by EM simulation software).
Converting SAR to electron volts as follows:
0.00151 W/kg * 2.63299907e-9 s (excitation duration) = 3.9758285957e-12 J
One joule is equal to 6.241509e18 eV
3.9758285957e-12 J = 24815171.327 eV
to calculate the energy absorbed per nanosecond
24815171.327 eV/ 2.63299907 nanoseconds = 9424679.108223156341639 eV/ nanosecond
to calculate the energy absorbed per picoseconds
9424679.108223156341639 eV / 1000 = 9424.679108223156341639 eV/ picosecond
The amount of absorbed energy per 1kg is 9424.679108223156341639 eV/ picosecond.
Does anyone have any comments on the above calculations?
Is it correct to use the above approach to convert SAR W/kg to eV/kg/ps?
Is there any modifications required to be done?
Looks right to me. I would change, "The amount of absorbed energy per 1kg is 9424.679108223156341639 eV/ picosecond." to "The rate at which energy is absorbed per 1kg is... eV/ps."
As watt is the unit of power and eV is used for energy then it is always possible to convert power into energy by multiplying it with time.
watt = Joule/s; 1.0 Joule = 6.24151 10^18 eV. Therefore,
0.00151 W/kg = 0.00151 Joule/s*kg = 0.00151*6.24151 *10^18 eV/s*kg
= 9.42468*10^15 eV/s*kg (dimension = power/unit mass)
If you want to have the energy/Kg exchanged during the time of excitation, you have to multiply this latter value with the duration of excitation, yielding
9.42468*10^15 eV/s*kg * 2.63299907*10^-9 s = 2.48151729*10^7 eV/Kg
Thank you, Todd J Desiato and Hemendra Tripathi for your comments
Vasile Surduca, thank you and i must comment on your comment.
I will start from the your last sentence. You said " The electromagnetic radiation is not absorbed in any material." this sentence has nothing to do with any true. Try to write a question on researchgate same as i did using the same sentence and you will get the answers.
You are wondering about the precision that I'm using in the calculations. Please remember that I'm dealing with numbers belongs to SAR that obtained from numerical technique. The validation of the EM simulation is not included in the question and it is another story.
Vasile Surducan, I'm not angry but I'm just wondering about your comment because what you are saying is not related to my question.You are taking about basics. My question related to numbers and a method to convert W/kg to eV/kg that I'm using.
I don't want to discuss and explain details about how EMF interacts with biological materials and water because it is basics too. The mechanisms such as dipole rotation and ionic conduction are well known mechanisms because the majority of the researchers approve both of them specially the dipole rotation because we use the microwave oven as a practical example of heating water using the dipole rotation if the water is pure and using both mechanisms (dipole rotation and ionic conduction) if we are using electrolyte solutions.
I know the non-thermal effects is a controversial topic. This is why I'm working on developing a new mechanism and i think you understood this from my calculations. This is why you commented about how to validate the calculations to make it practical. My calculations used to explain a theoretical point of view. I want to show you our research article that explained the details from theoretical point of view but the article is currently under-review and the journal does not accept the archived articles.
Frank Diederich, Thank you.
The result mentioned in my question is= 24815173.9 eV/kg
The result based on your calculations is= 24815172.9 eV/kg
Both of them are very similar.
You obtained very similar result but not exactly the same. I think because you multiplied the time by the eV and in the question the time was multiplied by W/kg or Joule/s*kg. I think the time should be multiplied by W/kg first because the power was calculated during the above excitation duration and then should be converted to eV. I think both of them should be OK.
It is not correct to ask to convert W/kg into eV/kg as Watt & eV is not units of same parameter (as is also clear from Abdelsamie's calculations and many comments above). Also it depends on the fact that for what purpose one wants to convert to smaller unit of different physical parameter. Accordingly one has to introduce time in seconds or smaller unit of time like micro, nano, pico etc.
P.P Pathak, watt is j/s and the possibility to convert joule to eV is well known.
http://www.rapidtables.com/convert/energy/joule-to-ev.htm
Regarding to the purpose of the calculations. Actually I have used the calculations in my submitted manuscript to explain my point of view about the results of EM simulation . My question is about the approach of the calculation.
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