Atomic excitation via photon absorption

[tyneoh]

Hi guys! I have come across a problem I can't seem to wrap my head around. I've learned that E.M. radiation can be propagated by discrete packets of energy , photons. Is the energy of each photon discrete, or can it have a continuous range of energies depending on its frequency? I would be inclined towards the latter, though. Then what does discrete actually mean in terms of energy propagation? For an atom to be excited, it has to absorb a photon whose energy is exactly equal to the energy needed for a quantum jump. Why can't the atom absorb what energy it needs from the photon(assuming that the photon has more energy than the atom needs) and leave the photon with the remainder? Whereas for the photoelectric effect, any photon above the threshold energy can be absorbed? Thank you for reading, have a great day.

 

[WannabeNewton]

A photon can only carry an energy $\hbar \omega$, nothing more and nothing less. In atomic transitions, electrons have to jump between energy states of the atom. Because these are bound states, the energy levels are discrete so an atom can only jump between energy levels if it has exactly enough energy gained to cover the difference. It canot take what it needs from a photon and leave the photon with the rest because the photon can only carry energy in the amount aforementioned. In the photoelectric effect, the electron is ejected from the atom so it goes from a bound state to a free state. The energy spectrum of free particles is continuous, unlike the discrete energy spectrum of bound particles, so the ejected electron can take on any energy it chooses beyond the minimum needed by the work function of the metal. Indeed in the photoelelctric scattering cross section we must take the electron as being ejected into a narrow continuous group of final free particle energy states.

Why do bound states have discrete energy spectra? It is just a consequence of the boundary conditions imposed on Schrodinger's equation. In classical EM if we have radiation bound to a cavity then Maxwell's equations together with these boundary conditions yeild a discrete spectrum of Fourier modes of the cavity radiation. With Schrodinger's equation the mathematical situation is almost entirely analogous (the PDEs differ in type) except now it applies to the spectrum of energy eigenstates of particles and as such also to the definite energies obtainable in this spectrum.

 

[tyneoh]

A photon can only carry an energy $\hbar \omega$, nothing more and nothing less. In atomic transitions, electrons have to jump between energy states of the atom. Because these are bound states, the energy levels are discrete so an atom can only jump between energy levels if it has exactly enough energy gained to cover the difference. It canot take what it needs from a photon and leave the photon with the rest because the photon can only carry energy in the amount aforementioned. In the photoelectric effect, the electron is ejected from the atom so it goes from a bound state to a free state. The energy spectrum of free particles is continuous, unlike the discrete energy spectrum of bound particles, so the ejected electron can take on any energy it chooses beyond the minimum needed by the work function of the metal. Indeed in the photoelelctric scattering cross section we must take the electron as being ejected into a narrow continuous group of final free particle energy states.

Why do bound states have discrete energy spectra? It is just a consequence of the boundary conditions imposed on Schrodinger's equation. In classical EM if we have radiation bound to a cavity then Maxwell's equations together with these boundary conditions yeild a discrete spectrum of Fourier modes of the cavity radiation. With Schrodinger's equation the mathematical situation is almost entirely analogous (the PDEs differ in type) except now it applies to the spectrum of energy eigenstates of particles and as such also to the definite energies obtainable in this spectrum.

What if the frequency of the photon decreases to the corresponding amount of the remainder energy? i.e. the frequency of the photon decreases after some of its energy is absorbed by the atom.

 

[WannabeNewton]

What if the frequency of the photon decreases to the corresponding amount of the remainder energy? i.e. the frequency of the photon decreases after some of its energy is absorbed by the atom.

The photon is absorbed by the electron. It doesn't exist anymore.

 

[tyneoh]

The photon is absorbed by the electron. It doesn't exist anymore.

Isn't the energy of a photon continuous, since its frequency can take on any value?

 

[WannabeNewton]

Isn't the energy of a photon continuous, since its frequency can take on any value?

A single photon has a fixed frequency by definition.

 

[tyneoh]

A single photon has a fixed frequency by definition.

OK so just to clarify, the quantum theory explains that radiation is carried by discrete packets of energy (photons) whose energy must also be discrete? Would that mean orange light from a sodium lamp cannot be changed in any way into other colours by altering the frequency of the "orange" photons?

 

[Nugatory]

OK so just to clarify, the quantum theory explains that radiation is carried by discrete packets of energy (photons) whose energy must also be discrete?

Not quite. The quantum theory says that when radiation interacts with anything, it always delivers its energy in discrete packets; the theory is pretty much silent about what is "carrying" the radiation when it's just traveling through space. Thus, you cannot think of photons as particles traveling through space like little bullets (although this is the mental model that most people form when they first hear about them).

 

[tyneoh]

Not quite. The quantum theory says that when radiation interacts with anything, it always delivers its energy in discrete packets; the theory is pretty much silent about what is "carrying" the radiation when it's just traveling through space. Thus, you cannot think of photons as particles traveling through space like little bullets (although this is the mental model that most people form when they first hear about them).

Then does the theory disallow the possibility of absorbing a portion the energy of an of an incoming say, violet photon and and render the photon into say, a green photon whose frequency corresponds to that of the remainder energy

 

[Nugatory]

Then does the theory disallow the possibility of absorbing a portion the energy of an of an incoming say, violet photon and and render the photon into say, a green photon whose frequency corresponds to that of the remainder energy

Absorb a photon of given energy, then emit another of different energy? Sure, that works fine. Whether that's what happens or not depends on the particular interaction you're considering. For example, in the photoelectric effect the photon is absorbed delivering all of its energy to the electron, and any energy beyond what's needed to kick the electron free goes into the kinetic energy of the emitted electron. Other scattering processes will behave differently; for example Compton scattering will release a photon of lower energy than the one that came in.

 

[tyneoh]

Absorb a photon of given energy, then emit another of different energy? Sure, that works fine. Whether that's what happens or not depends on the particular interaction you're considering. For example, in the photoelectric effect the photon is absorbed delivering all of its energy to the electron, and any energy beyond what's needed to kick the electron free goes into the kinetic energy of the emitted electron. Other scattering processes will behave differently; for example Compton scattering will release a photon of lower energy than the one that came in.

Right, how about an atom absorbing the excitation energy it needs from an incoming blue photon, and the photon leaves with the remainder of its original energy, except now that it has a lower frequency, say red?