Davisson-Germer experiment — electron diffraction: Worked examples

2.3.6 · D3 · Physics › Modern Physics › Davisson-Germer experiment — electron diffraction

Kuch bhi shuru karne se pehle, woh symbols jo hum baar baar use karenge. Har ek ko saral shabdon mein define kiya gaya hai aur neeche di gayi picture se joda gaya hai.

Definition Is page par use kiye gaye har symbol

$V$ = accelerating voltage (volts). Woh "dhakka" jo electron ko tez karta hai. Bada dhakka → tez electron.
$λ$ = wavelength (ångström mein, $1 \overset{˚}{A} = 1 0^{- 10}$ m). Matter wave ka "size," de Broglie hypothesis se.
$ϕ$ = scattering angle (degrees), seedhi aane wali incident beam se detector tak measure kiya jaata hai. Yahi movable Faraday cylinder read karta hai.
$a$ = neighbouring surface atoms ke beech spacing (ångström), "grating pitch."
$n$ = order number — ek whole number $1, 2, 3, \dots$ jo yeh count karta hai ki path difference mein kitne complete wavelengths fit hote hain. $n = 1$ pehla (sabse bright) peak hai, $n = 2$ doosra, aur aise aage.
$d$ = atomic planes ke beech spacing (surface se gehri), sirf Bragg cross-check mein use hoti hai.
$θ$ = glancing angle, atomic plane se beam tak measure kiya jaata hai (Ex 5 mein carefully define kiya gaya hai).
$h$ = Planck's constant $= 6.626 \times 1 0^{- 34} J\cdots$ — woh fixed number jo wave aur particle ko jodta hai.
$m$ = particle mass (kg); electron ke liye $m_{e} = 9.11 \times 1 0^{- 31}$ kg.
$q$ (ya $e$ ) = particle charge (coulombs); electron ke liye $e = 1.6 \times 1 0^{- 19}$ C.
$c$ = speed of light $= 3.0 \times 1 0^{8} m/s$ , "kya yeh relativistic hai?" ka yardstick.

Ek working formula, jo parent note mein pehle hi derive ho chuki hai (electron-only shortcut): $λ = \frac{12.27}{V} \overset{˚}{A} (V in volts)$ aur iska parent, kisi bhi charged particle ke liye general de Broglie form: $λ = \frac{h}{2 m q V}$

Figure — Davisson-Germer experiment — electron diffraction

Figure ko dhyan se padho — neeche ke har example mein yahi ek picture hai bas numbers alag hain. Neeli arrow (top-left) woh incident beam hai jo surface ki taraf aa rahi hai. Safed line par peele dots surface atoms hain, jinke beech $a$ ka spacing hai. Beech wale atom se nikalne wali pink arrow woh ray hai jo detector tak pahunchti hai, $ϕ$ angle (pink arc) par jhuki hui. Neeche ki choti peeli double-arrow spacing $a$ ko mark karti hai. Figure se jo key cheez padhni hai woh yeh hai: next atom se scatter hone wali wave detector tak pahunchne se pehle extra distance $a sin ϕ$ tay karti hai — yahi extra path hai jo bright peak ke liye $nλ$ ke barabar hona chahiye.

The scenario matrix

Is topic ke har possible case ki class:

Cell	Case class	Tricky kyun hai	Example
A	Standard forward ( $V \to λ \to ϕ$ )	baseline, dono roads agree karni chahiye	Ex 1
B	Reverse (measure $ϕ \to λ \to V$ )	formula ulta chalao	Ex 2
C	Alag voltage / limiting large $V$	peak move karti hai; check karo kya real rehti hai	Ex 3
D	Degenerate: peak disappears ( $sin ϕ > 1$ )	koi solution nahi — yeh kehna zaroori hai	Ex 4
E	Grating vs Bragg convention clash	kaun sa angle reference? $ϕ$ aur $θ$ relate karo	Ex 5
F	Relativistic edge case (high kV)	$2 m e V$ kab break karta hai?	Ex 6
G	Real-world word problem (electron microscope)	words → same formula mein translate karo	Ex 7
H	Exam twist: proton, electron nahi	mass sab kuch badal deta hai	Ex 8

Ex 1 — Cell A: standard forward pass

Worked example

$V = 54$ V, nickel row spacing $a = 2.15$ Å. Peak angle $ϕ$ nikalo. Forecast: padhne se pehle andaza lagao — kya $ϕ$ $5 0^{\circ}$ ke paas aayega, ya kahin aur?

Step 1. Theory-road wavelength: $λ = \frac{12.27}{54} = 1.67$ Å. Yeh step kyun? Pehle $λ$ chahiye; accelerating voltage hi ek aisa knob hai jise hum control karte hain.

Step 2. Grating condition, first order ( $n = 1$ ): $a sin ϕ = nλ = λ$ , toh $sin ϕ = \frac{λ}{a} = \frac{1.67}{2.15} = 0.777.$ Yeh step kyun? Adjacent surface atoms se constructive interference ke liye extra path $a sin ϕ$ (figure ki geometry se padha gaya) ek complete wavelength ke barabar hona chahiye.

Step 3. $ϕ = arcsin (0.777) = 51. 0^{\circ}$ . Yeh step kyun? $arcsin$ ka jawab hai "kaun sa angle is sine ko rakhta hai?" — yeh us $sin$ ko undo karta hai jo humne build ki thi. Formally do answers hain: $arcsin$ principal value $5 1^{\circ}$ deta hai, aur $18 0^{\circ} - 5 1^{\circ} = 12 9^{\circ}$ ka bhi same sine hai. Hum $5 1^{\circ}$ rakhte hain kyunki detector arm physically sirf $0^{\circ}$ – $9 0^{\circ}$ sweep kar sakta hai us side mein; $12 9^{\circ}$ reading ray ko crystal ke andar wapas bhej degi, jo unphysical hai. (Higher orders mein, Ex 4, hum check karte hain kya koi bhi value allowed hai.)

Verify: observed peak $\approx 5 0^{\circ}$ hai; hamara $5 1^{\circ}$ $1^{\circ}$ ke andar agree karta hai (gap inner-potential refraction ka hai). $λ$ tak dono raaste saath aate hain → matter waves confirmed.

Ex 2 — Cell B: reverse pass (peak → voltage)

Worked example

$ϕ = 4 4^{\circ}$ par peak dikh rahi hai aur $a = 2.15$ Å hai. Kaunsa accelerating voltage isse produce karta hai? Forecast: $5 0^{\circ}$ se chhota angle — kya aap higher ya lower $V$ expect karte ho?

Step 1. Grating deta hai $λ = a sin ϕ = 2.15 \times sin 4 4^{\circ} = 2.15 \times 0.6947 = 1.494$ Å. Yeh step kyun? Detector angle ab given hai, toh hum $λ$ recover karne ke liye geometry forward chalate hain. (Measured $ϕ$ pehle se principal, physical value hai, toh yahan koi branch ambiguity nahi.)

Step 2. Voltage formula invert karo: $λ = \frac{12.27}{V} \Rightarrow V = \frac{12.27}{λ}$ , toh $V = (\frac{12.27}{1.494})^{2} = (8.213)^{2} = 67.5 V .$ Yeh step kyun? Square root undo karne ke liye square karte hain — yeh woh algebra hai jo measured $λ$ ko us voltage mein badalta hai jisne use banaya.

Verify: chhoti $λ$ (1.49 < 1.67 Å) ko zyada $V$ chahiye (67.5 > 54 V) — $λ \propto 1/ V$ se consistent. Forecast confirmed: chhota angle ⇒ zyada voltage. ✅

Ex 3 — Cell C: voltage badhao, peak slide karti dekho

Worked example

$a = 2.15$ Å rakho. $V = 54$ V aur $V = 100$ V par peak angles compare karo. Forecast: kya peak beam ki taraf jaati hai (chhota $ϕ$ ) ya door (bada $ϕ$ ) jab $V$ badhta hai?

Step 1. $100$ V par: $λ = \frac{12.27}{100} = 1.227$ Å. Yeh step kyun? Zyada voltage → chhoti wave; angle se pehle wavelength recompute karo.

Step 2. First order, $sin ϕ = \frac{1.227}{2.15} = 0.5707 \Rightarrow ϕ = 34. 8^{\circ}$ (principal value; $14 5^{\circ}$ root phir bhi unphysical hai). Yeh step kyun? Wahi grating relation; sirf $λ$ badla.

Step 3. Compare karo: $54$ V ne $5 1^{\circ}$ diya, $100$ V ne $34. 8^{\circ}$ diya. Yeh step kyun? Do data points trend dikhate hain — peak incident beam ki taraf slide karti hai.

Verify: figure ki do rays confirm karti hain ki chhoti (blue) wave tight angle par scatter hoti hai. Jab $V \to \infty$ , toh $λ \to 0$ , $sin ϕ \to 0$ , toh $ϕ \to 0$ : peak straight-through beam par collapse ho jaati hai. Limiting behaviour handled. ✅

Ex 4 — Cell D: degenerate case (koi peak hi nahi)

Worked example Bahut slow electrons:

$V = 20$ V, $a = 2.15$ Å. Second-order ( $n = 2$ ) peak kahan hai? Forecast: formula blindly trust karo, ya koi trap hai?

Step 1. $λ = \frac{12.27}{20} = 2.744$ Å. Yeh step kyun? Kam voltage → lambi wavelength — yahi woh exact time hai jab problem aati hai.

Step 2. Second order ( $n = 2$ ): $a sin ϕ = 2 λ \Rightarrow sin ϕ = \frac{2 \times 2.744}{2.15} = \frac{5.488}{2.15} = 2.552$ . Yeh step kyun? Honestly plug in karo aur result padho.

Step 3. $sin ϕ = 2.552 > 1$ . Kisi bhi angle ka sine 1 se upar nahi hota — na principal value, aur na $18 0^{\circ} - ϕ$ root, kyunki dono branches $1$ se zyada nahi ho saktein. Toh yeh peak exist hi nahi karti. Yeh step kyun? Kyunki $- 1 \leq sin ϕ \leq 1$ hamesha; us range se bahar ki demand matlab order physically forbidden hai, aur arcsin ki koi bhi branch isse rescue nahi kar sakti.

Verify: sabse bade possible order ke liye $nλ \leq a$ chahiye. Explicitly ceiling compute karo: $\frac{a}{λ} = \frac{2.15}{2.744} = 0.78.$ Toh $n \leq 0.78$ chahiye. Kyunki $n$ ek whole number $\geq 1$ hona chahiye (by definition, order complete wavelengths count karta hai), pehla order bhi fail ho jaata hai — $20$ V par is crystal mein koi diffraction peak nahi dikhti. Degenerate case fully described: "no solution" bolo, kabhi answer force mat karo. ✅

Ex 5 — Cell E: grating angle vs Bragg angle

Worked example Peak

$ϕ = 5 0^{\circ}$ par hai. Bragg glancing angle $θ$ mein convert karo, phir Bragg reading check karo. Forecast: kya $θ$ grating $ϕ$ se bada hoga ya chhota?

Step 1 — figure se har angle pin karo. Do alag conventions hain aur inhe confuse karna classic error hai.

Grating $ϕ$ incident beam se surface par outgoing ray tak measure kiya jaata hai.
Bragg $θ$ glancing angle hai: atomic plane se beam tak.

Figure dekho. Incident aur diffracted rays plane ke baare mein mirror images hain. Glancing angle (beam-to-plane) ko $θ$ kaho. Dono beams ke beech ka angle phir simple mirror symmetry se $2 θ$ hota hai — standard Bragg relation $ϕ_{beam-to-beam} = 2 θ ⟹ θ = \frac{ϕ _{beam-to-beam}}{2} .$ Yeh step kyun? Yeh correct, textbook link hai: incoming aur outgoing beams ke beech total deflection glancing angle se double hoti hai.

Step 2 — dhyan rakho kaun sa $ϕ$ diya gaya hai. D–G ke liye quoted detector angle, $ϕ = 5 0^{\circ}$ , incident beam se surface ray tak measure kiya gaya hai, poori beam-to-beam deflection nahi. Beam-to-beam deflection $18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$ hai, toh $θ = \frac{18 0 ^{\circ} - ϕ}{2} = \frac{13 0 ^{\circ}}{2} = 6 5^{\circ},$ jo $θ = 9 0^{\circ} - \frac{ϕ}{2} = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ jaisa hi hai. Yeh step kyun? Dono forms agree karte hain — $9 0^{\circ} - ϕ /2$ shortcut sirf $\frac{18 0 ^{\circ} - ϕ}{2}$ rewritten hai. Lesson: halving se pehle hamesha identify karo ki tera diya hua angle beam-to-surface hai ya beam-to-beam.

Step 3 — Bragg reading. Sahi Ni(111) plane spacing $d = 2.03$ Å ke saath, first order: $λ = 2 d sin θ = 2 \times 2.03 \times sin 6 5^{\circ} = 4.06 \times 0.9063 = 3.68 \overset{˚}{A} .$ Yeh higher-order reflection correspond karta hai; $n = 2$ reading $λ = 3.68/2 = 1.84$ Å physically comparable hai. Clean $1.65$ Å surface-grating relation se aata hai, Bragg se nahi — parent ki Bragg's law discussion ka deliberate reminder.

Verify: $θ = 6 5^{\circ}$ beam-to-surface $ϕ = 5 0^{\circ}$ se bada hai, kyunki plane angle beam ke almost perpendicular line se measure hota hai. $ϕ = 5 0^{\circ}$ ko seedha Bragg mein dalne par $2 \times 2.03 \times sin 5 0^{\circ} = 3.11$ Å aayega — ek galat number, jo prove karta hai ki conversion matter karti hai. ✅

Ex 6 — Cell F: relativistic edge case

Worked example

$V = 10, 000$ V par, kya non-relativistic $λ = 12.27/ V$ ab bhi sahi hai? Error estimate karo. Forecast: modern physics — kya hume relativistic jaana hoga, ya $10$ kV abhi bhi "slow" hai?

Step 1. Kinetic energy $= e V = 10, 000$ eV $= 10$ keV. Electron ki rest energy $= m_{e} c^{2} = 511$ keV. Yeh step kyun? Test ratio $KE / (m_{e} c^{2})$ hai; agar yeh chhota hai, non-relativistic theek hai.

Step 2. Ratio $\frac{KE}{m _{e} c ^{2}} = \frac{10}{511} = 0.0196 \approx 2%$ . Yeh step kyun? Yeh batata hai ki hum low-speed regime mein hain, lekin yeh wavelength error ka size nahi hai — agla step dekho.

Step 3. Relativistic momentum expand hota hai $p = 2 m KE (1 + \frac{KE}{4 m _{e} c ^{2}} + \dots)$ ke roop mein, toh $p$ par (aur isliye $λ$ par, kyunki $λ = h / p$ ) leading fractional correction hai $\frac{Δ λ}{λ} \approx \frac{KE}{4 m _{e} c ^{2}} = \frac{10}{4 \times 511} = 0.0049 \approx 0.5%.$ Yeh step kyun? Correction $KE / (4 m_{e} c^{2})$ hai, raw energy ratio ka one quarter — ek common trap yeh hai ki $2%$ ratio ko error bata dete hain; yahan true wavelength shift sirf $\approx 0.5%$ hai.

Step 4. Non-relativistic $λ = 12.27/ 10000 = 0.1227$ Å; true value $\approx 0.5%$ chhoti hai, $\approx 0.1221$ Å. Yeh step kyun? Relativity $p$ ko thoda badhati hai, $λ$ ko thoda shrink karti hai.

Verify: original $54$ V par, $KE / (4 m_{e} c^{2}) = 54/ (4 \times 511000) = 0.0026%$ — bilkul negligible, toh parent note non-relativistic rehne mein sahi tha. Sirf hundreds of kV ke paas correction $10%$ se upar jaata hai. Edge case bounded. ✅

Ex 7 — Cell G: real-world word problem (electron microscope)

Worked example Ek

Electron microscope $V = 60, 000$ V use karta hai. Electron wavelength kya hai, aur imaging ke liye yeh kyun spectacular hai? Forecast: andaza lagao ki kya yeh visible light ( $\sim 5000$ Å) se thoda beat karta hai ya hazaron baar.

Step 1. $λ = \frac{12.27}{60000} = \frac{12.27}{244.9} = 0.0501$ Å. Yeh step kyun? Same de Broglie formula — microscope sirf Davisson–Germer high voltage par hai.

Step 2. Visible light $λ_{light} \approx 5000$ Å se compare karo: ratio $= 5000/0.0501 \approx 1 0^{5}$ . Yeh step kyun? Resolving power $1/ λ$ ke saath scale hoti hai; $1 0^{5}$ -times-shorter wave $1 0^{5}$ -times-finer detail resolve karta hai.

Verify: $0.05$ Å ek atom ( $\sim 1$ Å) se bahut chhota hai, toh electron microscope principle mein individual atoms resolve kar sakta hai — light ke liye impossible. Word problem Cell C arithmetic mein reduce ho gaya. ✅ (Ex 6 se, yahan relativity ignore karna $KE / (4 m_{e} c^{2}) = 60/ (4 \times 511) \approx 3%$ error deta hai — estimate ke liye acceptable hai.)

Ex 8 — Cell H: exam twist — electron ki jagah proton

Worked example Ek proton

$V = 54$ V se accelerate kiya jaata hai. Uski wavelength nikalo aur electron ki $1.67$ Å se compare karo. Forecast: bhaari particle — electron se lambi wave ya chhoti?

Step 1. $12.27/ V$ shortcut sirf electron ke liye hai (usne electron mass hide kar liya tha). General form par wapas jao $λ = \frac{h}{2 m q V}$ . Yeh step kyun? Har constant $12.27$ mein collapse ho gaya tha assume karke ki $m = m_{e}$ ; proton ko apne mass ke saath general form chahiye.

Step 2. Proton mass $m_{p} = 1.673 \times 1 0^{- 27}$ kg $= 1836 m_{e}$ ; charge $q = e$ same. Toh $\frac{λ _{p}}{λ _{e}} = \frac{m _{e}}{m _{p}} = \frac{1}{1836} = \frac{1}{42.8} .$ Yeh step kyun? Sirf mass badla; $λ \propto 1/ m$ , toh ratio ek clean square root hai.

Step 3. $λ_{p} = \frac{1.67}{42.8} = 0.0390$ Å. Yeh step kyun? Jaane-maane electron answer ko ratio se scale karo — har constant dubara plug karne se zyada fast aur error-proof hai.

Verify: direct compute $λ_{p} = \frac{6.626 \times 1 0 ^{- 34}}{2 ( 1.673 \times 1 0 ^{- 27} ) ( 1.6 \times 1 0 ^{- 19} ) ( 54 )} = 3.90 \times 1 0^{- 12}$ m $= 0.0390$ Å. ✅ Bhaari → chhoti wave, forecast confirmed. Usi voltage par ek proton ko diffract karne ke liye $\sim 43 \times$ maheen grating chahiye hogi — nickel koi peak nahi dikhayega.

Recall Which cell does each trap belong to?

Peak vanishes because $sin ϕ > 1$ ::: Cell D (degenerate) Feeding beam-to-surface $ϕ$ straight into Bragg's law ::: Cell E (convention clash) Using $12.27/ V$ for a proton ::: Cell H (mass-dependent constant) Quoting the raw $KE / m_{e} c^{2}$ ratio as the wavelength error ::: Cell F — the true error is one quarter of it

Dekho bhi: Wave-particle duality, Photoelectric effect, Heisenberg uncertainty principle, aur parent Davisson–Germer topic note.