2.3.4 · D3 · Physics › Modern Physics › Compton scattering — wavelength shift derivation
Intuition Ye page kis liye hai
Parent note ne ek formula derive kiya tha jo poora game chalata hai:
Δ λ = m e c h ( 1 − cos θ ) = λ C ( 1 − cos θ ) , λ C ≈ 2.43 pm .
Yahan hum har tarah ke questions dhundhte hain jo ye formula generate kar sakta hai — har angle, dono limits, "shift zero hai" wale traps, energy-transfer ka twist, aur ek exam-style curveball — aur har ek ko scratch se solve karte hain. Prerequisite ideas yahan hain:
Photon momentum and energy , Relativistic energy-momentum relation ,
Conservation of momentum aur Conservation of energy ; poori derivation parent note mein hai:
Compton scattering derivation .
Shuru karne se pehle, wo numbers pin down kar lo jo hum har jagah reuse karte hain:
Har Compton problem inhi cells mein se ek hoti hai. Neeche ke worked examples ko unki cell ke saath tag kiya gaya hai.
Cell
Case class
Kya cheez alag banati hai
Example
A
θ = 0 ∘ (degenerate, koi deflection nahi)
1 − cos θ = 0 ⇒ Δ λ = 0
Ex 1
B
θ = 9 0 ∘ (clean midpoint)
cos 9 0 ∘ = 0 ⇒ Δ λ = λ C
Ex 2
C
θ = 18 0 ∘ (backscatter, limiting max)
1 − cos θ = 2 ⇒ Δ λ = 2 λ C
Ex 3
D
General angle 0 < θ < 9 0 ∘
cos θ actually compute karna padega
Ex 4
E
General angle 9 0 ∘ < θ < 18 0 ∘
cos θ < 0 , shift λ C se zyada ho jaati hai
Ex 5
F
Electron ko diya gaya energy (recoil KE)
energy conservation use karo, sirf shift nahi
Ex 6
G
Fractional shift Δ λ / λ (kyun X-rays, light nahi)
same Δ λ , wildly alag fraction
Ex 7
H
Wrong-mass trap / unshifted peak
proton ya poora atom → bahut chhoti shift
Ex 8
I
Inverse problem / exam twist (Δ λ se θ dhundho)
cos θ backwards solve karo, range check karo
Ex 9
Worked example Ex 1 — Cell A: no-deflection degenerate case (
θ = 0 ∘ )
Ek 20 pm X-ray photon ek electron ko graze karta hai aur seedha aage chalta rehta hai (θ = 0 ∘ ).
Uski nayi wavelength λ ′ kya hai?
Forecast: Padhne se pehle andaza lagao — kya wavelength badhegi, ghateghi, ya wahi rahegi?
Formula likho: Δ λ = λ C ( 1 − cos θ ) .
Ye step kyun? Yahi ek relation hai angle aur wavelength change ke beech; sab kuch yahan se shuru hota hai.
θ = 0 ∘ daalo, toh cos 0 ∘ = 1 : Δ λ = λ C ( 1 − 1 ) = 0 .
Ye step kyun? θ = 0 ∘ par photon ne direction nahi badla, isliye usne koi momentum transfer nahi kiya — koi momentum transfer nahi matlab koi energy loss nahi, toh koi shift nahi.
Isliye λ ′ = λ + 0 = 20 pm .
Ye step kyun? λ ′ = λ + Δ λ shift ki definition se.
Verify: Ek photon jo bilkul miss kar jaaye wo unchanged rehna chahiye — λ ′ = λ physically sahi "kuch nahi hua" answer hai. Units: pm + pm = pm. ✓
Worked example Ex 2 — Cell B: clean midpoint (
θ = 9 0 ∘ )
λ = 30.0 pm wavelength ka ek X-ray exactly 9 0 ∘ par scatter hota hai. Δ λ aur λ ′ nikalo.
Forecast: Right angles par, kya shift ek Compton wavelength (2.43 pm) se kam, barabar, ya zyada hogi?
cos 9 0 ∘ = 0 , toh 1 − cos θ = 1 .
Ye step kyun? Ye woh angle hai jahan messy cos term gayab ho jaata hai — reference case hai.
Δ λ = λ C ⋅ 1 = 2.43 pm .
Ye step kyun? Bracket exactly 1 ke barabar hai, toh shift precisely ek Compton wavelength hai.
λ ′ = 30.0 + 2.43 = 32.43 pm .
Ye step kyun? Shift ko original wavelength mein add karo.
Verify: θ = 9 0 ∘ ko ek aisi shift deni chahiye jo θ = 0 ki value (0) aur θ = 18 0 ∘ ki value (4.86 pm) ke strictly beech ho. Sach mein 0 < 2.43 < 4.86 . ✓
Worked example Ex 3 — Cell C: backscatter, limiting maximum (
θ = 18 0 ∘ )
Ek 10.0 pm photon seedha wapas bounce karta hai (θ = 18 0 ∘ ). Δ λ aur λ ′ nikalo.
Forecast: Kya ye sabse badi possible shift hai, ya koi aur angle isse beat kar sakta hai?
cos 18 0 ∘ = − 1 , toh 1 − cos θ = 1 − ( − 1 ) = 2 .
Ye step kyun? cos θ 18 0 ∘ par sabse chhota (most negative) hota hai, isliye bracket 1 − cos θ wahan sabse bada hota hai — ye poore matrix ka extreme case hai.
Δ λ = 2 λ C = 2 ( 2.43 ) = 4.86 pm .
Ye step kyun? Ye maximum shift hai jo koi bhi Compton event kabhi produce kar sakta hai.
λ ′ = 10.0 + 4.86 = 14.86 pm .
Verify: Kyunki cos θ sirf [ − 1 , 1 ] ke upar range karta hai, bracket [ 0 , 2 ] ke upar range karta hai, toh koi bhi angle 2 λ C se zyada nahi de sakta. Backscatter confirmed ceiling hai. ✓
Worked example Ex 4 — Cell D: ek genuine acute angle (
θ = 6 0 ∘ )
λ = 50.0 pm, θ = 6 0 ∘ par scatter hota hai. Δ λ aur λ ′ nikalo.
Forecast: 6 0 ∘ , 9 0 ∘ tak jaane ke raaste mein do-tihai distance par hai. Kya shift lagbhag λ C ka do-tihai hogi, ya kuch aur?
cos 6 0 ∘ = 0.5 , toh 1 − cos θ = 0.5 .
Ye step kyun? Hume actual cosine compute karna hoga; general angle ke liye koi shortcut nahi hai.
Δ λ = λ C ( 0.5 ) = 2.43 × 0.5 = 1.215 pm .
Ye step kyun? Bracket exactly one-half hai, toh shift half Compton wavelength hai.
λ ′ = 50.0 + 1.215 = 51.215 pm .
Verify: Forecast trap note karo: shift λ C ka half hai, do-tihai nahi — kyunki shift 1 − cos θ track karta hai, jo angle ka ek curved (linear nahi) function hai. 0 < 1.215 < 2.43 , consistent with 0 < 6 0 ∘ < 9 0 ∘ . ✓
Worked example Ex 5 — Cell E: ek obtuse angle (
θ = 12 0 ∘ )
λ = 5.00 pm, θ = 12 0 ∘ par scatter hota hai. Δ λ nikalo.
Forecast: 9 0 ∘ ke baad, kya shift puri λ C se badi ya chhoti hogi?
cos 12 0 ∘ = − 0.5 , toh 1 − cos θ = 1 − ( − 0.5 ) = 1.5 .
Ye step kyun? 9 0 ∘ se aage cosine negative ho jaata hai, aur yahi wajah hai ki obtuse-angle shifts λ C se zyada hoti hain — sign flip hi is cell ka poora point hai.
Δ λ = λ C ( 1.5 ) = 2.43 × 1.5 = 3.645 pm .
Ye step kyun? Bracket 1.5 hai, toh shift derhh Compton wavelengths hai.
Verify: Obtuse angles 9 0 ∘ aur 18 0 ∘ ke beech hain, toh shift λ C aur 2 λ C ke beech honi chahiye: 2.43 < 3.645 < 4.86 . ✓
Worked example Ex 6 — Cell F: recoil electron ko diya gaya energy
Ex 3 ka backscatter case tha jahan λ = 10.0 pm → λ ′ = 14.86 pm. Electron ne kitni kinetic energy pick up ki?
Forecast: Andaza lagao — kya ye kuch eV hoga (jaise photoelectric effect mein) ya kai keV?
Energy conservation: electron ki KE photon ki energy loss ke barabar hai,
K E = E γ − E γ ′ = h c ( λ 1 − λ ′ 1 ) .
Ye step kyun? Shift formula sirf wavelengths deta hai; energy pane ke liye hume Conservation of energy par wapas jaana hoga — photon ki saari lost energy electron ko jaati hai (wo rest par shuru hua tha).
h c = 1240 eV·nm use karo aur pm ko nm mein convert karo (10.0 pm = 0.0100 nm, 14.86 pm = 0.01486 nm):
K E = 1240 ( 0.0100 1 − 0.01486 1 ) eV = 1240 ( 100 − 67.294 ) eV .
Ye step kyun? h c ko eV·nm mein aur wavelengths ko nm mein rakhne se answer seedha eV mein aata hai — ek units trick yaad rakhne layak hai.
K E = 1240 × 32.706 ≈ 4.055 × 1 0 4 eV ≈ 40.6 keV .
Verify: Ye photon ki apni ∼ 124 keV energy se comparable hai, jo sense banata hai: backscatter maximum-energy-transfer geometry hai. Photoelectric-scale eV energies se kaafi upar — X-rays ke liye expected hai. ✓ (Compare with Photoelectric effect scales.)
Worked example Ex 7 — Cell G: kyun X-rays aur visible light nahi
θ = 9 0 ∘ par fractional shift Δ λ / λ compute karo (a) λ = 24.3 pm wale X-ray ke liye aur (b) λ = 500 nm wali green light ke liye.
Forecast: Absolute shift Δ λ dono ke liye same hai (2.43 pm). Toh Compton scattering sirf X-ray phenomenon kyun hai?
Dono ke liye Δ λ = λ C = 2.43 pm hai (Cell B, θ = 9 0 ∘ ).
Ye step kyun? Formula incoming wavelength se andha hai — shift identical hai.
X-ray fraction: λ Δ λ = 24.3 2.43 = 0.100 = 10% .
Ye step kyun? 10% change ek spectrometer mein resolve karna aasaan hai.
Green-light fraction: 500 nm = 500000 pm, toh 500000 2.43 = 4.86 × 1 0 − 6 ≈ 0.000486% .
Ye step kyun? Wahi 2.43 pm ek 500 nm wave ka ek vanishing sliver hai — undetectable.
Verify: Absolute shifts equal hain lekin fractions 500000/24.3 ≈ 20576 factor se alag hain. Yahi precisely wajah hai ki Compton ne X-rays use kiye. ✓
Worked example Ex 8 — Cell H: wrong-mass trap (proton / bound atom)
θ = 9 0 ∘ case repeat karo lekin photon ko free electron ki jagah proton (m p ≈ 1836 m e ) se scatter karo. Shift kya hogi?
Forecast: Kya ye abhi bhi 2.43 pm hogi?
Compton wavelength recoiling particle ke mass use karta hai: λ C , p = m p c h = 1836 λ C .
Ye step kyun? Derivation dobara padho — h / ( m c ) mein m wo cheez ka mass hai jo recoil karti hai. Wrong mass use karna classic error hai jiske baare mein parent note warn karta hai.
Δ λ = λ C , p ( 1 − cos 9 0 ∘ ) = 1836 2.43 pm ≈ 1.32 × 1 0 − 3 pm .
Ye step kyun? Bhaari target → chhoti Compton wavelength → tinier shift.
Verify: Shift lagbhag 1836 guna chhoti hai — X-ray scale par essentially zero. Yahi unshifted peak ki origin hai: jo photons tightly-bound electrons se takrate hain wo poore bhaare atom ko recoil karaate hain, toh m ≫ m e aur Δ λ ≈ 0 . ✓
Worked example Ex 9 — Cell I: inverse problem / exam twist (angle nikalo)
Ek experiment mein Compton shift Δ λ = 3.645 pm measure hoti hai. Photon kis angle θ par scatter hua? Kya aisi shift possible bhi hai?
Forecast: Kya tum θ wapas nikaal sakte ho, aur kya ek se zyada answers ho sakte hain?
Formula ko cos θ ke liye rearrange karo:
Δ λ = λ C ( 1 − cos θ ) ⇒ cos θ = 1 − λ C Δ λ .
Ye step kyun? Hum relation ko invert karte hain unknown angle solve karne ke liye, pehle cos θ isolate karte hain kyunki θ sirf wahan appear hota hai.
cos θ = 1 − 2.43 3.645 = 1 − 1.5 = − 0.5 .
Ye step kyun? Measured shift aur known constant daalo.
θ = arccos ( − 0.5 ) = 12 0 ∘ .
Ye step kyun? arccos jawaab deta hai "is cosine wala kaunsa angle hai?" [ 0 ∘ , 18 0 ∘ ] range par — jo scattering angle ki sirf physical range hai — exactly ek hi answer hai, toh koi ambiguity nahi.
Verify: Pehle, feasibility: Δ λ / λ C = 1.5 ≤ 2 , toh ye allowed [ 0 , 2 λ C ] band mein hai — valid measurement. Doosra, ye Ex 5 ka ulta hai, jisne 12 0 ∘ → 3.645 pm bheja tha. Round trip close ho gaya. ✓
Common mistake Exam trap: feasibility check bhool jaana
Agar koi Δ λ = 6 pm report kare, toh cos θ = 1 − 6/2.43 = − 1.47 , jo impossible hai (cos θ ≥ − 1 ). Koi bhi electron-scattering angle isko produce nahi kar sakta — data mein error honi chahiye, ya target free electron nahi hai.
Recall Matrix ke across quick self-test
Shift at θ = 0 ∘ ? ::: 0 (Cell A — koi deflection nahi, koi momentum transfer nahi).
Shift at θ = 9 0 ∘ ? ::: λ C = 2.43 pm (Cell B).
Maximum possible shift aur kahan? ::: 2 λ C = 4.86 pm at θ = 18 0 ∘ (Cell C).
Kisi bhi λ ke liye θ = 12 0 ∘ par shift? ::: 1.5 λ C = 3.645 pm (Cell E — obtuse, λ C se zyada).
Agar Δ λ = 1.215 pm, toh kaun sa angle? ::: cos θ = 0.5 ⇒ θ = 6 0 ∘ (Cell D/I).
Visible light mein Compton scattering kyun nahi dikhti? ::: Same 2.43 pm shift ek 500 nm wave ka negligible fraction hai (Cell G).
9 0 ∘ par proton se photon? ::: ≈ 1.32 × 1 0 − 3 pm, yaani ~1836× chhoti (Cell H).
Largest valid Δ λ / λ C ? ::: 2 ; isse upar kuch bhi unphysical hai (Cell I).
Mnemonic Poora matrix ek line mein
"Zero at nose, one at side, two behind — angle decides."
(0 at θ = 0 ∘ , λ C at 9 0 ∘ , 2 λ C at 18 0 ∘ .)
Parent: Compton scattering derivation
Photon momentum and energy — har example mein use hone wale p = h / λ aur E = h c / λ supply karta hai.
Conservation of energy aur Conservation of momentum — shift aur recoil-energy example ke peeche ye do laws hain.
Relativistic energy-momentum relation — isliye Ex 6 mein electron recoil ko relativistically treat kiya gaya hai.
Photoelectric effect — wahan ke eV energy scale ko yahan ke keV recoil se compare karo.
X-ray production and Bremsstrahlung — incoming X-rays kahan se aati hain.
de Broglie wavelength — ek alag context mein same h / ( m c ) -style length scale.