Exercises — Davisson-Germer experiment — electron diffraction

Bump ek diffraction peak hai. Pure particles ka stream smoothly phail jaata (jaise kisi diwar par sand phenka jaaye), isliye ek sharp intensity peak sirf wave model hi produce kar sakta hai. Experiment ka poora point yehi hai: ye matter par wave picture ko impose karta hai. Dekho Wave-particle duality.

Doosra student. Davisson–Germer electrons (matter) fire karta hai aur wave behaviour paata hai. Ye Photoelectric effect ka mirror image hai, jo light (ek wave) ko particles ki tarah act karte dikhata hai. Dono milke duality ke do aadhe hain.

Shortcut use karo, kyunki usme pehle se $h$, $m$, $e$ bake in hain: $\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=\frac{12.27}{10}=1.227\text{A˚}$ Ye tool kyun? Hume voltage diya gaya hai aur wavelength chahiye — "theory road" formula inke beech exact bridge hai.

Shortcut ko invert karo. $\lambda =12.27/\sqrt{V}$ se shuru karo, to $\sqrt{V}=12.27/\lambda$, aur squaring karne par: $V={\left(\frac{12.27}{\lambda }\right)}^2={\left(\frac{12.27}{1.00}\right)}^2=150.6\text{V}$ Square kyun? $\lambda$ depend karta hai $1/\sqrt{V}$ par, isliye ise undo karne ke liye reverse order mein inverse operations chahiye: reciprocal, phir square.

$\sin \varphi =\frac{n\lambda }{a}=\frac{1\times 1.67}{2.15}=0.7767$ $\varphi =\arcsin (0.7767)=51.0^{\circ }$ Yahan $\arcsin$ kya kar raha hai? $\sin \varphi$ humein wo "height ratio" batata hai jo ye angle banata hai; $\arcsin$ reverse question hai — "kaun sa angle us sine ko rakhta hai?" — to ye humein $\varphi$ wapas de deta hai.

Experiment road: $\lambda =a\sin \varphi =2.15\times \sin 50^{\circ }=2.15\times 0.766=1.647\text{A˚}\approx 1.65\text{A˚}$ Theory road: $\lambda =\frac{12.27}{\sqrt{54}}=\frac{12.27}{7.348}=1.670\text{A˚}$ Do independent roads $1.65$ Å aur $1.67$ Å deti hain — lagbhag $1\%$ tak agreement. Ye match hi matter waves ka proof hai. Thoda jo residual gap hai wo crystal ki inner potential ke cross karte waqt electron ke refraction ki wajah se hai.

$\theta =90^{\circ }-\frac{50^{\circ }}{2}=90^{\circ }-25^{\circ }=65^{\circ }$ Reference matter karta hai: $\varphi$ ko incident beam se measure kiya jaata hai (wo straight line jis par electrons aaye the), jabki Bragg's law mein Bragg's $\theta$ ko crystal plane se measure kiya jaata hai. Neeche diya hua figure dikhata hai ki $\varphi$ ko bisect karke aur $90^{\circ }$ se subtract karke plane ka glancing angle kyun milta hai.

Causes ki chain follow karo:

• Bada $V$ → bada momentum $p=\sqrt{2meV}$ → chhoti $\lambda$ (kyunki $\lambda =h/p$).
• $a\sin \varphi =n\lambda$ mein, chhoti $\lambda$ ek chhote $\sin \varphi$ ko force karta hai, isliye same order $n$ ke liye chhota $\varphi$ milta hai.
• Limiting case: $V$ itna badha do ki $\lambda <a/n$ us order ke liye satisfy na ho sake — $\sin \varphi$ bas $\varphi \to 0$ ki taraf shrink hota rehta hai; lekin first-order peak incident beam ki taraf crowd hone lagta hai aur eventually order unresolvable ho jaata hai aur peak effectively disappear ho jaati hai. Convention hamesha naam karo: yahan $\varphi$ surface-grating incident-beam angle hai.

Step 1 — voltage se wavelength (theory road): $\lambda =\frac{12.27}{\sqrt{54}}=1.670\text{A˚}$ Step 2 — wavelength se angle (grating road): $\sin \varphi =\frac{\lambda }{a}=\frac{1.670}{2.15}=0.7767\Rightarrow \varphi =\arcsin (0.7767)=51.0^{\circ }$ Step 3 — reality se compare karo: observed $\varphi \approx 50^{\circ }$. $1^{\circ }$ ka gap inner-potential refraction hai. Prediction essentially confirm ho jaata hai. Ye akela chain end-to-end Davisson–Germer argument hai.

Shortcut ko invert karo: $V={\left(\frac{12.27}{\lambda }\right)}^2={\left(\frac{12.27}{0.10}\right)}^2=(122.7{)}^2=15,055\text{V}\approx 1.5\times 10^4\text{V}$ Interpretation: ek modest $\sim 15$ kV se wavelength visible light se $\sim 5000\times$ chhoti milti hai. Kyunki resolving power wavelength ke shrinkne se improve hoti hai, isliye electron microscopes optical ones se kahin zyada fine detail dekhte hain. (Is energy par hum abhi non-relativistic hain: $15$ keV $\ll 511$ keV rest energy.)

First order ($n=1$): $\lambda =\frac{2d\sin \theta }{n}=2(2.03)\sin 65^{\circ }=4.06\times 0.9063=3.679\text{A˚}$ Second order ($n=2$): $\lambda =\frac{2d\sin \theta }{2}=\frac{3.679}{2}=1.840\text{A˚}$ Koi bhi exactly $1.65$ Å nahi hai — kyunki clean $1.65$ Å surface-grating relation se aata hai row spacing $a=2.15$ Å ke saath, Bragg ke $d=2.03$ Å se nahi. Bragg picture confirm karti hai ki diffraction ek sensible order par ho rahi hai lekin (in numbers ke saath) surface value ko directly reproduce nahi karti. Lesson: wavelength quote karne se pehle state karo ki kaun sa model aur kaun sa spacing use kar rahe ho.

Heisenberg uncertainty principle se, $\Delta x\Delta p\ge \hslash /2$ jahan $\hslash =h/2\pi =1.055\times 10^{-34}$ J·s: $\Delta p\ge \frac{\hslash }{2\Delta x}=\frac{1.055\times 10^{-34}}{2\times 1.0\times 10^{-10}}=5.27\times 10^{-25}\text{kg⋅m/s}$ Electron ka apna momentum hai $p=h/\lambda =6.626\times 10^{-34}/1.67\times 10^{-10}=3.97\times 10^{-24}$ kg·m/s. Ratio: $\Delta p/p=5.27\times 10^{-25}/3.97\times 10^{-24}\approx 0.13$, yaani roughly $13\%$. To electron ko ek atom tak localise karne se uska momentum (aur isliye uska diffraction angle) roughly $13\%$ smear ho jaata hai — peak mein ek real, unavoidable width hoti hai. Peaks sharp hote hain kyunki beam ek saath bahut saare atoms ko sample karti hai, sirf ek ko nahi.

Wavelength: $\lambda =a\sin \varphi =2.15\times \sin 45^{\circ }=2.15\times 0.7071=1.520\text{A˚}$. Voltage: $V=(12.27/\lambda {)}^2=(12.27/1.520{)}^2=(8.072{)}^2=65.2\text{V}$. To historic $54$ V se thoda zyada voltage peak ko $50^{\circ }$ se $45^{\circ }$ tak kheench leta hai — bilkul wahi "higher $V$ ⇒ smaller $\varphi$" behaviour jo Problem 3.3 mein tha, ab by design use kiya gaya hai.