Energy Transfer, Photons, and Kinetic Energy
How does electromagnetic energy of photons convert itself
into internal energy of some object? For example if atoms of some
object absorb photons, their electrons jump to higher orbit, but how
does that affect kinetic energy of atoms?
Even though a photon has no mass, it still has momentum and kinetic
energy. The formulas we learn in introductory physics courses only
apply to objects with significant mass. When you get to the level of
single particles, such as electrons and photons, quantum physics takes
over. It turns out that day-to-day physics (i.e. Newtonian physics) is
an approximation of quantum physics for a set of many particles (e.g. a
The photon adds most of its energy to the kinetic energy of the
electron. The photon, electron, and nucleus interact. Energy and
momentum are conserved. Most of the energy enters the electron, raising
it to a higher state. This is a rotational effect. In order to
maintain momentum, a small amount of the energy enters the motion of the
nucleus, giving the atom a new velocity through space and a new kinetic
A simple model of this is a bullet entering a block of wood. After the
collision, the block is moving. Some of the bullet's energy entered the
motion of the block. A large portion of the bullet's energy was devoted
to changing the structure of the block.
Dr. Ken Mellendorf
Illinois Central College
OK, so your electron has jumped to a higher orbit (excitation).
It can return to the lower orbit (decay or de-excitation) two ways:
b) thermal de-excitation
Photoemission is the same process as the absorption, reversed.
You seem to understand that
photoemission does not turn the photon's energy into heat.
But thermal de-excitation is actually more common than photoemission
in dense solids.
(How many things glow like fluorescent paint or an LED?
Relatively few; only our intelligent manipulation makes them common to us.)
How does thermal de-excitation happen?
I have never heard anybody bother to explain or analyse it.
But I think it is easy to imagine that
if two free atoms gently bump each other,
and one is electronically excited, then
1) the change in local fields experienced by the excited atom
can trigger the de-excitation,
2) the energy of the excitation can be coupled into
pushing the atoms apart more forcefully than they came together.
It is like bumping into a spring mechanism
that suddenly unlatches and pushes you away strongly.
When atoms are already linked together in molecules,
often the energy of excitation finds its way out of the atom
by forcing a vibration into the molecular bonds.
Each atom has mass and each bond is spring-like.
Crystals do this too, whether covalent, ionic or metallic.
Ordinary silicon rectifier diodes do not emit light,
even though every electron of forward electric current
creates an excited electron in the crystal.
This lack of light is because every new excited electron
decays by emission of "phonons" (quanta of vibration in the mass of atoms)
rather than by emission of photons (light).
Phonons can be coherent and used as signals,
but these phonons are pretty randomly distributed,
in essence just more heat.
Most phonons have less energy than a photon of visible light.
one photon typically makes bunch of new phonons,
such that energy is conserved.
A visible photon is typically 2-3eV; the resulting phonons range from 0.01 to maybe 0.5 eV.
Phonon-mediated photoemission also happens more than pure photo-emission.
If a photon of blue is absorbed, and the atom re-emits
one photon of red and a some phonons carrying the energy difference,
then some of the energy has been turned into heat.
If deep in the solid, this can happen over and over until all used up.
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Update: June 2012