Compton scattering — wavelength shift derivation: Question bank

2.3.4 · D5 · Physics › Modern Physics › Compton scattering — wavelength shift derivation

Neeche use hone wale symbols, sab parent note se:

$λ$ = aane wale photon ki wavelength; $λ^{'}$ = scatter hue photon ki wavelength; $Δ λ = λ^{'} - λ$ = Compton shift (kitna lamba light hua).
$θ$ = woh angle jis se photon muda; $ϕ$ = woh angle jis par electron recoil karta hai.
$λ_{C} = h / m_{e} c \approx 2.43$ pm = electron ki Compton wavelength (sirf constants se bani ek fixed length).
$m_{e}$ = electron rest mass; $c$ = speed of light; $h$ = Planck's constant.

True or false — justify

True or false:

Δ λ

zyada ho jaata hai agar tum zyada intense (bright) X-ray beam use karo.

False. Intensity matlab hai zyada photons per second, har photon mein zyada energy nahi; har collision wahi

Δ λ = λ_{C} (1 - cos θ)

follow karta hai, isliye brightness change karti hai ki kitne shifted photons dikhte hain, kabhi nahi ki har ek kitna shift hoga.

True or false: Ek red laser jo free electrons par daali gayi ho, same angle par X-rays se bada

Δ λ

produce karta hai.

False. Absolute shift

Δ λ

sirf

θ

par depend karta hai, isliye dono ke liye identical hai; lekin red light ke liye

Δ λ / λ

bahut chhota hai, isliye undetectable hai — bilkul yehi wajah hai ki effect ke liye short-wavelength X-rays chahiye.

True or false:

θ = 0^{\circ}

par photon phir bhi thodi si energy kho deta hai.

False.

1 - cos 0^{\circ} = 0

, isliye

Δ λ = 0

exactly: ek photon jo seedha nikal jaata hai woh electron ko koi momentum transfer nahi karta aur apni poori energy rakhta hai.

True or false: Scattered photon chhoti wavelength ke saath aa sakta hai jaisi woh ander gayi thi.

False.

1 - cos θ

ka range

0

2

tak hai aur kabhi negative nahi hota, isliye

Δ λ \geq 0

hamesha — rest par pada free electron sirf photon se energy le sakta hai, usse energy de nahi sakta.

True or false: Maximum possible shift

Δ λ = λ_{C}

hai.

False. Maximum backscatter

θ = 18 0^{\circ}

par hota hai, jahaan

1 - cos 18 0^{\circ} = 2

, jo

Δ λ = 2 λ_{C} \approx 4.86

pm deta hai;

λ_{C}

akela

9 0^{\circ}

wala case hai.

True or false: Agar hum electron ki jagah proton use karein, toh Compton shift lagbhag

1836

guna chhota hoga.

True.

λ_{C} = h / m c

mass ke inversely proportional hai, aur proton

\approx 1836 \times

heavier hai, isliye uski Compton wavelength — aur isliye har shift — us factor se chhoti ho jaati hai.

True or false: Compton scattering aur Photoelectric effect ek hi phenomenon hain.

False. Photoelectric effect mein photon poori tarah absorb ho jaata hai aur uski saari energy ek bound electron ko eject karti hai; Compton scattering mein photon bachta hai, sirf redirect aur redden ho jaata hai. Dono phir bhi

E = h f

prove karte hain.

Spot the error

Error hunt: "Electron kinetic energy

m c^{2}

se shuru karta hai, isliye hum use Step 1 mein add karte hain."

Galat label:

m c^{2}

electron ki rest energy hai, uski energy rest par zero kinetic energy ke saath. Kinetic energy sirf collision ke baad

E_{e} - m c^{2}

ke roop mein aata hai.

Error hunt: "Hum recoil electron ke liye

K E = \frac{1}{2} m_{e} v^{2}

use karte hain algebra simple rakhne ke liye."

Recoil relativistically fast ho sakta hai, isliye tumhe

E_{e}^{2} = (p_{e} c)^{2} + (m c^{2})^{2}

use karna hi padega; non-relativistic formula

h^{2} / λ^{2}

terms ke exact cancellation ko kharab kar deta hai aur galat shift deta hai. Dekho Relativistic energy-momentum relation.

Error hunt: "Step 2 mein hum

ϕ

ko final answer mein rakhte hain kyunki electron ki direction matter karti hai."

Hum deliberately do momentum equations ko square karke add karke

ϕ

eliminate karte hain (

cos^{2} ϕ + sin^{2} ϕ = 1

); experiment sirf photon ka

λ^{'}

aur

θ

measure karta hai, isliye electron ki direction shift ke liye irrelevant hai.

Error hunt: "

λ_{C} = h / m c

mein poore atom ki mass use karo kyunki electron atom ke andar hai."

Sirf tightly-bound electrons ke liye jahan atom ek saath recoil karta hai — woh tiny/zero-shift unshifted peak deta hai. Shifted peak ke liye electron essentially free particle ki tarah behave karta hai, isliye

m_{e}

use karo.

Error hunt: "Photon ki energy

E = h f

hai lekin koi momentum nahi, kyunki momentum ke liye mass chahiye."

Photon massless hai phir bhi momentum

p = E / c = h / λ

carry karta hai; momentum ke liye energy aur motion chahiye, rest mass nahi. Photon momentum ke bina Step 2 mein conserve karne ke liye kuch hoga hi nahi. Dekho Photon momentum and energy.

Error hunt: "Kyunki energy conserved hai, scattered photon ki energy incoming wale jaisi hi hai."

Energy poore system ke liye conserved hai; photon apni energy ka kuch hissa recoiling electron ko deta hai, isliye scattered photon ki energy kam hai (

h c / λ^{'} < h c / λ

), isliye

λ^{'}

lamba hai.

Error hunt: "

Δ λ = λ_{C} (1 - cos θ)

, toh

θ

seedha degrees mein

cos

mein plug kar do."

Formula theek hai, lekin

cos

θ

us unit mein evaluate karna hoga jo tumhare calculator ka cosine expect karta hai; trap yeh hai ki degree/radian mode set karna bhool jaate hain, jaise

cos 9 0^{\circ} = 0

na ki

cos (90 rad)

Why questions

Shift

θ

sin θ

par nahi balki

1 - cos θ

par kyun depend karta hai?

cos θ

momentum components ko square karke add karne par dot-product-jaisi cross term

\frac{2 h ^{2}}{λ λ ^{'}} cos θ

se aata hai; "

1 -

" geometry ke baad jo bachta hai woh hai, aur yeh naturally

θ = 0

par

0

deta hai aur

θ = 18 0^{\circ}

par maximum.

Momentum ke sirf ek component ki jagah x- aur y- dono components kyun use karne padte hain?

Momentum ek vector hai, isliye woh direction by direction conserved hota hai; sirf x-equation use karne se

ϕ

aur

p_{e}

undetermined rahenge, aur hume dono equations chahiye square-and-add karne aur

ϕ

eliminate karne ke liye. Dekho Conservation of momentum.

Derivation mein relativistic energy relation ki zaroorat hi kyun parti hai?

Hamare paas

p_{e}

momentum se (Step 2) aur

E_{e}

energy se (Step 1) alag-alag facts ke roop mein hain; relation

E_{e}^{2} = (p_{e} c)^{2} + (m c^{2})^{2}

woh bridge hai jo unhe jodta hai taaki

E_{e}

aur

ϕ

dono drop ho jayein, sirf

λ

λ^{'}

θ

bachein.

Shift target material se independent kyun hai?

Final formula mein sirf

h

c

m_{e}

aur

θ

hain — atom ki koi property nahi bachti — kyunki shifted peak un electrons se aati hai jo free particles ki tarah behave karte hain chahe woh kisi bhi atom se aaye hon.

Rozmarra ki visible-light reflections mein Compton shifts kyun nahi dikhte?

Shift

\approx 2.43

pm ek fixed absolute length hai; visible wavelengths (

\sim 500, 000

pm) ke muqable mein yeh lakhon mein se ek part hai — bilkul dab jaata hai — jabki X-rays (

\sim 10

pm) ke muqable mein yeh ek bada fractional change hai.

Compton scattering light ki particle nature ka evidence kyun maana jaata hai?

Ek continuous wave electron ko shake karti aur same frequency par re-radiate karti, koi angle-dependent wavelength change nahi deti; sirf ek localized particle jo

p = h / λ

carry karta hai ek billiard-ball collision mein measured

λ^{'} (θ)

reproduce karta hai.

Electron ki recoil kinetic energy

h c (\frac{1}{λ} - \frac{1}{λ ^{'}})

kyun hoti hai?

Energy conservation se electron exactly wahi gain karta hai jo photon khota hai; photon ki energy

h c / λ

h c / λ^{'}

tak gir gayi, aur difference electron ki kinetic energy ban jaati hai. Dekho Conservation of energy.

Compton wavelength

h / m_{e} c

kyun likhi jaati hai, na ki

h / m_{e}

h / m_{e}

(units of momentum-length-per-mass) ko

c

se divide karne par yeh ek genuine length ban jaati hai; yeh woh length scale hai jo tumhe

h

m_{e}

c

se banana padega ek photon–electron collision characterize karne ke liye, de Broglie wavelength logic ki tarah.

Edge cases

Edge case: Formula bilkul

θ = 9 0^{\circ}

par kya predict karta hai?

cos 9 0^{\circ} = 0

, isliye

Δ λ = λ_{C} \approx 2.43

pm — sabse clean single-

λ_{C}

case, jis tarah Compton wavelength sabse aasani se measure ki jaati hai.

Edge case: Ek photon ke limit mein kya hoga jo

m_{e} c^{2}

se bahut kam energetic ho?

Absolute shift

Δ λ

unchanged hai (yeh kabhi bhi incoming energy par depend nahi karta), lekin fractional energy loss vanishingly small ho jaati hai, isliye collision near-elastic Thomson scatter ki tarah behave karta hai jisme essentially koi reddening nahi.

Edge case: Compton ke spectra mein shifted peak ke saath kya aur dikhta hai, aur kahan se?

Original

λ

par ek unshifted peak, un photons se produce hoti hai jo tightly-bound inner electrons se takraate hain isliye poora atom recoil karta hai (

m ≫ m_{e}

) aur

Δ λ \approx 0

Edge case: Agar target electron impact se pehle photon ki taraf pehle se move kar raha ho, kya

λ^{'} < λ

ho sakta hai?

Haan — standard formula ek rest par electron assume karta hai; ek moving electron photon ko energy de sakta hai (inverse Compton scattering), uski wavelength chhoti kar sakta hai. Hamari derivation ka

Δ λ \geq 0

sirf at-rest case ke liye hold karta hai.

Edge case: Photon

θ = 18 0^{\circ}

par backscatter kare toh electron ka recoil angle

ϕ

kya hoga?

Electron seedha incoming direction ke saath aage ud jaata hai (

ϕ = 0

), maximum momentum aur energy carry karta hai, kyunki saara transverse momentum cancel ho gaya.

Edge case: Compton ke experiment mein use hone wali X-rays khud kahan se aati hain?

X-ray production and Bremsstrahlung se — ek metal target mein decelerate hue fast electrons se — jo woh short wavelengths supply karte hain jo fixed

2.43

pm shift ko measurable banane ke liye chahiye.

Connections

Parent: Compton wavelength-shift derivation
Conservation of energy, Conservation of momentum — yahan har trap ke peeche yahi do laws hain.
Relativistic energy-momentum relation — kyun non-relativistic $K E$ fail karta hai.
Photon momentum and energy, de Broglie wavelength — $p = h / λ$ kahan se aata hai.
Photoelectric effect — contrast: absorption vs. scattering.