Classical or quantum electromagnetic theory?

[orlan2r]

In the last review of admission to a university in Peru (UNI) proposed a theoretical question true / false.

The sentence was: A beam of infrared radiation has less energy than a visible radiation of the same intensity.

This sentence is True or False?

According to classical electromagnetic theory the intensity of an electromagnetic wave is equal to the average energy density by the speed of light. Therefore, if both have the same intensity radiation will also have the same average energy, and therefore this statement is FALSE.

But there is another point of view logically valid.

According to quantum theory of radiation, the energy of a photon is proportional to the frequency of electromagnetic radiation, and the frequency of the infrared radiation is less than the average frequency of visible light, we conclude that the radiation energy infrared is less than that of visible light, and therefore the proposition is TRUE.

According to this view, the intensities of radiation are equal because there is greater amount of infrared radiation photons.

Who is right?

I would like to know your opinion about it.

Thanks in advance

 

[queenofbabes]

If the intensity is the same, and the beam sizes the same, the energy carried by each is the same...

Besides, even going by classical theory, how can you assume that energy densities of the two different beams are the same?

"According to this view, the intensities of radiation are equal because there is greater amount of infrared radiation photons."
Yup I guess that has to be the case.

I think they were really only trying to see if you knew the definitions of these various terms well o.0

 

[Redbelly98]

According to quantum theory of radiation, the energy of a photon is proportional to the frequency of electromagnetic radiation, and the frequency of the infrared radiation is less than the average frequency of visible light, we conclude that the radiation energy infrared is less than that of visible light, and therefore the proposition is TRUE.

True only if the beams have an equal number of photons. Is that the case here? What do they say is equal about the two beams?

 

[orlan2r]

True only if the beams have an equal number of photons. Is that the case here? What do they say is equal about the two beams?

From my point of view, if the intensity I of radiation is the same and the energy E of each photon infrared is less than photon visible light, we deduce that exist more photons infrared than photons visible light.

http://cepreuni.org/wp-content/uploads/2009/08/intensidad_onda1.gif

For you the original sentence is true?

Thanks for your response

 

[orlan2r]

If the intensity is the same, and the beam sizes the same, the energy carried by each is the same...

According to this paragraph the orginal sentence is FALSE.

Besides, even going by classical theory, how can you assume that energy densities of the two different beams are the same?

According to classical electromagnetic theory, if the intensity of radiation are equals then the energy densities are equals.

"According to this view, the intensities of radiation are equal because there is greater amount of infrared radiation photons."
Yup I guess that has to be the case.

According to this paragraph the orginal sentence is TRUE.

This is, true or false?

Greetings and thanks for your response

 

[queenofbabes]

Indeed the energy densities are the same precisely because the intensities are the same! The original sentence is false. The third paragraph you quoted does not imply in any sense that the statement is true, does it?

 

[orlan2r]

Indeed the energy densities are the same precisely because the intensities are the same! The original sentence is false. The third paragraph you quoted does not imply in any sense that the statement is true, does it?

I am of the opinion that the sentence is true.

My point of view the energy of infrared radiation is less than the average energy of a mixture of visible radiation (white light). The justification for this is the Planck's formula.

According to this, if the number of photons of both radiations are equals then intensity of radiation infrared will be less than the intensidad of visible radiation.

Then, if the intensities of these radiations are equal, there must be more photons of infrared light than visible light (see formula: http://cepreuni.org/wp-content/uploads/2009/08/intensidad_onda1.gif).

What is your opinion.

 

[Hootenanny]

According to this, if the number of photons of both radiations are equals then intensity of radiation infrared will be less than the intensidad of visible radiation.

But as Redbelly pointed out in the very first reply to this thread, are the number of photons in the two beams equal?

 

[Redbelly98]

But as Redbelly pointed out in the very first reply to this thread, are the number of photons in the two beams equal?

Another hint: the 3 possible answers to this are either yes, no, or we don't know.

 

[orlan2r]

But as Redbelly pointed out in the very first reply to this thread, are the number of photons in the two beams equal?

This question has led diverse opinions in my country (Peru) and I appeal to you to respond.

Is it possible that two of the radiaciación accepted theories of the classical electromagnetic theory and quantum theory of radiation reaching diametrically opposite conclusions? I think this is not possible.

According to classical electromagnetic theory this statement is FALSE.

According to quantum theory of radiation this statement is TRUE.

Who's right?

http://cepreuni.org/?p=143

 

[ideasrule]

This question has led diverse opinions in my country (Peru) and I appeal to you to respond.

Is it possible that two of the radiaciación accepted theories of the classical electromagnetic theory and quantum theory of radiation reaching diametrically opposite conclusions? I think this is not possible.

According to classical electromagnetic theory this statement is FALSE.

According to quantum theory of radiation this statement is TRUE.

Who's right?

http://cepreuni.org/?p=143

If you translated the problem correctly, the answer is most certainly FALSE: by definition, same intensity means same energy. The infrared beam has less energy per photon, but it has more photons. Quantum and classical theory do not disagree on the definition of the word "intensity".

 

[queenofbabes]

This question has led diverse opinions in my country (Peru) and I appeal to you to respond.

Is it possible that two of the radiaciación accepted theories of the classical electromagnetic theory and quantum theory of radiation reaching diametrically opposite conclusions? I think this is not possible.

According to classical electromagnetic theory this statement is FALSE.

According to quantum theory of radiation this statement is TRUE.

Who's right?

http://cepreuni.org/?p=143

In no way does the classical theory imply that the statement is false. As said above, quantum and classical theory do not disagree on the definition of the word "intensity".

 

[Redbelly98]

According to this, if the number of photons of both radiations are equals then intensity of radiation infrared will be less than the intensidad of visible radiation.

Then, if the intensities of these radiations are equal, there must be more photons of infrared light than visible light

The above statements are true. Since the two beams do have equal intensities, there are unequal numbers of photons in the two beams.

According to quantum theory of radiation, the energy of a photon is proportional to the frequency of electromagnetic radiation, and the frequency of the infrared radiation is less than the average frequency of visible light, we conclude that the radiation energy infrared is less than that of visible light, and therefore the proposition is TRUE.

This is incorrect reasoning. You are basing this on the energy per photon of infrared vs. visible. In essence, you are assuming equal numbers of photons in the two beams to reach your conclusion. But you have already correctly reasoned that there are unequal numbers of photons in the two beams.