Compton scattering — wavelength shift derivation

2.3.4 · Physics › Modern Physics

KYA ho raha hai

Photon carry karta hai:

Energy $E = \frac{h c}{λ}$ (kyunki $E = h f$ aur $c = f λ$ )
Momentum $p = \frac{E}{c} = \frac{h}{λ}$ (photon massless hai, isliye $E = p c$ )

Figure — Compton scattering — wavelength shift derivation

KAISE derive karte hain (scratch se)

Hum electron ko relativistically treat karte hain (woh tez recoil kar sakta hai). Symbols:

photon pehle: energy $\frac{h c}{λ}$ , momentum $\frac{h}{λ} \overset{x}{^}$
photon baad mein: energy $\frac{h c}{λ ^{'}}$ , momentum $\frac{h}{λ ^{'}}$ angle $θ$ par
electron pehle: rest mein, energy $m c^{2}$ , momentum $0$
electron baad mein: momentum $p_{e}$ , energy $E_{e} = (p_{e} c)^{2} + (m c^{2})^{2}$

Step 1 — Energy conserve karo

$\frac{h c}{λ} + m c^{2} = \frac{h c}{λ ^{'}} + E_{e}$

Yeh step kyun? Total energy conserve hoti hai; electron sirf rest energy $m c^{2}$ se shuru karta hai.

Toh: $E_{e} = m c^{2} + h c (\frac{1}{λ} - \frac{1}{λ ^{'}}) (⋆)$

Step 2 — Momentum conserve karo (vector!)

Maano photon angle $θ$ par scatter hota hai, electron angle $ϕ$ par.

x: $\frac{h}{λ} = \frac{h}{λ ^{'}} cos θ + p_{e} cos ϕ$

y: $0 = \frac{h}{λ ^{'}} sin θ - p_{e} sin ϕ$

Yeh step kyun? Momentum ek vector hai, isliye har component alag-alag conserve hota hai.

Electron ke terms isolate karo:

p_e\sin\phi = \frac{h}{\lambda'}\sin\theta$$ Square karo aur add karo taaki **$\phi$ eliminate ho jaye** (kyunki $\cos^2\phi+\sin^2\phi=1$): $$p_e^2 = \frac{h^2}{\lambda^2} - \frac{2h^2}{\lambda\lambda'}\cos\theta + \frac{h^2}{\lambda'^2} \quad(\star\star)$$ *$\phi$ kyun eliminate kiya?* Hume electron ki exact direction se matlab nahi — sirf photon ki wavelength shift se matlab hai. ### Step 3 — Relativistic energy relation use karo $E_e^2 = (p_e c)^2 + (mc^2)^2$, toh $p_e^2 c^2 = E_e^2 - (mc^2)^2$. $(\star)$ ko square karo: $$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$$ $(mc^2)^2$ subtract karo aur $c^2$ se divide karo: $$p_e^2 = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) + h^2\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right)^2 \quad(\star\star\star)$$ ### Step 4 — $(\star\star) = (\star\star\star)$ set karo $h^2(\tfrac1\lambda-\tfrac1{\lambda'})^2 = h^2(\tfrac1{\lambda^2} - \tfrac{2}{\lambda\lambda'} + \tfrac1{\lambda'^2})$ expand karo. Equal karo: $$\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}$$ Dono sides par $\frac{h^2}{\lambda^2}$ aur $\frac{h^2}{\lambda'^2}$ cancel ho jaate hain: $$-\frac{2h^2\cos\theta}{\lambda\lambda'} = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right) - \frac{2h^2}{\lambda\lambda'}$$ $h^2$ ke terms rearrange karo: $$\frac{2h^2}{\lambda\lambda'}(1-\cos\theta) = 2mhc\!\left(\tfrac1\lambda-\tfrac1{\lambda'}\right)$$ Note karo $\tfrac1\lambda - \tfrac1{\lambda'} = \tfrac{\lambda'-\lambda}{\lambda\lambda'}$. Dono sides ko $\frac{2h}{\lambda\lambda'}$ se divide karo: $$h(1-\cos\theta) = mc(\lambda'-\lambda)$$ > [!formula] Compton shift > $$\boxed{\;\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}\,(1-\cos\theta)\;}$$ > - $\dfrac{h}{m_e c} = \lambda_C \approx 2.43\times10^{-12}\text{ m} = 2.43\text{ pm}$ electron ki ==Compton wavelength== hai. > - Shift **incoming wavelength par depend nahi karti** aur material par bhi nahi — sirf $\theta$ matter karta hai. > - $\theta = 0$: koi shift nahi. $\theta = 90^\circ$: $\Delta\lambda = \lambda_C$. > $\theta = 180^\circ$ (backscatter): $\Delta\lambda = 2\lambda_C$ (maximum). --- ## Worked examples > [!example] 1) X-ray $90^\circ$ par scatter hoti hai > $\theta = 90^\circ$ ke liye $\Delta\lambda$ nikalo. > $$\Delta\lambda = \frac{h}{m_ec}(1-\cos 90^\circ) = \lambda_C(1-0) = 2.43\text{ pm}$$ > *Yeh step kyun?* $\cos90^\circ = 0$ hai, toh bracket exactly 1 ho jaata hai — sabse saaf case. > [!example] 2) Backscatter, nayi wavelength nikalo > $\lambda = 10.0$ pm, $\theta = 180^\circ$. > $1-\cos180^\circ = 1-(-1) = 2$, toh $\Delta\lambda = 2(2.43) = 4.86$ pm. > $$\lambda' = 10.0 + 4.86 = 14.86\text{ pm}$$ > *Yeh step kyun?* Backscatter **maximum** shift deta hai — photon sabse zyada momentum transfer karta hai, isliye sabse zyada energy electron ko milti hai. > [!example] 3) Electron ko di gayi energy > Example 2 ke liye, recoil electron ki kinetic energy $= E_\gamma - E'_\gamma = hc(\tfrac1\lambda - \tfrac1{\lambda'})$. > $$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}$$ > *Yeh step kyun?* Photon ne jo energy khoyi, woh electron ko milni chahiye (energy conservation). --- ## Common mistakes (Steel-manned) > [!mistake] "Shift incoming wavelength par depend karti hai." > **Kyun sahi lagta hai:** Optics mein intensity/colour usually matter karta hai, toh aap assume karte ho ki bada $\lambda$ → badi shift. **Fix:** Derivation mein saare sirf-$\lambda$ wale terms cancel ho jaate hain; *absolute* shift $\Delta\lambda$ sirf **$\theta$** aur $m_e$ par depend karti hai. (*Fractional* shift $\Delta\lambda/\lambda$ zaroor $\lambda$ par depend karti hai — isliye X-rays mein yeh clearly dikhta hai, visible light mein nahi.) > [!mistake] "Poore atom ka $m$ use karo." > **Kyun sahi lagta hai:** Electron atom se bound hai. **Fix:** Loosely-bound (valence/free) electrons ke liye **electron mass** $m_e$ use karo, jisse $\lambda_C = 2.43$ pm milta hai. Agar photon *tightly* bound electron se takraata hai, toh poora atom recoil karta hai ($m \gg m_e$) → shift ≈ 0 → yahi **unshifted peak** hai jo experiment mein bhi dikh ta hai. > [!mistake] "Electron ke liye non-relativistic $KE=\frac12 m v^2$ use karo." > **Kyun sahi lagta hai:** Mechanics ki aadat. **Fix:** Recoil tez ho sakta hai; $E_e^2 = (p_ec)^2+(mc^2)^2$ use karna zaroori hai. $\frac12mv^2$ use karne se saaf cancellation nahi hoti aur galat formula milta hai. --- > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Socho tum ek tiny super-bouncy ball (photon) ko ek marble (electron) par phenk rahe ho jo chup khadi hai. Jab ball bounce karti hai, woh marble ko thoda dhakka deti hai, toh ball **dheere** wapas aati hai — usne kuch energy kho di. Light ke liye, "kam energy" ka matlab hai **lambi wavelength** (zyada laal). Kitni lambi hogi yeh **angle** par depend karta hai jis par woh bounce karti hai: seedha aage = koi khaas change nahi, seedha peeche bounce = sabse bada change. Yahi poora story hai: light pool ke game mein ek choti si ball ki tarah behave karti hai. > [!mnemonic] Yaad rakho > **"One minus cos, Compton's the boss."** > $\Delta\lambda = \lambda_C(1-\cos\theta)$. Aur $\lambda_C = h/m_ec \approx$ **2.43 pm** > ("2-4-3, scatter free"). --- ## Active-recall flashcards #flashcards/physics Compton shift formula? ::: $\Delta\lambda = \dfrac{h}{m_e c}(1-\cos\theta)$ Electron Compton wavelength $h/m_ec$ ki value kya hai? ::: $\approx 2.43\times10^{-12}$ m = 2.43 pm Kya $\Delta\lambda$ incoming wavelength par depend karta hai? ::: Nahi — sirf scattering angle $\theta$ par (aur $m_e$ par). Kis angle par shift maximum hoti hai, aur woh kitni hoti hai? ::: $\theta=180^\circ$ (backscatter); $\Delta\lambda = 2\lambda_C \approx 4.86$ pm. $\theta=0$ par shift kya hoti hai? ::: Zero (photon undeflected, koi momentum transfer nahi). Photon momentum wavelength ke terms mein? ::: $p = h/\lambda$ ($E=pc$ aur $E=hc/\lambda$ se). Electron ko relativistically treat karna kyun zaroori hai? ::: Uski recoil speed badi ho sakti hai; $E_e^2=(p_ec)^2+(mc^2)^2$ use karo. Derivation mein kaun se conservation laws use hote hain? ::: Conservation of energy aur momentum (dono x aur y components). Compton ke data mein unshifted peak kya hai? ::: Photons jo tightly-bound electrons se scatter hote hain (poora atom recoil karta hai, $m\gg m_e$, shift ≈ 0). Compton scattering kyun prove karta hai ki light particle-jaisi hai? ::: Angle-dependent wavelength shift ek billiard-ball photon–electron collision se milti hai, jise wave theory explain nahi kar sakti. --- ## Connections - [[Photoelectric effect]] — dono photon energy $E=hf$ prove karte hain; photoelectric sirf energy use karta hai, Compton ko momentum bhi chahiye. - [[Photon momentum and energy]] — $E=pc$, $p=h/\lambda$. - [[Relativistic energy-momentum relation]] — $E^2=(pc)^2+(mc^2)^2$ Step 3 mein use hota hai. - [[de Broglie wavelength]] — particle ↔ wave duality, Compton ka ulta pahlu. - [[Conservation of momentum]] / [[Conservation of energy]] — derivation ke do pillars. - [[X-ray production and Bremsstrahlung]] — experiment mein use hone wali X-rays ka source. ## 🖼️ 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] ```