Exercises — De Broglie hypothesis — matter waves λ = h - p

KYA karein: master form directly use karo — kuch convert nahi karna, momentum already diya hua hai. $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{3.3\times 10^{-24}}.$ KYU yeh form: problem humein $p$ de raha hai, to $\lambda =h/p$ ko rewrite karne ki zaroorat nahi. $\lambda =2.008\times 10^{-10}\text{m}\approx 2.0\text{A˚}.$ Yeh roughly ek atom ki width hai — yehi wajah hai ki electrons crystals se diffract hote hain.

KYA karein: "accelerated electron" situation ko pehchano ⇒ ready-made shortcut use karo. KYU yeh form: electron ke liye voltage $V$ se, constants $12.27/\sqrt{V}$ mein pehle se bake ho jaate hain. $\lambda =\frac{12.27}{\sqrt{100}}=\frac{12.27}{10}=1.227\text{A˚}.$

KYA karein: photon massless hota hai, isliye $\lambda =h/(mv)$ forbidden hai (zero se divide ho jaayega). $p=E/c$ use karo. KYU: photon momentum $p=E/c$ ke through carry karta hai, jo $\lambda =hc/E$ deta hai. $\lambda =\frac{hc}{E}=\frac{(6.626\times 10^{-34})(3\times 10^8)}{3.0\times 10^{-19}}=6.626\times 10^{-7}\text{m}\approx 663\text{nm}.$ Visible red light — ek real photon ke saath consistent hai.

KYA: humein $K$ bataya gaya hai, $v$ nahi. Energy ko momentum mein convert karo $p=\sqrt{2mK}$ se. KYU $p=\sqrt{2mK}$: kyunki $K=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$, so $p^2=2mK$. $p=\sqrt{2(1.673\times 10^{-27})(1.6\times 10^{-16})}=\sqrt{5.35\times 10^{-43}}=7.32\times 10^{-22}\text{kg⋅m/s}.$ $\lambda =\frac{6.626\times 10^{-34}}{7.32\times 10^{-22}}=9.05\times 10^{-13}\text{m}\approx 9.05\times 10^{-3}\text{A˚}.$ (Kyunki $1\text{A˚}=10^{-10}\text{m}$, ångström paane ke liye $9.05\times 10^{-13}$ ko $10^{-10}$ se divide karte hain.)

KYA: speed aur mass diya hua hai, slow object ⇒ $p=mv$. KYU $p=mv$: grain light-speed se kaafi neeche move kar raha hai, isliye momentum sirf mass times velocity hai — koi relativistic correction nahi chahiye. $p=(1.0\times 10^{-9})(1.0\times 10^{-3})=1.0\times 10^{-12}\text{kg⋅m/s}.$ $\lambda =\frac{6.626\times 10^{-34}}{1.0\times 10^{-12}}=6.626\times 10^{-22}\text{m}.$ Comment: kisi bhi atom se kaafi kaafi chhota ($10^{-22}$ vs $10^{-10}$ m). Koi bhi wave behaviour kabhi observable nahi hogi — grain "classical" hai.

KYA: charge through voltage ⇒ $K=qV$, phir $\lambda =h/\sqrt{2mqV}$. Hum $12.27/\sqrt{V}$ shortcut use nahi kar sakte — woh sirf electron ke liye tha (specific $m$ aur $q$). KYU shortcut nahi: $12.27/\sqrt{V}$ mein $m_e$ aur $e$ andar chhupa hua hai; is particle ka alag mass aur charge hai. $2mqV=2(6.64\times 10^{-27})(3.2\times 10^{-19})(1000)=4.25\times 10^{-42}.$ $p=\sqrt{4.25\times 10^{-42}}=2.06\times 10^{-21}\text{kg⋅m/s},\lambda =\frac{6.626\times 10^{-34}}{2.06\times 10^{-21}}=3.21\times 10^{-13}\text{m}.$

KYA: fixed $v$ par, $\lambda =\frac{h}{mv}\propto \frac{1}{m}$. KYU ratio numbers nahi: $h$ aur $v$ cancel ho jaate hain, sirf pure mass dependence bachti hai. $\frac{{\lambda }_e}{{\lambda }_p}=\frac{m_p}{m_e}=\frac{1.673\times 10^{-27}}{9.11\times 10^{-31}}\approx 1836.$ Electron zyada wavy hai, $\approx 1836$ ke factor se.

KYA: fixed $K$ par, $\lambda =\frac{h}{\sqrt{2mK}}\propto \frac{1}{\sqrt{m}}$. KYU change aata hai: energy speed se alag tarike se share hoti hai — mass ab square root ke neeche aata hai. $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{1836}\approx \mathrm{42.8.}$ Electron ab bhi zyada wavy hai, lekin sirf $\approx 42.8$ se, $1836$ nahi.

KYA: $\lambda \propto 1/\sqrt{V}$, to $\lambda$ ko half karne ke liye $\sqrt{V}$ double hona chahiye, yaani $V$ chaar guna hona chahiye. KYU: $\lambda =12.27/\sqrt{V}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{V_1}{V_2}}$. $\frac{0.5}{1.0}=\sqrt{\frac{V_1}{V_2}}\Rightarrow \frac{V_1}{V_2}=0.25\Rightarrow V_2=4V_1.$ Concretely $V_1=(12.27/1.0{)}^2\approx 150.6\text{V}$, to $V_2\approx 602\text{V}$.

KYA: standing-wave condition kehti hai ki circumference mein exactly $n$ ripples fit hone chahiye. $n=1$ ke liye: $\lambda =2\pi r=2\pi (0.529\times 10^{-10})=3.32\times 10^{-10}\text{m}=3.32\text{A˚}.$ YEH aisa kyun dikhta hai (neeche figure dekho): ek wave jo poora number of times fit nahi hoti woh out of phase overlap ho jaayegi aur cancel ho jaayegi — sirf whole-number fits survive karte hain, jo exactly Bohr's quantisation ko wave interference ke roop mein rebirth deta hai. Cross-check momentum se, reduced constant $\hslash =h/2\pi$ use karke jo upar define kiya gaya: Bohr's rule $mvr=n\hslash$ deta hai $p=mv=\frac{n\hslash }{r}=\frac{1.055\times 10^{-34}}{0.529\times 10^{-10}}=1.994\times 10^{-24}$, to $\lambda =h/p=\frac{6.626\times 10^{-34}}{1.994\times 10^{-24}}=3.32\text{A˚}$ — same answer. ✓

KYA: temperature ⇒ energy $K=\frac{3}{2}{k}_{B}T$ se, phir $\lambda =h/\sqrt{2mK}$. KYU $\frac{3}{2}{k}_{B}T$: temperature $T$ par ek free particle motion ke har direction mein average $\frac{1}{2}{k}_{B}T$ kinetic energy carry karta hai, aur teen directions hote hain — to kul $\frac{3}{2}{k}_{B}T$. $K=\frac{3}{2}(1.38\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}.$ $p=\sqrt{2(1.675\times 10^{-27})(6.21\times 10^{-21})}=\sqrt{2.08\times 10^{-47}}=4.56\times 10^{-24}.$ $\lambda =\frac{6.626\times 10^{-34}}{4.56\times 10^{-24}}=1.45\times 10^{-10}\text{m}\approx 1.45\text{A˚}.$ Kyun matter karta hai: $\approx 1.4\text{A˚}$ ek crystal mein planes ke beech ki spacing hai — to crystal neutrons ke liye ek diffraction grating ki tarah kaam karta hai, bilkul jaise Davisson–Germer experiment mein electrons ke liye hota hai.

KYA: confinement ek minimum momentum spread $\Delta p\approx \hslash /(2\Delta x)$ set karta hai, reduced constant $\hslash =h/2\pi$ use karke. KYU $\hslash /(2\Delta x)$: uncertainty relation $\Delta x\Delta p\ge \hslash /2$ kehti hai ki position squeeze karna ($\Delta x$ chhota) momentum spread ko upar force karta hai; equality lene par minimum $\Delta p$ milta hai. $\Delta p=\frac{1.055\times 10^{-34}}{2(1.0\times 10^{-10})}=5.27\times 10^{-25}\text{kg⋅m/s}.$ Taking $p\sim \Delta p$: $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{5.27\times 10^{-25}}=1.26\times 10^{-9}\text{m}\approx 12.6\text{A˚}.$ Kyun sensible hai: wavelength box size ke comparable (kuch times) aati hai — ek wave just barely fit hoti hai, jo confined hone ka physical meaning hai. Dono frameworks scale mein agree karte hain.

KYA & KYU relativistic form: electron ki rest energy $m_e{c}^2=8.2\times 10^{-14}\text{J}=511\text{keV}$ hai. Yahan $K=100\text{keV}$ uska ek-paanchwa hissa hai — chhota nahi — to $p=mv$ true momentum ko under-count karta hai. Exact relation hai $E^2=(pc{)}^2+(m{c}^2{)}^2$ with total energy $E=K+m{c}^2$, jo deta hai $pc=\sqrt{(K+m{c}^2{)}^2-(m{c}^2{)}^2}=\sqrt{K^2+2Km{c}^2}.$

(a) Non-relativistic (yahan galat, contrast ke liye kiya gaya): $\lambda =\frac{12.27}{\sqrt{10^5}}=\frac{12.27}{316.2}=0.0388\text{A˚}=3.88\times 10^{-12}\text{m}.$

(b) Relativistic: with $K=1.6\times 10^{-14}\text{J}$, $m{c}^2=8.2\times 10^{-14}\text{J}$: $K^2=2.56\times 10^{-28},2Km{c}^2=2(1.6\times 10^{-14})(8.2\times 10^{-14})=2.62\times 10^{-27}.$ $pc=\sqrt{2.56\times 10^{-28}+2.62\times 10^{-27}}=\sqrt{2.88\times 10^{-27}}=5.37\times 10^{-14}\text{J},$ $p=\frac{5.37\times 10^{-14}}{3\times 10^8}=1.79\times 10^{-22}\text{kg⋅m/s},\lambda =\frac{6.626\times 10^{-34}}{1.79\times 10^{-22}}=3.70\times 10^{-12}\text{m}=0.0370\text{A˚}.$

(c) Compare: non-relativistic deta hai $0.0388\text{A˚}$; relativistic deta hai $0.0370\text{A˚}$ — shortcut lagbhag $5\%$ zyada lamba hai. Zyada voltage par error tezi se badhti hai.

KYA: equal $\lambda$ ⇒ equal momentum $p=h/\lambda$, lekin energy depend karti hai kaise har ek use store karta hai. KYU do cases mein split karo: photon ki energy $pc$ hai (linear in $p$) jabki slow electron ki $p^2/2m$ — same $p$ se do alag energy formulas milte hain. $p=\frac{6.626\times 10^{-34}}{1.0\times 10^{-10}}=6.626\times 10^{-24}\text{kg⋅m/s}.$ Photon: $E_{\gamma }=pc=(6.626\times 10^{-24})(3\times 10^8)=1.99\times 10^{-15}\text{J}=12.4\text{keV}$. Electron (non-relativistic, $K=p^2/2m$): $K_e=\frac{(6.626\times 10^{-24}{)}^2}{2(9.11\times 10^{-31})}=2.41\times 10^{-17}\text{J}=150\text{eV}$. Comment: same wavelength par photon kaafi zyada energy carry karta hai ($\sim 12\text{keV}$ vs $\sim 150\text{eV}$), kyunki photon ki energy $pc$ hai (linear in $p$) jabki slow electron ki $p^2/2m$ (aur $c$ enormous hai).

KYA: electron shortcut $\lambda =12.27/\sqrt{V}$ ($\lambda$ Å mein) ko invert karo. KYU yeh invert karo: hum woh electron voltage chahte hain jo target $\lambda$ deta hai, exactly yahi $12.27/\sqrt{V}$ encode karta hai — to $V$ ke liye solve karo. $\sqrt{V}=\frac{12.27}{{\lambda }_{\text{A˚}}}=\frac{12.27}{0.05}=245.4\Rightarrow V=245.4^2=6.02\times 10^4\text{V}\approx 60\text{kV}.$ Relativity check: $K=eV=60\text{keV}$, phir se $511\text{keV}$ rest energy ka ek hissa, to true voltage needed thoda $60\text{kV}$ se kam hai (relativistic momentum diye gaye $K$ ke liye bada hota hai). Real electron microscopes sach mein $\sim 60$–$300\text{kV}$ par chalte hain aur precision ke liye relativistic formula use karte hain.

$\lambda =h/p$ poora game hai ::: pehle $p$ sahi se nikalo, phir $h$ ko usse divide karo. Same speed vs same energy alag ratios dete hain ::: $1/m$ (speed) vs $1/\sqrt{m}$ (energy). Bohr orbit ::: whole number of ripples fit hote hain, $2\pi r=n\lambda$. $\hslash$ kya hai? ::: reduced Planck constant $\hslash =h/2\pi =1.055\times 10^{-34}$ J·s. Relativistic kab jaana chahiye ::: jab $K$, $m{c}^2$ ($=511\text{keV}$ electron ke liye) se $\ll$ nahi hota.