Planck Emission Spectrum

In summary, the conversation discussed the relationship between Planck's law and the spectral density of electromagnetic radiation emitted by a black body. It was determined that an ideal black body should have a horizontal line on its absorption spectrum, indicating that it absorbs all incident wavelengths. However, for an ideal black body, the emission spectrum would still have a characteristic shape due to internal equilibrium. The concept of emissivity and absorptivity being equal at a specific wavelength was also clarified, with Kirchhoff's law stating that absorptivity must be considered when calculating emission of thermal radiation. The difference between reflectivity and reflectance, as well as transmittance and emittance, was also discussed. In summary, Planck's function represents the power per unit
  • #1
fog37
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Hello Everyone,

I have some thoughts about Planck's law. The graph describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T.

I think the absorption (or absorbance?) spectrum of an ideal blackbody should be a horizontal line to indicate that the body absorbs, at each different wavelength, all in the incident wavelength falling on it, correct?
If a black body is irradiated with purely monochromatic light of any wavelength (sharply peaked spectrum), the blackbody emission spectrum would still looks broadband with its characteristic shape, correct? Why? Why does energy exist at wavelengths that are not present in the incident spectrum?

Assuming an opaque blackbody, i.e. transmittance = 0 and reflectance =0, all the energy that is incident gets absorbed and emitted. In general, it is always true that at, a given wavelength lambda,

emittance(lambda) = absorbance(lambda)

This means that emissivity and absorptivity must be equal at a specific wavelength: what gets absorbed at a certain wavelength should be integrally emitted at that same wavelength. But that does not seem to be the case in most cases. For example, for many materials, the absorption of incident energy occurs at visible wavelengths whereas the emission occurs in the IR so emittance and absorbance can actually be different at the same wavelength...
 
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  • #2
fog37 said:
I think the absorption (or absorbance?) spectrum of an ideal blackbody should be a horizontal line to indicate that the body absorbs, at each different wavelength, all in the incident wavelength falling on it, correct?

No. You can't know the absorption spectrum without knowing the external conditions; just knowing the characteristics of the body itself is not enough.

fog37 said:
If a black body is irradiated with purely monochromatic light of any wavelength (sharply peaked spectrum), the blackbody emission spectrum would still looks broadband with its characteristic shape, correct?

For an ideal black body, yes.

fog37 said:
Why does energy exist at wavelengths that are not present in the incident spectrum?

Because an ideal black body is made of matter that emits radiation at all wavelengths. It's not reflecting the incident radiation; it's absorbing it and converting it into emitted radiation, and that conversion process does not have to keep the wavelengths the same.

fog37 said:
In general, it is always true that at, a given wavelength lambda,

emittance(lambda) = absorbance(lambda)

No, this is not correct. The correct statement is that the total emitted power, integrated over all wavelengths, must be equal to the total absorbed power, integrated over all wavelengths, if the body is in thermal equilibrium with its environment.
 
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  • #3
fog37 said:
I think the absorption (or absorbance?) spectrum of an ideal blackbody should be a horizontal line to indicate that the body absorbs, at each different wavelength, all in the incident wavelength falling on it, correct?
Correct.

fog37 said:
If a black body is irradiated with purely monochromatic light of any wavelength (sharply peaked spectrum), the blackbody emission spectrum would still looks broadband with its characteristic shape, correct? Why? Why does energy exist at wavelengths that are not present in the incident spectrum?
Because internal equilibrium is assumed. That means that the emission process will be uncorrelated to the absorption. The light absorbed gets converted in thermal energy, and the emission is a function of temperature only, whatever the source of energy.

fog37 said:
Assuming an opaque blackbody, i.e. transmittance = 0 and reflectance =0, all the energy that is incident gets absorbed and emitted. In general, it is always true that at, a given wavelength lambda,

emittance(lambda) = absorbance(lambda)

This means that emissivity and absorptivity must be equal at a specific wavelength: what gets absorbed at a certain wavelength should be integrally emitted at that same wavelength.
What I underlined in red is incorrect. Absorptivity at a given wavelength will tell you what proportion of the incoming light of a given frequency is absorbed, the rest being reflected. Kirchhoff's law tells you that you will need to use that same factor when considering the emission of thermal radiation. Again, it doesn't matter what the energy source was.
 
  • #4
DrClaude said:
Correct.

Correct if the graph is just a graph of "percentage of incoming radiation absorbed by wavelength". But I don't think "absorption spectrum" is a good term for this graph; to me that term means the amount of energy (or power--energy per unit time) actually absorbed per unit wavelength, which, as I said, can't be known unless you know the external environment.
 
  • #5
Thank you.

So the parameters absorptivity, reflectivity, transmissivity and emissivity are all function of wavelength (and angle of incidence) and represent the proportion (i.e. the percentage) of the incoming radiation that gets absorbed, reflected, transmitted, and emitted at a particular wavelength respectively.

For an ideal opaque (zero transmissivity) blackbody, no energy gets reflected at any wavelength (so reflectivity is zero at all wavelengths) and absorptivity = emissivity = 1 at all wavelengths. My misunderstanding was to think that at a certain specific wavelength ##\lambda##, absorptivity should always be exactly equal to emissivity...

What is the difference between reflectivity and reflectance? is reflectance the integral quantity while reflectivity is just a parameter that is function of wavelength? Same goes for transmittance, emittance, etc.
 
  • #6
PeterDonis said:
Correct if the graph is just a graph of "percentage of incoming radiation absorbed by wavelength". But I don't think "absorption spectrum" is a good term for this graph; to me that term means the amount of energy (or power--energy per unit time) actually absorbed per unit wavelength, which, as I said, can't be known unless you know the external environment.
Fair enough. I was coming at it from the point of view of molecular spectroscopy, where absorption spectra are very often on a dimensionless 0 to 1 scale.
 
  • #7
Planck's function is ##B(\lambda, T)## and represents the power per unit wavelength emitted by a blackbody at temperature ##T## once at steady state. For a fixed temperature ##T##, the integral of ##B(\lambda, T)## gives the overall power emitted by the object at temperature T.

The formula ##B(\lambda, T)## does not say anything about the object and its characteristics. That means that different objects made of different materials, having different shapes, etc., if approximated as blackbodies, emit the same amount of energy if they have the same temperature ##T##. Is this an approximation? If so, how good is it?

Let's consider two objects, A and B at the same temperature ##T##. Do/can we know how much energy was required to bring each object to that same identical temperature? I think it is possible that object A may have absorbed more radiation than object B to reach the same steady state temperature ##T##.

In general, is there a way to know how much energy an object must absorb to reach a certain temperature ##T##?

Thanks!
 
  • #8
fog37 said:
The formula ##B(\lambda, T)## does not say anything about the object and its characteristics. That means that different objects made of different materials, having different shapes, etc., if approximated as blackbodies, emit the same amount of energy if they have the same temperature ##T##. Is this an approximation? If so, how good is it?
It is of course an approximation. Or to be more exact, an idealization. There is no perfect blackbody in nature. But the approximation can be quite good, especially for things like stars or the cosmic microwave background, see https://en.wikipedia.org/wiki/Cosmic_microwave_background#/media/File:Cmbr.svg

fog37 said:
Let's consider two objects, A and B at the same temperature ##T##. Do/can we know how much energy was required to bring each object to that same identical temperature? I think it is possible that object A may have absorbed more radiation than object B to reach the same steady state temperature ##T##.

In general, is there a way to know how much energy an object must absorb to reach a certain temperature ##T##?
https://chem.libretexts.org/Core/Ph...stry/Thermodynamics/Calorimetry/Heat_Capacity
 
  • #9
DrClaude said:
It is of course an approximation. Or to be more exact, an idealization. There is no perfect blackbody in nature.
I thought black holes were perfect blackbody.
 
  • #10
As far as emission and absorption go, is it always the true that if there is an absorption resonance (peak) at a certain wavelength ##\lambda##, there will automatically be a large radiation emission at that same wavelength?

For example, the human body strongly emits radiation peaking at a wavelength ~10 micron. Does it mean that the human body also strongly absorbs energy at a wavelength of 10 micron?

As far as emittance and reflectance, I have read that a wrench, being shiny and highly reflective, has a high reflectivity but becomes very hot if left in sunlight because it has low thermal emittance...What does that mean? The resource I was reading mentions that materials with high thermal emittance can get rid of the absorbed energy and remain cooler. Those with low thermal emittance "trap" energy and have a higher temperature..

From wikipedia: https://en.wikipedia.org/wiki/Thermal_emittance

"A roofing surface with high solar reflectance and high thermal emittance will reflect solar heat and release absorbed heat readily. High thermal emittance material radiates thermal heat back into the atmosphere more readily than one with a low thermal emittance."

Thanks.
 
  • #11
fog37 said:
As far as emission and absorption go, is it always the true that if there is an absorption resonance (peak) at a certain wavelength ##\lambda##, there will automatically be a large radiation emission at that same wavelength?

For example, the human body strongly emits radiation peaking at a wavelength ~10 micron. Does it mean that the human body also strongly absorbs energy at a wavelength of 10 micron?
I'm used to working with smaller systems, where this is true, but I see no reason why it wouldn't carry to something like a human body.

fog37 said:
As far as emittance and reflectance, I have read that a wrench, being shiny and highly reflective, has a high reflectivity but becomes very hot if left in sunlight because it has low thermal emittance...What does that mean? The resource I was reading mentions that materials with high thermal emittance can get rid of the absorbed energy and remain cooler. Those with low thermal emittance "trap" energy and have a higher temperature..

From wikipedia: https://en.wikipedia.org/wiki/Thermal_emittance

"A roofing surface with high solar reflectance and high thermal emittance will reflect solar heat and release absorbed heat readily. High thermal emittance material radiates thermal heat back into the atmosphere more readily than one with a low thermal emittance."
Notice in that last sentence the use of "high solar reflectance" and "high thermal emittance". You can have an object that will be reflective in the wavelengths where the Sun is ost active, but still have high emittance in the wavelengths where its peak blackbody radiation will be. This is similar to the greenhouse effect: low absorptivity in the visible part of the spectrum, where sunlight is, but it significant absorption in the IR, where the Earth is radiating.
 
  • #12
DrClaude said:
I'm used to working with smaller systems, where this is true, but I see no reason why it wouldn't carry to something like a human body.Notice in that last sentence the use of "high solar reflectance" and "high thermal emittance". You can have an object that will be reflective in the wavelengths where the Sun is ost active, but still have high emittance in the wavelengths where its peak blackbody radiation will be. This is similar to the greenhouse effect: low absorptivity in the visible part of the spectrum, where sunlight is, but it significant absorption in the IR, where the Earth is radiating.

Thanks Dr. Claude.

I see how an object/material can efficiently reflect most of the incident energy at all the wavelengths of sunlight (290nm to 2500nm). The portion of energy in the 295-2500nm range that gets absorbed is then emitted at much longer wavelengths. My confusion point is the following: in the example of the metal wrench, most of the energy from sunlight is reflected (hence high reflectance). But the wrench is said to have low thermal emittance and therefore remains hot at the touch.
The source I was reading says that materials with low thermal emittance trap energy on a molecular level at longer wavelengths between 5 and 40micron. These wavelengths correspond to low temperatures. What is special about that wavelength range (5-40nm) and why does this range determine if an object is hot or less hot? I am not clear why an object that weakly emits in that range must "trap" energy and remain hot.
 
  • #13
If you look at Wien's displacement law, you will find that the range of 5 to 40 μm corresponds to the peak blackbody wavelength for temperatures in the range 72 to 580 K (-200 to 300 °C). An object with low thermal emittance in that region of the spectrum will therefore lose very little thermal energy through radiation if it is at the kind of temperature we can expect it to be in everyday life. It will thus essentially cool only through conduction, which is much slower. The object will thus stay warmer longer.

(Note that the object will also take longer to heat up, but on a very sunny day it can still achieve temperatures hot enough to burn you.)
 
  • #14
Ok, I am close to getting this. Let me see if I understand it correctly:

For a blackbody, an emission spectrum peaking at 5 micron, the temperature is 300C (super hot). When the peak wavelength is instead 40 micron, the temperature T= -200 C (super cold). This is a very wide range of temperatures if we pick the two temperatures as extreme. This is clear.

As you mention, if an object emits little radiation in that spectral region (5-40micron), it gets called a low emittance object. The temperature range between -200C and 300C cover all everyday common temperatures.

But the object could be a good emitter, i.e. have large thermal emittance, in a different region of the spectrum and lose energy at those wavelengths instead. I guess I still don't get why an object with a temperature between -200C and 300C that loses little energy via radiation in the range 5-40micron has to remain hot. It can lose energy via radiation in a different spectral range and cool down.
 
  • #15
fog37 said:
But the object could be a good emitter, i.e. have large thermal emittance, in a different region of the spectrum and lose energy at those wavelengths instead. I guess I still don't get why an object with a temperature between -200C and 300C that loses little energy via radiation in the range 5-40micron has to remain hot. It can lose energy via radiation in a different spectral range and cool down.
Yes, but the amount emitted will be small. The overall Planck's law remains valid.

One way to see this is the following. Planck's law for blackbody radiation will give you the power emitted at different wavelengths for a given temperature. For something else than a perfect blackbody, you have to multiply by the emissivity, ##\epsilon(\lambda)##. For an object with low thermal emission, ##\epsilon(\lambda) \ll 1## for all ##\lambda## where there is significant emission. The resulting power emitted will thus be very low overall.
 
  • #16
Thanks DrClaude.

a low thermal emissivity ##\epsilon (\lambda) << 1## scales the emission spectrum down (hence less energy is radiated per unit time). Energy is being emitted slowly ( the emitted power is small). That does not mean that the energy is being "trapped" inside the object but it is just being released more slowly hence the temperature remains high for a longer time. For example, at steady state, the hot wrench left in sunlight continues to absorb energy but releases it slowly via radiation and in part due to conduction with the surrounding air remaining extremely hot for a while even if it has high reflectance in the sunlight spectrum which means it absorbs very little energy per unit time. Nonetheless, its steady state temperature can become very high even if the absorption power is low.
 
  • #17
fog37 said:
That does not mean that the energy is being "trapped" inside the object but it is just being released more slowly hence the temperature remains high for a longer time.
Depends on what one means by "trapping energy." Since the steady-state temperature is higher than the equilibrium temperature would be, one can say that energy is trapped.
 
  • #18
Thank you again.

As far as emissivity being equal to absorptivity at every wavelength, I found these articles that talks about Kirchhoff's law:

https://apatruno.files.wordpress.com/2015/09/kirchhoffs_law.pdf

http://lprl.org/resources/Kirchhoffs_law.pdf

As PeterDonis, mentioned, or an opaque material, it does not seem correct to state ##emissivity(\lambda)= absorptivity(\lambda)## since the energy absorbed at one wavelength is usually emitted at different wavelength but the articles (which I am trying to understand) seem to present a different view...
 
  • #19
fog37 said:
As PeterDonis, mentioned, or an opaque material, it does not seem correct to state ##emissivity(\lambda)= absorptivity(\lambda)## since the energy absorbed at one wavelength is usually emitted at different wavelength but the articles (which I am trying to understand) seem to present a different view...
That's not what @PeterDonis was saying. Kirchhoff's law is of course valid. You are confusing emissivity and absorptivity with emission and absorption.
 
  • #20
Hello again,

I am still wrestling with Kirchhoff's law. Thanks for the patience.

It is clear that at steady-state (thermal equilibrium) as much energy is absorbed as much as it is emitted. This is true for anybody (blackbody or not). For a blackbody with zero reflectivity and transmissivity, we have that ##absorptivity (\lambda) = emissivity(\lambda)##.

Does the equation ##absorptivity (\lambda) = emissivity(\lambda)## state that for a perfect blackbody, the emission efficiency at a particular wavelength is equal to the absorption efficiency at the same wavelength, correct? Is that the same thing as saying that all the energy absorbed at a particular wavelength ##\lambda_1## will be emitted as radiation specifically and only at wavelength ##\lambda_1##?

Example: Let's assume a perfectly monochromatic radiation beam of power 100 W= 100 J/s and wavelength ##\lambda_1## incident on the blackbody. Once thermal equilibrium is reached, every second, 100 Joule of energy at the specific wavelength ##\lambda_1## are integrally absorbed by the body. The emission spectrum will always be broadband which means that the absorbed energy, at a single wavelength, will be emitted and distributed over many wavelengths. Does that contradict the equation ##absorptivity (\lambda_1) = emissivity(\lambda_1)##? If not, why not?

Thanks!
 
  • #22
fog37 said:
Example: Let's assume a perfectly monochromatic radiation beam of power 100 W= 100 J/s and wavelength ##\lambda_1## incident on the blackbody. Once thermal equilibrium is reached, every second, 100 Joule of energy at the specific wavelength ##\lambda_1## are integrally absorbed by the body. The emission spectrum will always be broadband which means that the absorbed energy, at a single wavelength, will be emitted and distributed over many wavelengths. Does that contradict the equation ##absorptivity (\lambda_1) = emissivity(\lambda_1)##? If not, why not?
No, that does not contradict Kirchhoff's law. If the body absorbs all light at ##\lambda_1##, then ##a(\lambda_1) = 1##. When you look at the spectrum of the body and compare it to a blackbody at the same ##T##, you will find that it emits at ##\lambda_1## the same as the blackbody would, so ##e(\lambda_1)=1##.
 
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  • #23
Thank you. I see what you are saying and sorry for maybe hitting a dead horse here but I really want to understand this topic.

From what I am reading, emissivity, reflectivity, absorptivity and transmissivity are dimensionless parameters defined independently of the concept of blackbody. Conservation of energy, at a very specific wavelength lambda, implies that

incident energy = absorbed energy + reflected energy + transmitted energy.

By dividing each side by the incident energy, we obtain the following dimensionless parameters:
  • Absorptivity is the ratio between the absorbed energy at wavelength lambda and the total incident energy at wavelength lambda.
  • Emissivity is the ratio between the emitted energy at wavelength lambda and the total incident energy at wavelength lambda.
  • Reflectivity is the ratio between the reflected energy at wavelength lambda and the total incident energy at wavelength lambda.
  • Transmissivity is the ratio between the transmitted energy at wavelength lambda and the total incident energy at wavelength lambda.
and the above equation becomes absorptivity(lambda) + reflectivity(lambda) + transmissivity(lambda) = 1.

At steady-state and in the case of a blackbody, reflectivity(lambda)=0 and transmissivity(lambda)=0 at any wavelength. This leads to the equation

absorptivity(lambda) = emissivity(lambda) = 1


which seems to imply that all the energy at wavelength lambda incident on the body is absorbed (hence absorptivity equal to one) and also completely emitted as radiation having the very same exact wavelength lambda (hence emissivity equal to one).But that is not true because when the incident radiation is purely monochromatic with wavelength lambda, the blackbody emission spectrum is always broadband and not just a line emission spectrum at that wavelength lambda...
 
  • #24
Kirchhoff's law applies in thermodynamic equilibrium. The case you are considering is a steady-state, out-of-equilibirum condition.

You can consider also the inverse condition:
Joos - Theoretical Physics said:
According to Kirchhoff's Law the thermal emission of any other body is obtained, for any spectral region, by multiplying the black body value by the spectroscopically determined absorptive power. The emissivity of a non-black body must be less than that of a black body for all colours, and can attain the black body value only in the regions of strong selective absorption. In instances where the emission exceeds that of a black body, e.g. for a gas mantle, the excitation is not purely thermal. Here, for example, the gas in the region of combustion is by no means in thermal equilibrium, and the glow is due in part to chemiluminescence. The same is true of "cold" flames, e.g. the zone of reaction of a sodium vapour jet in chlorine, which gives the yellow sodium line with considerable intensity as temperatures as low as 500° C.
 

Related to Planck Emission Spectrum

What is the Planck Emission Spectrum?

The Planck Emission Spectrum is a graph that shows the distribution of electromagnetic radiation emitted by a blackbody at different temperatures.

Who developed the concept of the Planck Emission Spectrum?

The concept of the Planck Emission Spectrum was developed by German physicist Max Planck in the early 20th century.

What is a blackbody?

A blackbody is an idealized object that absorbs all incoming electromagnetic radiation and emits radiation at all wavelengths according to its temperature.

Why is the Planck Emission Spectrum important?

The Planck Emission Spectrum is important because it helped to explain the observed distribution of electromagnetic radiation from objects in the universe and laid the foundation for the development of quantum mechanics.

How does the Planck Emission Spectrum relate to the study of stars and galaxies?

The Planck Emission Spectrum is used to analyze the radiation emitted by stars and galaxies, providing valuable information about their temperatures and compositions.

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