# The color of an apple (light absorption and emission)

1. Jul 19, 2012

### kenewbie

Emission of light from a solid object

From my textbook (explaining why an apple is red):

We imagine that the red apple in the picture is illuminated by daylight, IE light which contains all the wavelengths of the visible spectrum. The apple is red because the main part of the light it reflects is in the red area of the visible spectrum.

This is all good, and is what I expected. My understanding of this is that the atom is exited by a photon of a certain wavelength, and then when it returns to its original energy level it emits a photon of the same wavelength again. This is why it reflects the light.

However the next sentence goes on to say:

Molecules in the skin of the apple absorbs the photons in the blue and the green part of the spectrum. If we look at it with a strictly blue light, the apple will appear black.

This threw me off. My understanding is that matter will only interact with photons of certain wavelengths, and the wavelengths an atom can absorb are the exact same which it can emit. Other wavelengths should not interact with it at all?

So if the atoms in the skin of the apple absorbs the blue light, it should also emit it once the atoms return to their original energy level?

It seems to me that the the first paragraph is enough to explain the red color of the apple, and I am at a loss of what they mean with the second one.

Is my understanding of this way off?

Any and all help appreciated.

k

2. Jul 20, 2012

### DrDu

As a matter of fact, an apple does not consist of isolated atoms but of molecules which are themselves interacting with each other. While an atom has in deed very sharp absorption lines, molecules absorb in broad frequency ranges (bands). Furthermore, as they have internal degrees of freedom, namely vibrations, the energy absorbed can be transformed into vibrational energy and finally into heat. Thats called internal conversion (IC). Hence molecules will re-emit (fluorescence) only a small fraction of the light they absorbed and even this usually not at the same wavelength, but at longer wavelengths.

3. Jul 20, 2012

### krd

No. That's not correct - that does not explain reflection. What you're thinking of is absorption and re-emission of light along the spectral lines of the atoms. This is not what's happening in reflection.

I have my own fuzzy theory on reflection - but maybe someone could give the standard and accepted version.

4. Jul 20, 2012

### kenewbie

Ok, that makes sense. My only question then is if absorption and conversion to heat is an emergent behavior in molecules, or if this happen at the layer of individual atoms as well?

5. Jul 20, 2012

### sophiecentaur

That is an oxymoron. Either the atoms are individual or there is a layer of them on the surface of a 'condensed object'. There is a great difference between the interaction of photons with individual atoms and with large numbers of atoms in close proximity.
When an isolated atom absorbs a photon, under normal circumstances, it will re-radiate light at the same frequency that it observed. But there is a possibility of the decay being in two or more jumps if the appropriate energy level in the atom. (Look at the operation of a Laser, for instance).
When an atom or molecule on the surface of an object (dense material) absorbs an optical photon there are many more possible energy levels involved (bands, rather than discrete levels, in fact) and the incident photon's energy can be redistributed in the material of the object in many different ways and a broad range of frequencies will be absorbed, as a consequence. (No line absorption spectrum)

So your apple will absorb all sorts of green(ish) and blue(ish) wavelength photons but not the red(ish) ones. Under white light illumination, the apple will look red. But if there is no red in the incident light, the apple will look black(ish).

6. Jul 20, 2012

### DrDu

It is quite astonishing that already a hand full of atoms forming a molecule is sufficiently complex a system to justify a thermodynamical description. So yes, it is an emergent thermodynamical property of molecules.
Just yesterday there was a related question in the chemistry forum. Maybe you want to have a look at the classic article by Bixon and Jortner I cited there:

7. Jul 20, 2012

### DrDu

Still it came as a big surprise in the 1960's that essentially irreversible behaviour was found already for rather small isolated molecules like benzene in the gas phase.

8. Jul 20, 2012

### sophiecentaur

I guess it's just a matter of nCm, which involves Factorials. It's very easy to get large numbers from a molecule with three or more atoms in it and when several of the electrons in each atom are interacting with those in the other atoms. And then there are the vibrational modes etc.

The old Hydrogen Atom stuff we start with is just not enough to deal with anything more complicated - but, deep down, we want it to.

9. Jul 20, 2012

### DrDu

That's the point. In a molecule like benzene you have already 31 vibrational modes. Given that the electronic excitation energy corresponds to an order of magnitude $\sqrt{m/M}\approx 100$ (with m electron and M nuclear mass) vibrational quanta, the number of vibrational states to decay into is already astronomical.

10. Jul 20, 2012

### TSny

Re: Emission of light from a solid object

Hi kenewbie. I think your question is a very good one. Since no one else has yet replied, I will try to give some explanation as I understand it.

You are right that an atom (molecule) that jumps to an excited state by absorbing a photon of a specific frequency could "de-excite" by emitting a photon of the same frequency. But, when atoms are crowded together in a solid or liquid, an excited atom is more likely to pass the energy to its neighboring atoms in the form of motional energy of the atoms (heat) rather than re-emit the energy as a photon. So, the original photon that excited the atom has become absorbed by the material with no light being re-emitted. That's why if you shine only blue light onto an apple, the apple will appear dark.

But then why is it that when you shine red light on an apple, the apple reflects the red light? Even though red light does not have the right energy of photons to excite the atoms or molecules to higher energy states, the light nevertheless does interact with the electrons in the molecules and causes the light to be scattered. The reflection of the red light from the apple is this scattered light.

The best non-mathematical discussion on this topic that I have ever seen is the article How Light Interacts with Matter by Victor Weisskopf which was published in the book Lasers and Light, Readings From Scientific American 1968. Could be hard to find, but well worth the effort.

11. Jul 20, 2012

### kenewbie

I had to re-read this a few times but I think i have found that my choice of the word "layer" was poor, as it seems to have a specific meaning. What I meant to ask was if individual atoms sometimes absorb photons without emitting photons, or if this is behavior which only happens in molecules.

And you did answer my question, so thanks :)

12. Jul 20, 2012

### kenewbie

Thanks, I'll read that. The question now is, of course, HOW can this behavior arise, what is happening which makes this possible. But I'll read that thread and do a little digging on my own, it is far ahead of what I am supposed to be looking at anyway.

Thanks again.

13. Jul 20, 2012

### Simon Bridge

Re: Emission of light from a solid object

I'd go along with that ... I found: