Davisson-Germer experiment — electron diffraction: Visual walkthrough

2.3.6 · D2 · Physics › Modern Physics › Davisson-Germer experiment — electron diffraction

Step 1 — Electron ko ek dhakka do (energy)

KYA. Hum ek electron ko rest mein shuru karte hain aur ek voltage $V$ usse kheenchne dete hain. "Voltage" bas electric energy ki ek pahaadi hai; electron usse neeche fisal kar tez ho jaata hai.

KYUN. Humein electron ki speed ko precisely control karna hai, kyunki speed se momentum set hota hai, aur momentum se woh wavelength set hoti hai jo hum measure karna chahte hain. Voltage ek aisa knob hai jise hum exactly dial kar sakte hain.

PICTURE. Electron voltage ki pahaadi se neeche fisal kar speed $v$ ke saath pahunchta hai.

Figure — Davisson-Germer experiment — electron diffraction

Yahan $e = 1.6 \times 1 0^{- 19}$ C ek fixed charge hai, $m = 9.11 \times 1 0^{- 31}$ kg ek fixed mass hai — sirf $V$ aur $v$ change hote hain.

Step 2 — Speed ko momentum mein badlo

KYA. Hum energy equation ko rewrite karte hain taaki momentum $p$ ("quantity of motion", $p = m v$ ) akela nazar aaye.

KYUN. de Broglie ka rule $p$ ke terms mein likha gaya hai, $v$ ke nahi. Isliye humein translate karna hoga. Momentum ek wave-particle ki natural currency hai, kyunki ek bhaari-lekin-dheema aur ek halka-lekin-tez particle jinka $p$ same ho, yahan identically behave karte hain.

PICTURE. Wahi electron, ab apni motion ki direction mein $p$ arrow se label kiya gaya.

Step 3 — Wavelength attach karo (de Broglie)

KYA. Hum finally woh hypothesis invoke karte hain jo trial par hai: har momentum $p$ apne saath wavelength $λ = h / p$ lekar chalta hai.

KYUN. Yeh woh claim hai jiske liye yeh poora experiment exist karta hai. Hum isse abhi prove nahi kar rahe — hum woh number compute kar rahe hain jo yeh predict karta hai, taaki baad mein isse measurement se compare kar sakein. Tool hai de Broglie's relation kyunki yahi ek formula hai jo particle ke momentum ko wave ki wavelength se jodata hai.

PICTURE. Momentum arrow ek chhoti si wave ban jaati hai jiska spacing $λ$ hai: tez electron $\to$ chhoti wave, dheema electron $\to$ lambi wave.

Convert-to-Å

12.27

mein already

h

m

e

aur Å conversion bundle ho chuki hai, isliye tumhe bas

V

volts mein dena hai.

Step 4 — Crystal ruler se milao

KYA. Hum yeh wave ek nickel single crystal par fire karte hain. Iske surface atoms evenly spaced rows mein baithe hain, ek doosre se $a$ distance par. Yeh spacing hi hamari ruler hai.

KYUN. Wave measure karne ke liye tumhe koi cheez chahiye jo wavelength ke size ki ho — jaise comb use karna, fence nahi, daanton ko feel karne ke liye. Nickel ki row spacing $a \approx 2.15$ Å $\approx λ$ hai, isliye crystal ek natural diffraction grating ki tarah kaam kar sakti hai. Koi bhi bahut moti cheez kuch nahi dikhayegi.

PICTURE. Aane wale wavefronts (flat lines) $a$ spacing wale atoms ki ek row par pahunchte hain; har atom ek chhoti wave re-radiate karta hai.

Step 5 — Extra path count karo (peak kahan se aati hai)

KYA. Angle $ϕ$ par jaane wali wave dekhte hain (incident beam se measure kiya gaya). Ek atom ka ripple apne padosi ki tulna mein $a sin ϕ$ extra distance travel karta hai.

KYUN. Do ripples brightly tab add hote hain jab woh in step pahunchein — crest pe crest. Aisa exactly tab hota hai jab extra path wavelengths ki ek poori sankhya ho. Yeh woh ek geometric fact hai jo diffraction peak produce karta hai. Hum $sin ϕ$ use karte hain ( $cos$ ya $tan$ nahi) kyunki extra path ek right triangle ki woh side hai jo angle $ϕ$ ke opposite hai, spacing $a$ hypotenuse ka base hai — "opposite over hypotenuse" exactly $sin$ hai.

PICTURE. Do paas wale atoms ka zoom; amber segment extra path $a sin ϕ$ hai.

Nickel ke liye $a = 2.15$ Å, observed $ϕ = 5 0^{\circ}$ par, $n = 1$ : $λ = a sin ϕ = 2.15 \times sin 5 0^{\circ} = 1.65 \overset{˚}{A} .$

Step 6 — Dono raaste milte hain

KYA. Dono independent numbers ko side by side rakhte hain.

KYUN. Yahi verdict hai. Road 1 (voltage $\to$ de Broglie) aur Road 2 (angle $\to$ grating) ne kabhi ek doosre se baat nahi ki, phir bhi:

Road	Input	$λ$
Theory ( $h / 2 m e V$ )	$V = 54$ V	$1.67$ Å
Experiment ( $a sin ϕ$ )	$ϕ = 5 0^{\circ}$	$1.65$ Å

PICTURE. Dono arrows essentially ek hi wavelength mark par land karte hue.

Step 7 — Edge cases: agar hum knob badlein to kya?

KYA. Hum degenerate aur limiting situations check karte hain taaki koi reader ambush na ho.

KYUN. Ek derivation jise tum extremes tak push nahi kar sakte, woh samjhi nahi gayi. Dekho $a sin ϕ = nλ$ boundaries par kya karta hai.

PICTURE. Teen panels — high voltage (peak andar slide karti hai), low voltage, aur "no solution" wall.

Ek-picture summary

Yeh ek blueprint poori story ko thread karta hai: voltage $V$ left par electron ko drive karta hai, jo momentum $p = 2 m e V$ acquire karta hai, jise de Broglie wavelength $λ = h / p$ mein convert karta hai (top road). Wahi wave crystal ruler $a$ se takraati hai aur geometry $a sin ϕ = nλ$ peak angle se seedha $λ$ padhti hai (bottom road). Dono raaste ek hi number par milte hain — matter waves ka proof.

Recall Feynman retelling — apne words mein batao

Humne ek electron ko voltage ki pahaadi se neeche roll kiya taaki hum jaanein woh exactly kitna tez gaya. Tez matlab "bahut zyada momentum". De Broglie whisper karta hai ki jis bhi cheez ke paas momentum ho woh secretly ek wave carry karta hai, aur tez electron ke liye ek chhoti si wave. Humne woh wave nickel par fainka, jiske atoms ek comb ke daanton ki tarah ek line mein laage hain aur bilkul sahi spacing par hain. Paas wale atoms se bounce hone wale ripples sirf tabhi ek saath cheer karte hain — bright bump banate hain — jab unke beech ka extra distance wavelengths ki ek poori sankhya ho. Humne us bump ka angle measure kiya, "extra path = $a sin ϕ$ " ka arithmetic kiya, aur $1.65$ Å ki wavelength nikli. Isi beech voltage-aur-momentum arithmetic ne $1.67$ Å predict kiya tha. Do bilkul alag calculations, same answer. Woh "same answer" hi poora experiment hai: electrons waves hain.

Recall Self-test

Extra-path term mein $sin ϕ$ kyun hai, $tan ϕ$ kyun nahi? ::: Path difference ek right triangle ki woh side hai jo $ϕ$ ke opposite hai, atom spacing reference length hai — opposite/hypotenuse $sin$ hota hai. Peak bilkul gayab kab ho jaati hai? ::: Agar $λ > a$ ho, to $sin ϕ = λ / a > 1$ ka koi solution nahi — koi bhi order constructively interfere nahi kar sakta. Matter waves "prove" hone ke liye kaunsi do independent quantities agree karni chahiye? ::: Theory-road wavelength $h / 2 m e V$ aur experiment-road wavelength $a sin ϕ$ .

Connections: de Broglie hypothesis · Bragg's law · Wave-particle duality · Photoelectric effect (complementary half) · Electron microscope (yeh effect kaam mein lagaya gaya) · Heisenberg uncertainty principle.