A collision with three participants

Write the collision with all three participants present:

$$hf = hf' + K_{\text{electron}},$$

with momentum balancing as vectors in two dimensions alongside it.

Describing the photon as absorbed and gone leaves nothing to explain what a detector placed off-axis is registering, and it produces an energy equation with the $hf'$ term simply missing. It also predicts the wrong outcome: if the photon vanished, the electron would take all its energy, and there would be no wavelength shift to measure at all.

Some energy always leaves with the scattered photon, which is why the shift is a finite amount rather than the photon ceasing to exist. Absorption and scattering are different processes, and this one is scattering.

$$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}\left(1 - \cos\theta\right).$$

That is a change, so the scattered wavelength is

$$\lambda' = \lambda + \Delta\lambda.$$

Computing the shift and submitting it as $\lambda'$ is the error. For a $0.100$ nm X-ray scattered through $90^\circ$ it means reporting $0.0024$ nm instead of $0.1024$ nm, which is off by a factor of forty.

The tell is easy to spot: a scattered wavelength shorter than the incident one would say the photon gained energy from a stationary electron, which is not available to it. So after computing, check the direction: $\lambda'$ must always come out longer and the photon less energetic.

The shift is a small correction sitting on top of a larger number, not a replacement for it.

Read the formula for what it actually contains: $h$, $m_e$, $c$ and $\theta$. The incident wavelength appears nowhere.

So the shift at $90^\circ$ is the same $0.0024$ nm whether the incoming photon is an X-ray or green light. Scaling $\Delta\lambda$ by the incident wavelength, or expecting a more energetic photon to shift more, reads a dependence into the formula that is not there.

The angle runs the whole range:

1. $\theta = 0$ (straight ahead): $1 - \cos 0 = 0$, so no shift and no interaction to speak of.
2. $\theta = 90^\circ$: $\Delta\lambda = h/(m_ec) \approx 0.0024$ nm, the Compton wavelength.
3. $\theta = 180^\circ$ (straight back): the maximum, $2h/(m_ec)$.

What does depend on the incident wavelength is the fractional change, and that is what decides whether the effect is measurable at all. On a $0.1$ nm X-ray, $0.0024$ nm is a detectable $2\%$; on a $500$ nm photon it is five parts per million. That is why the experiment is done with X-rays.

The photoelectric effect showed that light delivers energy in lumps. Compton scattering showed that a photon also carries momentum, and that a photon and an electron collide like two particles, obeying both conservation laws together.

That is a stronger claim than the photoelectric result and harder to explain away. A classical wave shaken by an electron would re-radiate at the same frequency it arrived with, so the measured shift with angle has no wave account at all.

It also explains the size of the effect. The constant $h/(m_ec)$ contains the electron's mass, so a heavier target shifts the photon less: scattering off a tightly bound electron, effectively off the whole atom, produces essentially no shift, which is the unshifted peak that appears alongside the shifted one in the real data.