2
$\begingroup$

We know that whenever white light falls on an object, photons of particular wavelength(de broglie's wavelength) gets absorbed by atoms which causes excitation of electron and then electron releases same energy photon to decrease its energy. This emitted photon we see as the color of the object. What happens with the remaining photons that atom doesn't absorbs?

$\endgroup$
2
$\begingroup$

When white light falls on a red object the red photons are not absorbed, we say they are reflected. The blue and green photons are absorbed by electrons, most of these electrons will release the energy as IR photons (heat).

$\endgroup$
  • $\begingroup$ You are saying that red photons are not absorbed but reflected. Can you please explain the phenomenon of reflection of photon. $\endgroup$ – Vasu Goyal Sep 30 '18 at 8:29
  • $\begingroup$ What comes to mind is 2 types of reflection. 1) a mirror with metallic coating, actually the photon does become immediately absorbed and immediately re-radiated by a lone electron, the electron is free to jump up and down in energy. 2) Scattering (most common) - the electrons are not able too jump to an energy equal to the photons so the photon's path gets deflected by an interaction with its EM field and the EM field of an electron in the substance, there will be few interactions one after the other until the photon finds its way out of the material in a random direction. $\endgroup$ – PhysicsDave Sep 30 '18 at 12:57
1
$\begingroup$

The color of an object is in general determined by the light that it reflects. Of course, light sources such as the sun or lamps are exceptions. Moreover, in general the absorption of light does not result in emission of light. Fluorescence is an exception.

The reflection and absorption of light in a dielectric material or a metal cannot be described in terms of interaction with single atoms. Rather, collective excitations involving many atoms are involved. Light propagates in a dielectric as a mixture involving these excitation and consequently obeys a modified, effective medium equation in which the propagation speed is altered. At the boundary partial reflection occurs because of the abrupt change in propagation properties.

$\endgroup$
0
$\begingroup$

When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and changes angle

  2. inelastic scattering, the photon gives part of its energy to the atom and changes angle

  3. absorption, the photon gives all its energy to the atom and the absorbing electron moves to a higher energy level as per QM

Now you are saying that the color of the object is the photons that are being re-emitted. This is only partially true:

  1. some objects could glow even in the dark if no light shines on them, like heated metals, so they do not have to re-emit light. Or the Sun, does not re-emit light, it only emits light because of internal processes, that cause emission. These objects have their own color regardless if light shines on them or not.

  2. not all objects have their own color, like metals, do not have their own color. metals are usually silver, because. This is because as per QM, metals usually have available d orbits, and s electrons move to d orbits when certain light shones on them. Now this certain light in the case of metals is so high energy, that visible wavelength photons cannot do the job. All visible photons are reflected, and none absorbed by metals. So metals like silver do not have their own color. They just reflect all visible light. And they usually do not absorb nor re-emit any visible light.

  3. some objects do not absorb (or rarely) visible light, they just reflect it, like mirrors (this is because of 2.)

You are of course right, that some objects do have their own color, meaning that they absorb some visible photons and re-emit some visible photons.

So you are asking what happens with all the photons that do not get absorbed. Some photons get elastically scattered and some get inelastically scattered.

  1. elastic scattering is what happens in mirrors surfaces, that is reflection (this usually happens to visible light)

  2. inelastic scattering is usually what happens with non-visible wavelength light, these photons transfer vibrational energy to the molecules of the material, heating them up. (this usually happens to non-visible light)

  3. absorbed light (this can happen to both visible and non-visible light) can be re-emitted, but the relaxation of the electron is no always the same way as the excitation. relaxation can happen in multiple steps, called cascades, and so the electron might absorb a blue wavelength photon, and might re-emit two different wavelength photons, so the visible light might be re-emitted as a different color.

It is a common misconception that as you say a wall is white just because it reflects white light. Physically there is no such thing as white wavelength light. White light is a combination of all visible wavelength photons. Our eyes perceive white light as white because we have receptors for red, green, blue, and white light uses all these combinations, and uses all receptors at about the same ratio.

Now a white wall seems white because the Sunlight is not yellow, but the Sunlight is a white light. As white light from the Sun shines on a white object, that object absorbs and re-emits all wavelength photons, and we perceive it as white. If you shine red light on a white object, it will seem red, and it works with all colors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.