If a photon of wavelength $\lambda = 2\,\unicode{x212B}$ hits a stationary electron, and scatters at $60^{\circ}$, its energy loss is about $0.6\,\%$. On the other hand, if its wave length is $\lambda = 0.032\,\unicode{x212B}$, its energy loss is about $28.7\,\%$.

Why does this make sense? I would think that a more energetic particle will care less than a less energetic particle. But maybe this is because I haven't yet the intuition behind photons as light particles.


There are two separate ideas to understand here. The second one, which your numbers relate to, is the energetics of the collision, determined by conservation of energy and momentum. But the first one is the probability of a collision in the first place (and the probability of scattering a photon to a particular angle).

Scattering Probability

This is described by the Klein-Nishina forumla which gives the scattering cross section (essentially, the probability) of scattering to a given angle for a photon of a given energy. The key point is that if the photon energy is much lower than the rest mass of an electron, the photon has more or less equal probability to scatter in all directions (give or take polarisation effects). This is the case for a photon of wavelength $2\,\unicode{x212B}$ which has an energy of $6.2\,\text{keV}$, compared to the electon rest mass of $511\,\text{keV}$.

For more energetic photons, the probability of scattering to large angles (large changes in momentum for both particles) decreases rapidly while the cross section for tiny changes in angle remains very high.

angle dependence for photon scattering

You can see for the $10\,\text{MeV}$ line, the photon, if it interacts at all, is very likely to be deflected by an angle very close to $0^{\circ}$. You $\lambda=0.032\,\unicode{x212B}$ case corresponds to an energy of ${}\approx 390\,\text{keV}$

So basically, the $390\,\text{keV}$ photon is less likely than the $6\,\text{keV}$ photon to scatter through $60^{\circ}$ (by a factor of something around $1/2$)


The is basically conservation of energy and momentum. If the scattering does happen, then actually the most energetic photon is always going to lose more energy at a given angle. The electron is initially at rest (since we are working in the electron's reference frame). During the scattering, the photon transfers some energy to the electron.

If we are talking about a photon initially travelling along the $x$-axis, and scattering through $60^{\circ}$, then it should be obvious that the momentum transferred to the electron is going to increase as it is hit by higher energy photons. If the photon scatters through $60^{\circ}$ and loses only a few percent of its energy, then it has lost roughly half its $x$ momentum, so the $x$ momentum transferred to the electron is roughly proportional to the energy of the photon. There is also some a $y$ component to the momentum obviously, but in the end the final energy of the electron is going to increase with increasing photon energy. So necessarily, the change in photon energy must be greater for more initially energetic photons.

So in the final analysis, by looking at the result of a $60^{\circ}$ scatter, it is no surprise that the highest energy photon loses the most energy. And your intuition is sort of right: the higher energy photon is less likely to interact in the first place, and less likely to be scattered through large angles if it does interact with the electron. But your intuition is misleading you because it is irrelevant if you are already talking about a $60^{\circ}$ scatter - you are implicitly ignoring the parts of the Compton scattering scenario where the higher energy photon is less likely to end up with a $60^{\circ}$ scatter in the first place.


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