Topic 15.6: Compton Scattering

What Topic 15.6 requires

Topic 15.6 sits in Unit 15, Modern Physics, which the CED weights at 12-15% of the multiple-choice section and estimates at about 14 to 22 class periods. It is the shorter of the unit's two photon-and-matter topics: 15.5 makes the case for photons from energy, and 15.6 makes it again from momentum.

15.6.A, describe the interaction between photons and matter using Compton scattering. One learning objective, three numbered essential knowledge statements.

• 15.6.A.1 In Compton scattering, a photon interacts with a free electron. The Compton effect is when the photon emerging from the interaction has a lower energy and longer wavelength than the incoming photon, and the magnitude of the change is related to the direction of the photon after the collision.
• 15.6.A.2 Compton scattering provides evidence that light is "a collection of discrete, quantized energy packets called photons".
• 15.6.A.2.i It can be explained by treating a photon as a particle and applying conservation of energy and conservation of momentum to the collision between the photon and electron.
• 15.6.A.2.ii The transfer of a photon's energy to an electron changes the photon's energy, momentum, frequency and wavelength. Relevant equations: $E = hf$ and $\lambda = h/p$.
• 15.6.A.3 The change in wavelength a photon experiences after colliding with an electron is related to how much the photon's direction changes. Relevant equation:

The boundary statement, whole: "AP Physics 2 includes full quantitative and qualitative treatments of conservation of momentum in two dimensions."

Read that one carefully, because it runs the opposite way from most boundary statements. It does not fence anything off. It opens something up: two-dimensional momentum conservation, quantitatively, is inside this course. So a Compton question can legitimately ask you to resolve momentum into components and solve for the electron's direction, which is what the third worked example below does.

Four suggested skills, one fewer than 15.5 and 15.7 carry.

• 1.A Create diagrams, tables, charts, or schematics to represent physical situations.
• 2.B Calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway.
• 2.C Compare physical quantities between two or more scenarios or at different times and locations in a single scenario.
• 3.C Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 1.A is listed first for a reason: this topic is a collision diagram. Draw the incoming photon, the scattered photon at angle $\theta$, and the recoil electron, and most of the reasoning becomes visible. Skill 2.C is the angle comparison, and 3.C is the "why does this prove photons exist" argument.

A photon that carries momentum

Everything in this topic follows from one move, and 15.6.A.2.i names it: treat the photon as a particle and apply the two conservation laws you already know from mechanics.

A particle has momentum. Essential knowledge 15.6.A.2.ii lists $\lambda = h/p$ as a relevant equation, and the sheet prints it in the Modern Physics group, so rearranged the photon's momentum is

That single relation is what makes a photon behave like something you can bounce off an electron. It is worth pausing on how strange it is, because the mechanics definition on the same sheet is $\vec{p} = m\vec{v}$, and a photon has no mass to put in it. The photon's momentum is not $mv$; it is set by wavelength alone. A shorter wavelength is a more energetic photon and a higher-momentum photon at the same time, because $E = hf$ and $p = h/\lambda$ move together.

Now run a collision. Momentum and energy are both conserved, exactly as in Topic 4.3 and Topic 3.4, and the electron starts at rest and ends up moving. It gained kinetic energy, and that energy came from the only other thing in the system:

Since the photon's energy dropped, $E = hf$ says its frequency dropped, and $\lambda = c/f$ says its wavelength grew. That is 15.6.A.1 in three steps, and it also explains why 15.6.A.2.ii bothers to list all four quantities: energy, momentum, frequency and wavelength are not four separate facts about a photon but four names for one number.

Two details in the CED's wording are worth holding on to.

"A free electron." 15.6.A.1 specifies it. A free electron is not bound in an atom, so no energy is spent breaking it loose, and the whole exchange is a clean two-body collision. This is precisely where 15.6 parts company with 15.5, where the electron is bound and the work function is the price of freeing it.

"Lower energy and longer wavelength." The CED writes the direction into the essential knowledge. The scattered photon can never come off with a shorter wavelength in this process. The equation enforces it, as the next section shows.

The Compton shift equation, term by term

Four things to read off it, and the fourth is the one worth an exam mark.

$\Delta \lambda$ is the increase, $\lambda' - \lambda$. The scattered wavelength is $\lambda' = \lambda + \Delta\lambda$. Never subtract.

$\theta$ is the photon's scattering angle, measured from its original direction of travel, and in degrees. It is the angle the photon turned through, not the electron's recoil angle. Mixing those two up is the standard way to get a plausible wrong number.

$(1 - \cos\theta)$ runs from 0 to 2 and is never negative. At $\theta = 0$ the photon carried straight on, nothing happened, and $\Delta\lambda = 0$. At $\theta = 180^\circ$ the photon came straight back and $\cos\theta = -1$, giving the largest possible shift, $2h/(m_e c)$. Because this factor cannot go negative, the wavelength cannot decrease, which is the CED's "lower energy and longer wavelength" written as algebra.

$h/(m_e c)$ is a fixed length made of three constants, and it does not appear on the sheet as a number. The AP Physics 2 sheet prints the combination symbolically inside the equation and never evaluates it, and the CED never uses the phrase "Compton wavelength" at all. So compute it from the constants block rather than recalling it:

That is $2.43 \, \text{pm}$, or $0.00243 \, \text{nm}$. Doing this once at the top of a Compton question and keeping the number is faster than carrying three constants through every line.

Here is the whole angular dependence in one table, using the cosine values the CED's own Table of Information prints for common angles.

Two readings of that table. First, the shift at $90^\circ$ is exactly $h/(m_e c)$, which makes it the easiest case to check an answer against. Second, the growth is not proportional to the angle: doubling $60^\circ$ to $120^\circ$ triples the shift, because $1 - \cos\theta$ goes from 0.5 to 1.5. Any 2.C comparison question in this topic is a question about $1 - \cos\theta$ and nothing else.

The shift does not depend on the incoming wavelength

Look at the right-hand side of the equation again. Planck's constant, the electron mass, the speed of light, and the angle. No $\lambda$. The absolute increase in wavelength is the same whether the photon arrived as a radio wave or a gamma ray, and that is the single most surprising statement in this topic.

It is also the source of the topic's most useful practical fact, because "the same absolute shift" and "the same noticeable shift" are very different things. What an experiment sees is the fractional change, $\Delta\lambda / \lambda$, and that does depend on the incoming wavelength, through the denominator.

Take the maximum shift, $\Delta\lambda = 4.85 \times 10^{-12} \, \text{m}$ at $180^\circ$, and compare two beams.

• Green light, $\lambda = 500 \, \text{nm}$. The fraction is $\dfrac{4.85 \times 10^{-12}}{5.00 \times 10^{-7}} = 9.7 \times 10^{-6}$, about one part in a hundred thousand. Nothing you could measure with a school spectrometer, and nothing that changes the colour.
• An X-ray, $\lambda = 0.0100 \, \text{nm}$. The fraction is $\dfrac{4.85 \times 10^{-12}}{1.00 \times 10^{-11}} = 0.485$, a 48.5% increase. Impossible to miss.

So Compton scattering is an X-ray and gamma-ray phenomenon, not because the physics changes but because a fixed shift of a few picometres is only a large fraction of a picometre-scale wavelength. If a question hands you visible light and asks whether the Compton shift matters, that ratio is the answer, and it is a clean skill 3.C justification.

The same reasoning explains why the photon's energy loss depends on the incoming wavelength even though the wavelength shift does not. A short-wavelength photon that loses a fixed absolute amount of wavelength has lost a large fraction of its wavelength, and therefore a large fraction of its energy. Written out, the electron's kinetic energy is

with $\lambda$ and $\lambda'$ multiplied in the denominator. That form is worth writing down, because it is also numerically safer than subtracting two nearly equal photon energies, a point the first worked example returns to.

Conservation of energy and momentum, in two dimensions

Essential knowledge 15.6.A.2.i does not say Compton scattering is consistent with conservation of energy and momentum. It says the effect can be explained by applying them to the collision. The conservation laws are the mechanism, not a check afterwards, and the boundary statement then confirms that two-dimensional momentum conservation is fully in scope for this course.

Set the collision up the way skill 1.A wants it drawn. The incoming photon travels along the $+x$ axis. Declare that axis first and hold it: $+x$ is the incoming photon's direction of travel, and $+y$ is the side the scattered photon goes to. The electron starts at rest.

Energy. The electron's rest energy is unchanged by the collision, so it cancels from both sides and the accounting is exactly

Momentum, $x$ component. Before, only the photon moves: $h/\lambda$. After, the photon contributes $(h/\lambda')\cos\theta$ and the electron the rest.

Momentum, $y$ component. Before, zero. After, the photon's $(h/\lambda')\sin\theta$ must be cancelled by the electron's $y$ momentum, so the electron recoils on the opposite side of the axis from the scattered photon. That is the qualitative result to say out loud in any 1.A answer: if the photon goes up, the electron goes down.

Three limiting cases fall straight out and are worth recognising on sight.

• $\theta = 0^\circ$. No shift, no energy transferred, no recoil. The photon missed, in effect.
• $\theta = 90^\circ$. The photon keeps no $x$ momentum, so the electron carries the entire original $h/\lambda$ forward, plus enough $y$ momentum to cancel the photon's.
• $\theta = 180^\circ$. Maximum shift, maximum energy transfer, and the electron is driven straight along $+x$ with more forward momentum than the photon originally had, because the photon's momentum reversed sign.

That last case looks wrong until you write it: $p_{\text{electron}} = h/\lambda + h/\lambda'$, a sum rather than a difference, exactly as a ball bouncing back off a wall delivers more impulse than one that stops dead. The same argument appears in Topic 4.4 and in the conservation of momentum guide.

One honest limitation. The exact Compton relation comes out of relativistic energy and momentum for the electron, and AP Physics 2 is algebra-based, so the CED hands you the finished formula rather than a derivation. What you can do at this level is check consistency, and the third worked example does: at X-ray energies the electron's kinetic energy is a tiny fraction of its rest energy, so the non-relativistic $p = \sqrt{2 m_e K}$ agrees with the momentum found by adding components to about a tenth of a percent. That agreement is evidence, not proof, and it stops being close at higher photon energies.

Why this counts as evidence for photons, and how it differs from 15.5

Both 15.5.A.2.ii and 15.6.A.2 end with the same phrase, "evidence that light is a collection of discrete, quantized energy packets called photons". The CED asks for the same conclusion twice, from two different experiments, which is a strong hint that a question can ask you to distinguish them.

A classical wave scattering off charges predicts something quite specific: the charge is driven back and forth at the wave's frequency, and it therefore re-radiates at that same frequency. A wave model has no mechanism for the scattered light to come back at a different wavelength, and certainly not at a wavelength that depends on the direction you look. The observed shift is not a small correction to that picture; it is a result the picture cannot produce at all.

The particle model produces it in one line, because a collision between two particles necessarily redistributes energy, and how much depends on the geometry.

Two rows in that table are the ones students most often get backwards. Compton scattering has no threshold and no work function, because 15.6.A.1 specifies a free electron and a free electron costs nothing to move. And the photon is not absorbed, it is deflected, which is why the topic has a scattering angle at all.

The clean one-sentence version of the 3.C answer: the photoelectric effect shows that light delivers energy in fixed packets, and Compton scattering shows that those packets also carry momentum and collide like particles.

Traps, and how 15.6 gets tested

Pair each suggested skill with the mistake it catches.

1.A, draw it. The trap is putting the electron on the same side of the axis as the scattered photon. Transverse momentum started at zero, so they must go to opposite sides.

2.B, calculate. Two traps. First, forgetting that $\Delta\lambda$ is added, giving a scattered wavelength shorter than the incident one, which contradicts 15.6.A.1 on its face. Second, subtracting two photon energies that agree to three or four digits and keeping the difference to three figures. Use $K = hc\Delta\lambda / (\lambda\lambda')$ instead.

2.C, compare. Comparing the wrong quantity. The absolute shift is the same for every incoming wavelength; the fractional shift and the energy transferred are not. Say which one you mean.

3.C, justify. Asserting that Compton scattering proves photons without saying what the wave model fails to predict. Name the failure: a classical wave re-radiates at the frequency that drove it, so it cannot produce an angle-dependent wavelength change.

Five errors worth naming directly.

• Using the electron's recoil angle in the formula. The $\theta$ in $\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)$ is the photon's turn.
• Putting the incoming wavelength on the right-hand side. It is not there. If your shift changed when the incident wavelength changed, you invented a term.
• Quoting a remembered value for $h/(m_e c)$. It is not printed on the AP Physics 2 sheet as a number. Build it from $h$, $m_e$ and $c$, all three of which are in the constants block.
• Treating $1 - \cos\theta$ as proportional to $\theta$. It is not, as the table above shows.
• Expecting a threshold. There is none. Every scattering angle gives some shift except $0^\circ$, and no minimum photon energy is required.

On exam shape: the AP Physics 2 exam runs 3 hours, with 42 multiple-choice questions and 4 free-response questions, weighted 50/50. The unit overview notes that the first free-response question, the Mathematical Routines question, asks students to calculate or derive an expression and to create and justify claims from a representation, and that it can draw on any of the seven units. A Compton question fits that shape well, since it has a formula, a diagram and a comparison in it.

For the photon model underneath this topic and for $\lambda = h/p$ applied to matter rather than light, see Topic 15.1. For the other photon-and-matter interaction, see Topic 15.5. For the mechanics this topic borrows, see Topic 4.1 on momentum itself and the guide on what is conserved in a collision. The Unit 15 overview sets the eight topics in order and the AP Physics 2 equation sheet holds the ten Modern Physics entries.