electronsenergyphotonsquantum mechanics
I am studying quantum physics and I have a question: what is the physical explanation for electrons having less energy than photons with the same wavelength?
Energy of a photon : $E = h c/\lambda$.
Energy of an electron: $E = h^2/(2m\lambda^2)$
More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and harmonic generation at metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
Okay, so just to be clear I am going to consider processes in which a photon and an atom at some energy level go in, and the photon and atom exchange energy (and momentum) such that a photon with a shifted (either higher or lower) energy comes out, while the atom ends up in a different internal electronic state than it started in. A general diagram looks like:
where non-resonant light comes in, and causes a transition between states 1 and 2 with the outgoing photon's energy also shifted in the corresponding way. I have not showed the momentum change, but that will be determined by the energies and the geometry of the situation. I've drawn this where $E_2>E_1$, but the reverse process is also possible in which the photon gains energy.
First of all, you are certainly right to wonder why this process shouldn't be allowed. As I've mentioned in a different context, a useful way to think about many physical processes, attributed to Gell-Mann, is that "everything not forbidden is mandatory." So, when the internal atomic transition + photon shift can be accomplished in a way that conserves energy, and angular and linear momentum (and also obeys some other selection rules such as those involving parity), we should expect that it is possible. And it is!
As mentioned, these processes are normally called "Raman scattering," and are an important tool in materials science for the study of vibrational levels of materials. However, to directly address the question of charles boyant, the idea of a Raman transition is more general than this. For example, in atomic physics, Raman transitions (in a slightly different form known as stimulated Raman) are often used to go between two spin states of an atom. In this case, the polarization of the photon must change along with its energy so that all conservation laws are obeyed.
Okay, so if this can happen why did you learn that atoms can only absorb light at certain frequencies that correspond to atomic transitions? There were probably two motives behind this simplification:
Although Raman processes are allowed, they generally occur with very low probability compared to absorption near a resonance, and also compared to scattering of photons without a change in photon energy. So in many cases they only have a very small impact on the overall atom-light interaction.
Because the photon never fully disappears, Raman scattering (as the name suggests) is normally thought of as an inelastic scattering process, instead of as a "partial absorption."
This way of distinguishing between absorption and inelastic scattering is particularly useful when comparing Raman processes to processes where the light is on resonance with an atomic transition. This would be the case, for example, in which the light is resonant with the 1->E transition, and the atoms can then decay both to states 1 and 2. This has a similar result to the Raman process, in the sense that photons of one energy come in and photons with a shifted energy corresponding to the difference between atomic levels come out. However, since the absorption is a resonant process, the transition strength, wavelength dependence, and actual atomic state during the process is different in these two cases.
Best Answer
For the photon we have $$E_\gamma = \frac{hc}{\lambda}$$ and for the electron $$E_e = \frac{h^2}{2m\lambda^2} =\frac{hc}{\lambda} \frac{h}{2mc\lambda} = E_\gamma \frac{h}{2mc\lambda}. $$ You can check that the proportionality factor is dimensionless. So what you are asking is why this quantity is less than unity. But recall that $$\frac{h}{\lambda} =p$$ where $p$ is the momentum. What we are looking at is really (one half) the ratio $$\frac{pc}{mc^2} = \frac{mvc}{mc^2}$$ where I assumed that $v \ll c$, that is, we have a non-relativistic electron. Then we get the result you stated in your question. On the other, hand if we don't make this approximation we have the ratio $$\frac{pc}{mc^2} =\frac{mv\gamma c}{mc^2} = \frac{v\gamma}{c}$$ which is unbounded when $v \to c$.
You could also argue from Einstein's $$E^2 = m^2 + p^2$$ (in units where $c = 1$). For $m = 0$ we have of course $E = p$. If you make a Taylor expansion of $E$ for $m\neq 0$, $$E = m + \frac{p^2}{2m} + \ldots$$ you see that the kinetic energy, compared to the energy of a massless particle has a factor $p/m$ (as we found above). The non-relativistic regime is precisely when this quantity is small, and if it is not, we have to include terms proportional to $p^4/m^3$ and higher, and again that the energy can be larger for a massive particle than for a massless particle with the same momentum. So the answer to your question really is: because you are considering non-relativistic particles.