Compton scattering — wavelength shift derivation: Exercises

2.3.4 · D4 · Physics › Modern Physics › Compton scattering — wavelength shift derivation

Neeche, $θ$ photon ka scattering angle hai (apni original direction se measure kiya), aur $ϕ$ electron ka recoil angle hai. Shuru karne se pehle figure ko ek map ki tarah padho: yeh woh ek picture hai jis par is page ke har problem ka reference hota hai.

Intuition Collision figure padho (Figure s01)

Pale-yellow arrow follow karo: yoh incoming photon hai, $x$ -axis ke saath left-se-right move kar raha hai, momentum $p = h / λ$ ke saath. Yeh blue dot se takraata hai — electron, origin par bilkul still baitha hai. Hit ke baad, origin se do nayi arrows nikalti hain:

pink arrow scattered photon hai, angle $θ$ se upar muda hua, ab lamba wavelength $λ^{'}$ ke saath (toh chota momentum $h / λ^{'}$ );
blue arrow recoiling electron hai, angle $ϕ$ se neeche kick kiya gaya, momentum $p_{e}$ ke saath.

Dashed grey line photon ki original direction hai — jisse hum $θ$ measure karte hain. Dhyaan do photon upar muda aur electron neeche recoil kiya: unke sideways ( $y$ ) kicks cancel hone chahiye, kyunki electron ne zero $y$ -momentum se shuru kiya tha. Woh cancellation bilkul wahi equation hai jo hum L4·Q2 mein solve karte hain.

Figure — Compton scattering — wavelength shift derivation

Figure s01 — Compton collision. Yellow = incoming photon, pink = scattered photon (angle $θ$ ), blue arrow = recoil electron (angle $ϕ$ ), blue dot = electron at rest, dashed line = photon ki original axis.

Level 1 — Recognition

Recall Solution

$Δ λ = \frac{h}{m _{e} c} (1 - cos θ) .$

$Δ λ = λ^{'} - λ$ : wavelength mein badhav (hamesha $\geq 0$ ).
$h$ : Planck's constant. $m_{e}$ : electron rest mass. $c$ : speed of light.
$θ$ : woh angle jo scattered photon apni incoming direction se banata hai.
$\frac{h}{m _{e} c} = λ_{C} \approx 2.43 pm = 2.43 \times 1 0^{- 12} m$ .

Recall Solution

Factor $(1 - cos θ)$ sab kuch control karta hai.

$θ = 0^{\circ}$ : $cos 0 = 1$ , toh $1 - cos θ = 0$ → $Δ λ = 0$ (photon seedha nikal jaata hai, koi momentum transfer nahi).
$θ = 18 0^{\circ}$ : $cos 18 0^{\circ} = - 1$ , toh $1 - cos θ = 2$ → maximum $Δ λ = 2 λ_{C} \approx 4.86 pm$ .

Level 2 — Application

Recall Solution

$cos 6 0^{\circ} = 0.5$ , toh $1 - cos θ = 0.5$ . $Δ λ = λ_{C} (0.5) = 2.43 \times 0.5 = 1.215 \approx 1.22 pm .$ $λ^{'} = 20.0 + 1.215 = 21.215 \approx 21.2 pm .$

Recall Solution

$cos 12 0^{\circ} = - 0.5$ , toh $1 - cos θ = 1.5$ . $Δ λ = 2.43 \times 1.5 = 3.645 \approx 3.65 pm .$ Dhyaan do yeh $9 0^{\circ}$ shift ( $2.43$ pm) se bada hai lekin backscatter max ( $4.86$ pm) se chota — jaisa hona chahiye.

Recall Solution

Set karo $λ_{C} (1 - cos θ) = λ_{C}$ , toh $1 - cos θ = 1$ , jaisse $cos θ = 0$ milta hai. $θ = 9 0^{\circ} .$ Isliye $9 0^{\circ}$ textbook ka "clean" angle hai — shift exactly ek Compton wavelength ke barabar hoti hai.

Level 3 — Analysis

Recall Solution

$9 0^{\circ}$ par, $Δ λ = λ_{C} = 2.43$ pm dono ke liye (shift $λ$ par independent hai).

X-ray: $\frac{Δ λ}{λ} = \frac{2.43}{50.0} = 0.0486 = 4.86%$ — aasani se measurable.
Visible: $λ = 500 nm = 5.00 \times 1 0^{5} pm$ , toh $\frac{Δ λ}{λ} = \frac{2.43}{5.00 \times 1 0 ^{5}} = 4.86 \times 1 0^{- 6} = 4.86 \times 1 0^{- 4} %$ — bilkul invisible. Conclusion: shift ka absolute size same hai, lekin matter sirf tab karta hai jab yeh wavelength ke relative ho. Short-wavelength X-rays mein fixed $2.43$ pm ek badi fraction ban jaati hai, toh effect clearly nazar aata hai.

Recall Solution

Energy conservation kehta hai electron ki kinetic energy exactly utni hai jitni photon ne khoyi. Photon ki energy pehle $E = \frac{h c}{λ}$ aur baad mein $E^{'} = \frac{h c}{λ ^{'}}$ hai (har ek sirf photon energy $h c / wavelength$ hai). Toh electron ko diya gaya energy hai

= hc\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right).$$ Kyunki $\lambda' > \lambda$, bracket positive hai — photon ne sach mein energy khoyi. Pehle $\lambda' = 50.0 + 2.43 = 52.43$ pm. Phir $$KE = 1.240\times10^{6}\left(\frac{1}{50.0} - \frac{1}{52.43}\right) = 1.240\times10^{6}(0.0200000 - 0.0190730)$$ $$= 1.240\times10^{6} \times 9.270\times10^{-4} \approx 1.149\times10^{3}\text{ eV} \approx 1.15\text{ keV}.$$

Level 4 — Synthesis

Recall Solution

Energy ko wavelength mein convert karo: $λ = \frac{h c}{E} = \frac{1.240 \times 1 0 ^{6}}{1.00 \times 1 0 ^{5}} = 12.40$ pm. $9 0^{\circ}$ par shift: $Δ λ = 2.43$ pm, toh $λ^{'} = 12.40 + 2.43 = 14.83$ pm. Wapas energy mein: $E^{'} = \frac{h c}{λ ^{'}} = \frac{1.240 \times 1 0 ^{6}}{14.83} \approx 8.36 \times 1 0^{4} eV \approx 83.6 keV .$ Electron ne isliye $100 - 83.6 = 16.4$ keV gain kiya.

Recall Solution

Pehle, symbol: == $p_{e}$ recoil electron ke momentum vector $p_{e}$ ka magnitude hai== — ek single positive number (same units mein jaise $h / λ$ ), Figure s01 mein $x$ -axis se neeche angle $ϕ$ par blue arrow ki direction mein.

Ab axes par har arrow ko decompose karke do momentum equations banao. Momentum ek vector hai, isliye $x$ -total aur $y$ -total alag alag conserve hote hain.

Pehle: sirf incoming photon move kar raha hai — momentum $h / λ$ purely $+ x$ ke saath, aur $y$ mein $0$ (electron rest par hai).
Baad mein: scattered photon ka momentum $h / λ^{'}$ angle $θ$ par hai, toh uske components hain $\frac{h}{λ ^{'}} cos θ$ ( $x$ ke saath) aur $\frac{h}{λ ^{'}} sin θ$ ( $+ y$ ke saath). Electron ka momentum $p_{e}$ angle $- ϕ$ par hai, toh uske components hain $p_{e} cos ϕ$ ( $x$ ke saath) aur $- p_{e} sin ϕ$ ( $y$ ke saath, yani neeche).

Har direction mein before $=$ after set karke electron ke terms isolate karo:

\qquad \underbrace{p_e\sin\phi = \frac{h}{\lambda'}\sin\theta}_{y\text{-momentum}}.$$ $y$ equation kehta hai electron ka downward kick exactly photon ke upward kick se match karta hai — wahi cancellation jo humne figure mein dekhi thi. Doosre ko pehle se divide karo (common factor $h$ cancel ho jaata hai; $1/\text{pm}$ ki units mein kaam karo): $$\tan\phi = \frac{\dfrac{1}{\lambda'}\sin\theta}{\dfrac{1}{\lambda} - \dfrac{1}{\lambda'}\cos\theta}.$$ $\theta=90^\circ$ ke saath ($\sin=1,\ \cos=0$): $$\tan\phi = \frac{1/\lambda'}{1/\lambda} = \frac{\lambda}{\lambda'} = \frac{12.40}{14.83} = 0.8362.$$ $$\phi = \arctan(0.8362) \approx 39.9^\circ.$$ Electron forward-aur-sideways recoil karta hai, photon ke sideways kick ke opposite — exactly wahi jo $y$-momentum ka conservation demand karta hai.

Level 5 — Mastery

Recall Solution

Derivation mein kabhi assume nahi kiya tha ki recoiling mass electron ki hai — usne sirf use $m$ kaha tha. Toh same formula $m_{e}$ ki jagah recoiling mass $M$ ke saath kaam karta hai: $Δ λ = \frac{h}{M c} (1 - cos θ) .$ Us number ko reuse karne ke liye jo hum pehle se jaante hain, electron ka Compton wavelength factor out karo. Kyunki

= \frac{\lambda_C}{M/m_e},$$ zyada bhari mass simply electron shift ko mass ratio $M/m_e$ se *divide* kar deti hai. $M/m_e = 22032$ aur $(1-\cos 180^\circ)=2$ ke saath: $$\Delta\lambda = \frac{2.43}{22032}\times 2 = 2.206\times10^{-4}\text{ pm} \approx 2.2\times10^{-4}\text{ pm}.$$ Yeh free-electron shift se lagbhag $22000$ guna chota hai — experimentally **zero** se practically indistinguishable. Toh jo photons tightly-bound electrons se bounce karte hain woh *original* wavelength par appear hote hain: yeh wahi ==unshifted peak== hai jo real spectra mein shifted peak ke saath baithta hai.

Recall Solution

Photon ki fractional energy loss:

= \frac{\lambda'-\lambda}{\lambda'} = \frac{\Delta\lambda}{\lambda + \Delta\lambda}.$$ Yeh $\Delta\lambda$ ke saath badhti hai, jo $\theta = 180^\circ$ par sabse badi hoti hai ($\Delta\lambda = 2\lambda_C$). Toh backscatter maximum energy transfer deta hai. $\lambda = 2.43$ pm $= \lambda_C$ ke liye: $$\Delta\lambda = 2\lambda_C = 4.86\text{ pm}, \qquad \lambda' = 2.43 + 4.86 = 7.29\text{ pm}.$$ $$\frac{\Delta E}{E} = \frac{4.86}{7.29} = 0.6667 = 66.7\%.$$ Is photon ki two-thirds energy head-on backscatter mein electron ko de di jaati hai.

Recall Solution

(a) $θ \to 0$ : $1 - cos θ \to 0$ toh $Δ λ \to 0$ — ek grazing (undeflected) photon koi momentum transfer nahi karta, apna wavelength rakhta hai. ✔ (b) $θ = 9 0^{\circ}$ : $1 - cos 9 0^{\circ} = 1$ , toh $Δ λ = λ_{C} = 2.43$ pm exactly. ✔ (c) Sabhi $θ \in [0^{\circ}, 18 0^{\circ}]$ ke liye, $cos θ \in [- 1, 1]$ , toh $(1 - cos θ) \geq 0$ . Kyunki $λ_{C} > 0$ hai, hume $Δ λ \geq 0$ milta hai: scattered wavelength kabhi incoming wavelength se choti nahi hoti. Ek photon resting electron ko sirf energy de sakta hai, kabhi le nahi sakta — woh energy conservation violate karta. ✔

Connections

Photon momentum and energy — har problem $E = h c / λ$ aur $p = h / λ$ use karta hai.
Conservation of energy aur Conservation of momentum — L4 ke peeche ke do twin laws.
Relativistic energy-momentum relation — kyun recoil electron $E_{e}^{2} = (p_{e} c)^{2} + (m_{e} c^{2})^{2}$ follow karta hai.
X-ray production and Bremsstrahlung — yeh X-rays scatter hone se pehle kahaan se aate hain.
Parent: the full derivation.