De Broglie hypothesis — matter waves λ = h - p: Worked examples

2.3.5 · D3 · Physics › Modern Physics › De Broglie hypothesis — matter waves λ = h - p

Kuch bhi karne se pehle, har symbol ko pin down kar lete hain jo hum use karenge, taaki koi bhi symbol bina define kiye andar na aa jaye.

Definition Symbols ka cast

$λ$ (Greek "lambda") — wavelength: matter wave ki ek poori ripple ki lambaai, metres mein naapi jaati hai. Picture karo ek wave-crest se doosre tak ka distance.
$p$ — momentum: "kitni motion" koi cheez carry karti hai. Dheemi objects ke liye $p = m v$ (mass times velocity). Units: $kg\cdotm/s$ .
$m$ — kilogram mein mass; $v$ — metres per second mein speed.
$K$ — kinetic energy, motion ki energy, joules (J) mein.
$E$ — kisi particle ki total energy, bhi joules (J) mein. Ek photon (light particle) ke liye iske paas sirf yahi $E$ hoti hai; alag "rest" part nahi hota kyunki uski mass zero hai. Hum $E$ ko photon example (Cell H) mein milenge.
$c = 3 \times 1 0^{8} m/s$ — vacuum mein speed of light, universe ki fixed top speed. Yeh tab aata hai jab light ya relativity enter karti hai (Cells H aur I).
$h = 6.626 \times 1 0^{- 34} J\cdots$ — Planck's constant, woh fixed number jo pure quantum world ko scale karta hai.
$q$ ya $e$ — electric charge; electron ke liye $e = 1.6 \times 1 0^{- 19} C$ (coulombs).
$V$ — accelerating voltage volts mein.
$γ$ (Greek "gamma") — relativistic factor $γ = 1/ 1 - v^{2} / c^{2}$ , hamesha $\geq 1$ ; yeh measure karta hai ki kitna fast motion (near $c$ ) momentum ko inflate karta hai. Cell I mein build hoga.
$1 \overset{˚}{A}$ ("angstrom") $= 1 0^{- 10} m$ — ek handy tiny ruler, roughly atom ke size ka.

Hum exactly chaar "faces" use karenge ek hi formula ke. Kaunsa face choose karoge yeh depend karta hai problem kya deti hai usse.

Scenario matrix

Har de Broglie problem in cells mein se kisi ek mein aati hai. Baaki page har ek ko fill karta hai.

Cell	Kya diya gaya hai	Kaunsa face	Dhyan raho	Example
A	mass + speed, chota particle	$h / m v$	SI units rakho	Ex 1
B	mass + speed, macroscopic object	$h / m v$	answer absurdly tiny hoga — yeh sahi hai	Ex 2
C	electron + voltage $V$	$12.27/ V$ Å	yeh shortcut sirf electrons ke liye hai	Ex 3
D	kinetic energy $K$ (koi bhi particle)	$h / 2 m K$	agar sirf $K$ pata ho toh kabhi $h / m v$ use mat karo	Ex 4
E	temperature $T$ (thermal particle)	$K = \frac{3}{2} k_{B} T$ phir face D	$K$ average hai, exact nahi	Ex 5
F	ratio / comparison, same $K$	$λ \propto 1/ m$	kaunsi quantity fixed rakhi gayi hai?	Ex 6
G	ratio / comparison, same $v$	$λ \propto 1/ m$	cell F se alag!	Ex 7
H	photon (massless) — degenerate case	$λ = h / p$ , $p = E / c$	$h / m v$ forbidden hai ( $m = 0$ )	Ex 8
I	relativistic (speed near light) — limiting case	$p = γ m v$	$h / m v$ fail karta hai, correct karna padega	Ex 9
J	exam twist: $λ$ given, $V$ ya $K$ find karo	ek face ko invert karo	algebra ulta chalta hai	Ex 10

Do "danger cells" hain — H (mass zero hai, toh $h / m v$ face literally zero se divide karta hai) aur I (speed itni zyada hai ki $p \neq = m v$ ). Hum dono ko seedha aankh mein dekhenge.

Cell A — chota particle, speed given

Worked example Ex 1 · Ek electron

$2 \times 1 0^{6} m/s$ pe chal raha hai Uska de Broglie wavelength find karo. Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

Forecast: electrons famous hain "wavy" hone ke liye. Guess karo: kya $λ$ atom ke size ke around aayega ( $\sim 1 \overset{˚}{A} = 1 0^{- 10} m$ ), ya bahut hi chota? Guess pakad ke rakho.

Momentum compute karo $p = m_{e} v = (9.11 \times 1 0^{- 31}) (2 \times 1 0^{6}) = 1.822 \times 1 0^{- 24} kg\cdotm/s$ . Yeh step kyun? Hame mass aur speed pata hai, aur speed light se bahut chhoti hai, toh cell A ka face $h / m v$ apply hoga — aur uska dil hai $p = m v$ .
Divide karo $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{1.822 \times 1 0 ^{- 24}} = 3.64 \times 1 0^{- 10} m$ . Yeh step kyun? Yeh $λ = h / p$ khud hai — master face.
Angstroms mein: $3.64 \times 1 0^{- 10} m = 3.64 \overset{˚}{A}$ .

Verify karo: units check — $\frac{J\cdots}{kg\cdotm/s} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{kg\cdotm/s} = m$ . ✓ Aur $3.6 \overset{˚}{A}$ kuch atom-widths hai — bilkul wahi "wavy on the atomic scale" jo hamara forecast chahta tha. ✓

Cell B — macroscopic object

Cell C — ek electron ko voltage ke through accelerate kiya gaya

Cell D — kinetic energy given (koi bhi particle)

Worked example Ex 4 · Ek proton jisme

$K = 1.0 MeV$ hai $λ$ find karo. Proton mass $m_{p} = 1.673 \times 1 0^{- 27} kg$ ; $1 MeV = 1.6 \times 1 0^{- 13} J$ .

Forecast: protons electrons se ~1836× bhaari hote hain, toh ek given energy pe woh zyada momentum carry karte hain aur electrons se kam wavy hone chahiye. $λ$ ko $1 \overset{˚}{A}$ se kaafi neeche expect karo.

Energy convert karo $K = 1.0 MeV = 1.6 \times 1 0^{- 13} J$ . Yeh step kyun? Formula ko joules chahiye; MeV sirf ek nickname hai.
Momentum $p = 2 m_{p} K = 2 (1.673 \times 1 0^{- 27}) (1.6 \times 1 0^{- 13})$ . Andar: $2 (1.673 \times 1 0^{- 27}) (1.6 \times 1 0^{- 13}) = 5.35 \times 1 0^{- 40}$ , toh $p = 5.35 \times 1 0^{- 40} = 2.31 \times 1 0^{- 20} kg\cdotm/s$ . Yeh step kyun? Cell D ka face — energy in, momentum out, $v$ ke through koi detour nahi.
Wavelength $λ = \frac{6.626 \times 1 0 ^{- 34}}{2.31 \times 1 0 ^{- 20}} = 2.87 \times 1 0^{- 14} m$ .

Verify karo: $2.87 \times 1 0^{- 14} m = 2.87 \times 1 0^{- 4} \overset{˚}{A}$ — ek atom se hazaron times chota, jaise ek bhaari, energetic proton ke liye forecast tha. ✓ Kya $1 MeV ≪ m_{p} c^{2} \approx 938 MeV$ ? Haan, toh non-relativistic $2 m K$ yahan valid hai (cell I se contrast karo). ✓

Cell E — thermal particle (temperature given)

Worked example Ex 5 · Ek thermal neutron

$T = 300 K$ pe (room temperature) $λ$ find karo. Neutron mass $m_{n} = 1.675 \times 1 0^{- 27} kg$ .

Forecast: neutron diffraction ek real lab technique hai, jo tab hi kaam karti hai jab $λ \approx$ crystal spacing $\approx 1 \overset{˚}{A}$ . Toh hum expect karte hain $1 \overset{˚}{A}$ ke paas ka answer.

Average KE $K = \frac{3}{2} k_{B} T = 1.5 (1.38 \times 1 0^{- 23}) (300) = 6.21 \times 1 0^{- 21} J$ . Yeh step kyun? Temperature ko kinetic energy mein convert karo jo formula chahti hai.
Momentum $p = 2 m_{n} K = 2 (1.675 \times 1 0^{- 27}) (6.21 \times 1 0^{- 21})$ . Andar: $2.08 \times 1 0^{- 47}$ , toh $p = 4.56 \times 1 0^{- 24} kg\cdotm/s$ . Yeh step kyun? Ab yeh exactly cell D hai.
Wavelength $λ = \frac{6.626 \times 1 0 ^{- 34}}{4.56 \times 1 0 ^{- 24}} = 1.45 \times 1 0^{- 10} m = 1.45 \overset{˚}{A}$ .

Verify karo: $\approx 1.5 \overset{˚}{A}$ — bilkul atomic-spacing scale pe, exactly isi liye neutron diffraction room temperature pe ki jaati hai. ✓ Parent note ke $\approx 1.4 \overset{˚}{A}$ se match karta hai (rounding). ✓

Cells F & G — comparisons (yahan students sabse zyada confuse hote hain)

Sabse common exam trap: kya energy same hai, ya speed same hai? Proportionality flip ho jaati hai.

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

Worked example Figure — do panels padhna

Upar ka blueprint figure ek electron (cyan) aur ek proton (amber) ki actual wave-ripples draw karta hai, stack karke taaki tum crest-to-crest spacing seedha compare kar sako.

Left panel ("Same kinetic energy $K$ "): dono particles equal energy carry karte hain. Electron ki cyan wave clearly zyada lambi crest-to-crest distance ke saath draw ki gayi hai proton ki amber wave se. Double-headed white arrows har ek ki ek full wavelength mark karte hain — electron ka arrow kaafi wider hai. Yeh $λ \propto 1/ m$ visualize karta hai: kyunki $m_{p} / m_{e} = 1836 \approx 42.8$ , electron ki ripple lagbhag 43× longer hai.
Right panel ("Same speed $v$ "): ab speed lock hai instead. Proton ki amber wave left panel se bhi zyada tight squeeze ho gayi hai, kyunki yahan $λ \propto 1/ m$ hai (no square root), jo $1836$ ka factor deta hai. Electron ki cyan wave unchanged hai, toh dono waves ke beech gap dramatically bada hai. Takeaway jo tumhe picture se padhnee chahiye: electron dono cases mein wavier hai, lekin kitna wavier hai yeh depend karta hai completely iss baat pe ki kya fixed rakha gaya hai — right panel ka gap left se kaafi bada hai. Yahi difference Cells F aur G ka poora lesson hai.

Cell H — degenerate case: ek massless photon

Worked example Ex 8 · Ek photon jisme energy

$E = 2.0 eV$ hai (visible light) Uska wavelength find karo. $1 eV = 1.6 \times 1 0^{- 19} J$ ; $c = 3 \times 1 0^{8} m/s$ .

Forecast: $2 eV$ roughly red-orange light hai, toh $λ$ lagbhag $600 nm = 6000 \overset{˚}{A}$ expect karo — upar ke kisi bhi matter wave se enormously bada, kyunki photons bahut kam momentum carry karte hain.

Energy in joules $E = 2.0 (1.6 \times 1 0^{- 19}) = 3.2 \times 1 0^{- 19} J$ .
Photon momentum $p = \frac{E}{c} = \frac{3.2 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.067 \times 1 0^{- 27} kg\cdotm/s$ . Yeh step kyun? $m = 0$ mass face block karta hai; $p = E / c$ photon ka proper momentum hai.
Wavelength $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{1.067 \times 1 0 ^{- 27}} = 6.21 \times 1 0^{- 7} m = 621 nm$ .

Verify karo: $621 nm$ orange visible light hai — exactly wahi jo $2 eV$ photons hote hain. ✓ Aur yeh upar ke har matter wavelength ko dwarf karta hai, kyunki tiny $p$ ⇒ long $λ$ . Mass face crash ho jaata; momentum face sail through kar gaya. ✓

Cell I — limiting case: relativistic speed

Worked example Ex 9 · Ek electron

$v = 0.9 c$ pe $λ$ find karo, aur galat (naive) answer se compare karo. Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

Forecast: naive $h / m v$ $p$ ko underestimate karta hai (woh $γ$ stretch drop karta hai), toh $λ$ overestimate karta hai. Isliye correct $λ$ naive se shorter hona chahiye — roughly $γ$ factor se.

Gamma factor $γ = \frac{1}{1 - 0. 9 ^{2}} = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294$ . Yeh step kyun? $0.9 c$ pe hum firmly relativistic hain; $γ$ measure karta hai ki correction kitna bada hai momentum touch karne se pehle.
Speed in SI $v = 0.9 c = 0.9 \times 3 \times 1 0^{8} = 2.7 \times 1 0^{8} m/s$ . Yeh step kyun? Formula ko metres per second chahiye, na " $0.9 c$ ."
Correct momentum $p = γ m_{e} v = (2.294) (9.11 \times 1 0^{- 31}) (2.7 \times 1 0^{8}) = 5.64 \times 1 0^{- 22} kg\cdotm/s$ . Yeh step kyun? Cell I ka face — woh $γ$ stretch insert karo jo naive formula bhool jaati hai.
Correct wavelength $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{5.64 \times 1 0 ^{- 22}} = 1.17 \times 1 0^{- 12} m$ .
Naive (galat) momentum $p_{naive} = m_{e} v = (9.11 \times 1 0^{- 31}) (2.7 \times 1 0^{8}) = 2.46 \times 1 0^{- 22}$ , jo deta hai $λ_{naive} = \frac{6.626 \times 1 0 ^{- 34}}{2.46 \times 1 0 ^{- 22}} = 2.69 \times 1 0^{- 12} m$ . Yeh step kyun? Hum galti deliberately compute karte hain, taaki error ka size dekh sakein.

Verify karo: naive answer ( $2.69 pm$ ) exactly $γ = 2.294$ times too big hai correct $1.17 pm$ se compare karte hue: $2.69/1.17 = 2.30 \approx γ$ . ✓ Relativity ignore karne se tumhara wavelength double ho jaata — pehle hamesha $K$ vs $m c^{2}$ check karo. ✓

Cell J — ulta exam twist

Worked example Ex 10 · Tumne ek electron ka wavelength

$λ = 0.5 \overset{˚}{A}$ measure kiya. Kis voltage $V$ se use accelerate kiya gaya?

Forecast: chhota wavelength ( $0.5 \overset{˚}{A}$ shorter hai $100 V$ pe $1.23 \overset{˚}{A}$ se) matlab high momentum matlab high voltage. $V$ ko $100 V$ se kaafi upar expect karo.

Electron face se shuru karo $λ = \frac{12.27}{V} \overset{˚}{A}$ aur $V$ ke liye solve karo: rearrange karo $V = \frac{12.27}{λ}$ , phir square karo: $V = (\frac{12.27}{λ})^{2}$ jahan $λ$ Å mein hai. Yeh step kyun? Forward formula $V \to λ$ jaati hai; hum simply algebra reverse karte hain taaki $λ \to V$ ho jaye.
Plug in karo $V = (\frac{12.27}{0.5})^{2} = (24.54)^{2} = 602 V$ . Yeh step kyun? Measured $λ$ ko inverted formula mein substitute karo.

Verify karo: $602 V > 100 V$ , jaise forecast ne demand ki thi. Aur forward check karo: $λ = 12.27/ 602 = 12.27/24.5 = 0.50 \overset{˚}{A}$ . ✓ Round trip close hoti hai — hamesha apna answer original formula mein plug back karo. ✓

Recall Kaun sa face use karna hai? (self-test)

Sirf ek temperature diya gaya ho, toh kaunsa face aur pehla step kya hoga? ::: Face D ( $h / 2 m K$ ); pehle $K = \frac{3}{2} k_{B} T$ compute karo. Photon $λ = h / m v$ kyun nahi use kar sakta? ::: Uski mass zero hai, toh $h / (m v) = h /0$ undefined hai; $λ = h / p$ use karo jahan $p = E / c$ . Equal kinetic energy pe, $λ \propto$ ? ::: $1/ m$ (toh electron/proton ratio $1836 \approx 42.8$ hai). Equal speed pe, $λ \propto$ ? ::: $1/ m$ (ratio $1836$ , na ki $1836$ ). $p = m v$ kab fail karta hai? ::: Light speed ke paas; tab $p = γ m v$ jahan $γ = 1/ 1 - v^{2} / c^{2}$ . $λ$ given hone pe electron voltage kaise find karte ho? ::: Face invert karo: $V = (12.27/ λ)^{2}$ jahan $λ$ Å mein ho.

Connections

Parent topic — woh theory jin examples ko yeh exercise karte hain.
Davisson–Germer experiment — Ex 3 ka $54 V$ electron exactly uska setup hai.
Photoelectric effect — photon momentum $p = E / c$ ka origin jo Ex 8 mein use hua.
Compton effect — photon momentum ka ek aur test.
Heisenberg Uncertainty Principle — kyun yeh tiny wavelengths matter karti hain.
Bohr model — standing matter waves orbits ko quantize karte hain.
Wave–particle duality — woh umbrella idea jo sab das cells ke peeche hai.