Compton scattering — wavelength shift derivation

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WHAT is happening

The photon carries:

• Energy $E = \dfrac{hc}{\lambda}$ (since $E = hf$ and $c = f\lambda$)
• Momentum $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$ (a photon is massless, so $E = pc$)

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HOW to derive it (from scratch)

We treat the electron relativistically (it can recoil fast). Symbols:

• photon before: energy $\frac{hc}{\lambda}$, momentum $\frac{h}{\lambda}\hat x$
• photon after: energy $\frac{hc}{\lambda'}$, momentum $\frac{h}{\lambda'}$ at angle $\theta$
• electron before: rest, energy $m c^2$, momentum $0$
• electron after: momentum $\vec p_e$, energy $E_e = \sqrt{(p_e c)^2 + (mc^2)^2}$

Step 1 — Conserve energy

$\frac{hc}{\lambda} + mc^2 = \frac{hc}{\lambda'} + E_e$

Why this step? Total energy is conserved; the electron starts with only rest energy $mc^2$.

So: $E_e = mc^2 + hc\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) \quad(\star)$

Step 2 — Conserve momentum (vector!)

Let the photon scatter at angle $\theta$, electron at angle $\phi$.

x: $\dfrac{h}{\lambda} = \dfrac{h}{\lambda'}\cos\theta + p_e\cos\phi$

y: $0 = \dfrac{h}{\lambda'}\sin\theta - p_e\sin\phi$

Why this step? Momentum is a vector, so each component is separately conserved.

Isolate the electron terms:

p_e\sin\phi = \frac{h}{\lambda'}\sin\theta$$ Square and add to **eliminate $\phi$** (because $\cos^2\phi+\sin^2\phi=1$): $$p_e^2 = \frac{h^2}{\lambda^2} - \frac{2h^2}{\lambda\lambda'}\cos\theta + \frac{h^2}{\lambda'^2} \quad(\star\star)$$ *Why eliminate $\phi$?* We don't care about the electron's exact direction — only the photon's wavelength shift. ### Step 3 — Use the relativistic energy relation $E_e^2 = (p_e c)^2 + (mc^2)^2$, so $p_e^2 c^2 = E_e^2 - (mc^2)^2$. Square $(\star)$: $$E_e^2 = (mc^2)^2 + 2mc^2\,hc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) + h^2c^2\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right)^2$$ Subtract $(mc^2)^2$ and divide by $c^2$: $$p_e^2 = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) + h^2\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right)^2 \quad(\star\star\star)$$ ### Step 4 — Set $(\star\star) = (\star\star\star)$ Expand $h^2(\tfrac1\lambda-\tfrac1{\lambda'})^2 = h^2(\tfrac1{\lambda^2} - \tfrac{2}{\lambda\lambda'} + \tfrac1{\lambda'^2})$. Equate: $$\frac{h^2}{\lambda^2} - \frac{2h^2\cos\theta}{\lambda\lambda'} + \frac{h^2}{\lambda'^2} = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) + \frac{h^2}{\lambda^2} - \frac{2h^2}{\lambda\lambda'} + \frac{h^2}{\lambda'^2}$$ The $\frac{h^2}{\lambda^2}$ and $\frac{h^2}{\lambda'^2}$ cancel on both sides: $$-\frac{2h^2\cos\theta}{\lambda\lambda'} = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) - \frac{2h^2}{\lambda\lambda'}$$ Rearrange the $h^2$ terms: $$\frac{2h^2}{\lambda\lambda'}(1-\cos\theta) = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right)$$ Note $\tfrac1\lambda - \tfrac1{\lambda'} = \tfrac{\lambda'-\lambda}{\lambda\lambda'}$. Divide both sides by $\frac{2h}{\lambda\lambda'}$: $$h(1-\cos\theta) = mc(\lambda'-\lambda)$$ > [!formula] Compton shift > $$\boxed{\;\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}\,(1-\cos\theta)\;}$$ > - $\dfrac{h}{m_e c} = \lambda_C \approx 2.43\times10^{-12}\text{ m} = 2.43\text{ pm}$ is the > ==Compton wavelength== of the electron. > - Shift is **independent of incoming wavelength** and of material — only $\theta$ matters. > - $\theta = 0$: no shift. $\theta = 90^\circ$: $\Delta\lambda = \lambda_C$. > $\theta = 180^\circ$ (backscatter): $\Delta\lambda = 2\lambda_C$ (maximum). --- ## Worked examples > [!example] 1) X-ray scattered at $90^\circ$ > Find $\Delta\lambda$ for $\theta = 90^\circ$. > $$\Delta\lambda = \frac{h}{m_ec}(1-\cos 90^\circ) = \lambda_C(1-0) = 2.43\text{ pm}$$ > *Why this step?* $\cos90^\circ = 0$, so the bracket is exactly 1 — the cleanest case. > [!example] 2) Backscatter, find new wavelength > $\lambda = 10.0$ pm, $\theta = 180^\circ$. > $1-\cos180^\circ = 1-(-1) = 2$, so $\Delta\lambda = 2(2.43) = 4.86$ pm. > $$\lambda' = 10.0 + 4.86 = 14.86\text{ pm}$$ > *Why this step?* Backscatter gives the **maximum** shift — the photon transfers the most > momentum, hence the most energy, to the electron. > [!example] 3) Energy given to the electron > For example 2, kinetic energy of recoil electron $= E_\gamma - E'_\gamma = hc(\tfrac1\lambda - \tfrac1{\lambda'})$. > $$KE = (1240\text{ eV·nm})\!\left(\frac{1}{0.0100} - \frac{1}{0.01486}\right)\text{nm}^{-1} > \approx 124000 - 83446 \approx 4.05\times10^4\text{ eV} \approx 40.5\text{ keV}$$ > *Why this step?* Energy lost by the photon must go to the electron (energy conservation). --- ## Common mistakes (Steel-manned) > [!mistake] "Shift depends on the incoming wavelength." > **Why it feels right:** Intensity/colour usually matters in optics, so you assume bigger > $\lambda$ → bigger shift. **Fix:** The derivation cancels all $\lambda$-only terms; the > *absolute* shift $\Delta\lambda$ depends **only on $\theta$** and $m_e$. (The *fractional* > shift $\Delta\lambda/\lambda$ does depend on $\lambda$ — that's why X-rays, not visible > light, show it clearly.) > [!mistake] "Use $m$ of the whole atom." > **Why it feels right:** The electron is bound to an atom. **Fix:** For loosely-bound > (valence/free) electrons use the **electron mass** $m_e$, giving $\lambda_C = 2.43$ pm. If > the photon hits a *tightly* bound electron, the whole atom recoils ($m \gg m_e$) → shift > ≈ 0 → that's the **unshifted peak** also seen in experiment. > [!mistake] "Use non-relativistic $KE=\frac12 m v^2$ for the electron." > **Why it feels right:** Habit from mechanics. **Fix:** The recoil can be fast; you must use > $E_e^2 = (p_ec)^2+(mc^2)^2$. Using $\frac12mv^2$ breaks the clean cancellation and gives the > wrong formula. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine throwing a tiny super-bouncy ball (the photon) at a marble (the electron) sitting > still. When the ball bounces off, it gives a little push to the marble, so the ball comes > away **slower** — it lost some energy. For light, "slower energy" means a **longer > wavelength** (redder). How much longer depends on the **angle** it bounces: straight ahead > = barely any change, bounced straight back = biggest change. That's the whole story: > light acts like a little ball in a game of pool. > [!mnemonic] Remember it > **"One minus cos, Compton's the boss."** > $\Delta\lambda = \lambda_C(1-\cos\theta)$. And $\lambda_C = h/m_ec \approx$ **2.43 pm** > ("2-4-3, scatter free"). --- ## Active-recall flashcards #flashcards/physics Compton shift formula? ::: $\Delta\lambda = \dfrac{h}{m_e c}(1-\cos\theta)$ Value of the electron Compton wavelength $h/m_ec$? ::: $\approx 2.43\times10^{-12}$ m = 2.43 pm Does $\Delta\lambda$ depend on the incoming wavelength? ::: No — only on scattering angle $\theta$ (and $m_e$). At what angle is the shift maximum, and what is it? ::: $\theta=180^\circ$ (backscatter); $\Delta\lambda = 2\lambda_C \approx 4.86$ pm. At $\theta=0$ what is the shift? ::: Zero (photon undeflected, no momentum transferred). Photon momentum in terms of wavelength? ::: $p = h/\lambda$ (from $E=pc$ and $E=hc/\lambda$). Why must the electron be treated relativistically? ::: Its recoil speed can be large; use $E_e^2=(p_ec)^2+(mc^2)^2$. Which conservation laws are used in the derivation? ::: Conservation of energy and of momentum (both x and y components). What is the unshifted peak in Compton's data? ::: Photons scattering off tightly-bound electrons (whole atom recoils, $m\gg m_e$, shift ≈ 0). Why does Compton scattering prove light is particle-like? ::: Angle-dependent wavelength shift follows from a billiard-ball photon–electron collision, which wave theory can't explain. --- ## Connections - [[Photoelectric effect]] — both prove photon energy $E=hf$; photoelectric uses energy only, Compton needs momentum too. - [[Photon momentum and energy]] — $E=pc$, $p=h/\lambda$. - [[Relativistic energy-momentum relation]] — $E^2=(pc)^2+(mc^2)^2$ used in Step 3. - [[de Broglie wavelength]] — particle ↔ wave duality, the flip side of Compton. - [[Conservation of momentum]] / [[Conservation of energy]] — the two pillars of the derivation. - [[X-ray production and Bremsstrahlung]] — source of the X-rays used in the experiment. ## 🖼️ Concept Map ```mermaid flowchart TD W[Wave model fails] -->|motivates| P[Photon particle model] P -->|photon has| E[Energy hc over lambda] P -->|photon has| M[Momentum h over lambda] C[Photon-electron collision] -->|conserve| EN[Energy conservation] C -->|conserve| MO[Momentum conservation] EN -->|gives star| Ee[Electron energy expr] MO -->|square and add| PE1[p_e squared star star] PE1 -->|eliminates phi| PE1 Ee -->|relativistic E-p| PE2[p_e squared star star star] R[Relativistic energy relation] -->|links E and p| PE2 PE1 -->|set equal| SHIFT[Compton shift delta lambda] PE2 -->|set equal| SHIFT SHIFT -->|depends on| TH[Scattering angle theta] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, Compton scattering ka core idea simple hai: light ko sirf wave maano toh X-ray > electron se takkar ke baad **lambi wavelength** kyun ban jaati hai, ye samajh nahi aata. Lekin > agar light ko **photon** (chhota particle) maano jiska energy $E=hc/\lambda$ aur momentum > $p=h/\lambda$ hai, toh ye bilkul carrom ki tarah ek collision ban jaata hai — photon electron > se takraata hai, thodi energy electron ko de deta hai, isliye photon ki energy kam, matlab > wavelength badi. > > Derivation me bas do cheezein lagti hain: **energy conservation** aur **momentum conservation** > (x aur y dono components). Electron tez bhag sakta hai isliye usko relativistic treat karte hain: > $E_e^2=(p_ec)^2+(mc^2)^2$. Jab tum momentum equations ko square karke add karte ho, electron ka > angle $\phi$ gayab ho jaata hai — kyunki humein sirf photon ki wavelength shift chahiye. Sab > algebra ke baad clean answer milta hai: > $\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta)$. > > Yaad rakhne wali baat: ye shift **incoming wavelength par depend nahi karta**, sirf **angle** > $\theta$ par. $\theta=0$ par koi shift nahi, $\theta=180^\circ$ (seedha peechhe) par maximum > shift $=2\times2.43=4.86$ pm. Constant $h/m_ec=2.43$ pm ko **Compton wavelength** kehte hain. > > Exam tip: agar electron tightly bound hai toh poora atom recoil karta hai (mass bahut badi), > isliye shift almost zero — yahi "unshifted peak" hai jo data me bhi dikhta hai. Galti se atom ki > mass mat use karna free electron wale shift ke liye; aur electron ke liye $\frac12mv^2$ mat > lagana, relativistic formula hi lagao. ![[audio/2.3.04-Compton-scattering-—-wavelength-shift-derivation.mp3]]