When an atom of an element is irradiated with a monochromatic light, what will happen in the case that the frequency of the light does not match the frequency needed for any of the several possible electronic transitions? Will the photon just pass through the atom? Or will it collide with the nucleus and get deflected or reflected? The nucleus only occupies a tiny space as compared to the atomic volume and so will a whole collection of such atoms be transparent to the respective monochromatic light as most photons just pass through the empty space in the atom?

  • $\begingroup$ I'll comment on the part of the question about the size and empty space. The quantum world is different from classical. Atoms and other particles are not little balls. There is no such thing as a "direct hit". Size is not well defined there either. So the size of the nucleus and empty space are irrelevant. Even if the nucleus were the size of an atom, if a photon does not interact, it would just pass through regardless. No such thing there as "solid matter", but rather waves of probability of interactions. $\endgroup$ – safesphere Nov 28 '17 at 13:55
  • $\begingroup$ @safesphere I do have some very basic information about how the quantum world behaves and the fact that particles like the electron are defined by a superposition of wavefunctions and not by classical properties like position and momentum. However, in this analogy what would it actually mean for a photon to not "interact" with a particle, say an electron for example? What is an inherent property that a photon of particular wavelength has that allows it to "interact" with certain particles but not others? Thanks! $\endgroup$ – Rutwik Nov 28 '17 at 15:07
  • $\begingroup$ Two questions in your comment. (1) To not interact means the photon (or wave, if you prefer) will pass as if through the empty space. (2) What propertirs define photon interactions? Photons interact only with electrically charged particles (ignoring higher orders for now). Energy is important, but may vary in scattering. There also always is probability (even with the ideal parameters the photon sometimes may just pass through). I will leave the rest to the experts (like spin, etc.) to say if it matters or not. $\endgroup$ – safesphere Nov 28 '17 at 15:23

first of all, you need to understand the use of frequency of light in abstracting elctrons from the metallic surfaces.we have a fix threshold frequency for a metallic surface, for which the incident light should have the frequency equal or greater than the threshold frequency.if any incident light have the same frequency as threshold frequency,then electrons will come out the attractive forces of electrons but never escape from the metal surface(this term is call work function of metal).elcetrons required a frequency above the threshold fruency because to overcome the attractive forces and remaining frquency will use to escape from metallic surfaces by gaining kinetic energy.

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  • $\begingroup$ Apologies but I don't see how the OP refers implicitly or explicitly to metallic surfaces. $\endgroup$ – ZeroTheHero Nov 28 '17 at 17:09

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