White and coloured bodies have enough energy so reflect out excess while Black bodies have low energy and so have enough room to absorb all and reflect no light.
Is my hypothesis correct?
First of all, white is not a color, it is a mixture of all the colors. The Sun light is close to white, and we can see its composition when passing a beam of Sun light through a prism.
Second, color of objects is due to the wavelengths that these objects reflect.
When an object is illuminated with the Sun light, some wavelengths are reflected, others absorbed, and others are partly absorbed and partly reflected, with different absorption and reflection coefficients.
Black bodies have an absorption coefficient close to 1 (at least theoretically) for all the wavelengths.
The color of the body is therefore given by the mixture of reflected wavelengths, each wvelength with the respective reflected intensity. Thus, so far, the question "how much energy contain the atoms of the body" is not connected with the phenomenon examined.
However, if the atoms of the object are highly excited and emit photons, e.g. in an incandescent object, the color of the body changes according to the emitted wavelength. So happens with stars. They emit light, the reflected light is negligible.
No doubt that the photons in the blue part of the spectrum are more energetic than photons in the red part. However, to say something about the energy of the atom, is not simple, because the energy of an emitted photon is the difference between two levels. An atom may be excited to a high level, but the highest emission probability may be to a very close level. To the contrary, an atom excited to a level not so high may emit a photon and fall to the ground level, s.t. the enery of the photon may be bigger than in the former case.
Max,
>Is my hypothesis correct?
No.
The energy of an atom (the aggregate electron and nuclear energy states) has very little bearing on the colour of an object.
Colours arise principally through molecular and atomic transitions - atoms in a slab of polished iron are identical (and of the same 'energy') as atoms in rust - the difference being that crystals of iron in the pure metal are free to move, and reflect light admirably well. In rust you have particular bonds with oxygen that have characteristic transition energies.
No really. As Mr.James Garry said, colour of a material is due to transitions.
First of all, white is not a color, it is a mixture of all the colors. The Sun light is close to white, and we can see its composition when passing a beam of Sun light through a prism.
Second, color of objects is due to the wavelengths that these objects reflect.
When an object is illuminated with the Sun light, some wavelengths are reflected, others absorbed, and others are partly absorbed and partly reflected, with different absorption and reflection coefficients.
Black bodies have an absorption coefficient close to 1 (at least theoretically) for all the wavelengths.
The color of the body is therefore given by the mixture of reflected wavelengths, each wvelength with the respective reflected intensity. Thus, so far, the question "how much energy contain the atoms of the body" is not connected with the phenomenon examined.
However, if the atoms of the object are highly excited and emit photons, e.g. in an incandescent object, the color of the body changes according to the emitted wavelength. So happens with stars. They emit light, the reflected light is negligible.
No doubt that the photons in the blue part of the spectrum are more energetic than photons in the red part. However, to say something about the energy of the atom, is not simple, because the energy of an emitted photon is the difference between two levels. An atom may be excited to a high level, but the highest emission probability may be to a very close level. To the contrary, an atom excited to a level not so high may emit a photon and fall to the ground level, s.t. the enery of the photon may be bigger than in the former case.
It is a very interesting question indeed. I'll only try to help Max to himself figure out the answer. Let's compare the two extreme cases i.e. Black and White.
The physics of reflection and absorption is the same for both black and white objects. In both cases, the reflection follows the absorption.
The white and black objects might absorb the same portion of the spectrum of sunlight, which extends over UV to Visible to IR, but emit/reflect entirely different parts of the spectrum. The white objects emit over the visible spectrum while the black objects emit on other-than-visible range of the spectrum. In principle, however, both objects emit as much light as they absorb so no change in atoms energy.
The black objects, which might emit longer wavelengths in the Far Infrared spectrum, are likely to be 'hotter' than the white objects emitting shorter at wavelengths in the visible.
If this were true, then we should expect a temperature difference between the two otherwise identical objects (named here as black and white). A simple demo can be helpful.
Take two identical aluminum rods, painted one as black and other as white. Fix the two thermometers respectively to the two rods and place them in the sun. Take the temperature readings of the two thermometers with time (after every minute) for 20 minutes.
First outcome: There should be no temperature difference. Both rods are same thermodynamically. Not very interesting. As much energy is emitted as much energy is absorbed.
Second Outcome: There is a non-zero temperature difference. It is quite interesting. This temperature difference can be used to sustain a heat flow, from higher temperature to a lower temperature, by placing the two rods in contact. This kind of heat flow can make a Carnot Heat Engine to perform some work.
For example, take an Aluminuml rod, paint half of it as black and other half of it as white, and place it in the sunlight. What are your expectations?
Let's all keep thinking.
Dear Muhammad
i saw your very explicit answer. I just have some objection. You say,
==> "The black objects, which might emit longer wavelengths in the Far Infrared spectrum, are likely to be 'hotter' than the white objects emitting shorter at wavelengths in the visible."
No! Objects are hotter or colder, depending on what they absorb, not on what they emit. The very explanations and examples that you give in your post, say this.
Best regards!
Dear Sofia
Thanks for your concern. Let me explain again..
The sunlight spectrum consists of IR, Visible and UV spectra, but only IR spectra produce heating. Why?
If we use window pans which reflect 'heat' (IR photons) and only transmit the visible light, then lesser air conditioning will be required to cool it.
Assuming that a body absorbs a photon of higher frequency (visible or UV), then if the object emits the photon at the same frequency (visible or UV) then the object will be a 'white' object. However, alternatively, if the object emits the absorbed photon energy in the form of several cascaded photons of smaller energy (in the IR), then the object will be a 'black' object. Since the IR photons have more heating effect as compared with visible photons, therefore, the 'black' objects will appear slightly 'hotter'.
I hope this reasoning works!
You might complement it.
I agree with Sofia and Muhammad. The basic points are: 1) The absorption - reflection process has little if anything to do with the energy of participating atom / molecule. 2) Each reflection is actually a 2-staged process (absorption ---> emission).
If the emitted photon has the same frequency as the absorbed one, we have a special case (elastic scattering) of the scattering process in the light-matter interaction. The object can in this case be called "white" in the given frequency range. If the same happens for all other ranges of the IRUV spectrum, the object is generally white. Otherwise, the scattering is elastic within some narrow range (the object has the "color" of this range) and inelastic in other ranges (the same object is black - it does not emit the same frequency). But in terms of the luminescence theory any black body is luminous in some other (usually lower) frequency range and accordingly has the color of that range.
No object can be totally black when in thermodynamic equilibrium with environment. E.g., a pile of soot within a uniformly heated container is luminous with a prevailing color characteristic for given temperature. As the latter raises, the color of soot will change from black through read to blue (actually, to the mixture called white).
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