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Author Topic: What is the energy imparted by light when it interacts with matter?  (Read 1897 times)

Offline mxplxxx

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When a photon impinges on something it, presumably, transfers its energy to that something. Is this energy given by E=hf where h is Planck's constant and f is the frequency of the photon? Can you give me examples of research where this has been measured to be the case?


 

Offline damocles

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The maximum energy a photon can transfer is indeed given by E = hf as you say. This is what happens when the photon is 'absorbed' (i.e. disappears). It is also possible for a photon to be 'inelastically scattered' and transfer only part of its energy. In this case the photon (or another one; photons have no identity tags) comes out of the encounter with a lower energy, such that E = h (f {in} - f {out} ).

There are many ways to establish this experimentally. Perhaps the simplest is obtained via photoionization: The absorption of light by an atom occurs at several very sharp and specific frequencies, seen as lines in the absorption spectrum. These lines get closer together at higher frequencies until they reach a 'series limit' where they merge into a continuous absorption.

The lines mean that photons of a particular frequency have been absorbed to produce a higher allowed energy level of the atom from its normal state. The 'series limit' is where a photon has transferred enough energy to remove an electron completely from an atom, and form a positive ion.

This same 'ionization energy' can also be measured electrically. So the frequency of the photon can be directly related to the voltage of the electrical pulse required to produce the same effect, and the electrical energy is simply the voltage multiplied by the charge on an electron.
 

Offline mxplxxx

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Thanks for your reply. I was looking for a simple experiment where the photon is fully absorbed and the energy of the absorbing body is measured before and after.
 

Offline Bored chemist

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The classic example is where a photon hits and electron and the energy is measured.
http://en.wikipedia.org/wiki/Photoelectron_spectroscopy
 

Offline mxplxxx

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Re "inelastic scattering", are we talking classical or quantum physics? If classical, then the equation E=hf would not be applicable I would have thought.
 

Offline damocles

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We are (or at least I am) definitely talking quantum, not classical, but non-relativistic quantum theory, not QED.

(1) The photon is a quantum particle; light is not quantised in classical theory, so E = hf is fairly meaningless.
(2) read the posts carefully -- there is a very important distinction between "classic" and "classical"
 

Offline mxplxxx

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Ok, that all makes sense. Thanks a lot for your replies.
 

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