Quantum tunneling — concept, transmission coefficient: Worked examples

2.3.11 · D3 · Physics › Modern Physics › Quantum tunneling — concept, transmission coefficient

Yeh parent note ka worked-example workshop hai. Wahan humne transmission coefficient derive kiya tha. Yahan hum ise use karenge — har tarah ke input par jo yeh formula handle kar sakta hai — thin barriers, thick barriers, heavy particles, woh moment jab $E$ upar $V_{0}$ ki taraf creep karta hai, degenerate width-zero case, aur kuch exam-style twists.

Shuru karne se pehle, teen symbols ko re-anchor karte hain taaki aapko scroll back na karna pade. Ek particle (ek choti si wave) ko imagine karo jo left se aa rahi hai, ek wall se takra rahi hai jiska height $V_{0}$ aur width $L$ hai.

Do formulas jinpar hum poore page lean karenge:

Wall ki picture jinhe hum refer karte rehte hain:

Figure — Quantum tunneling — concept, transmission coefficient

Scenario matrix

Is topic ka har problem exactly inhi cells mein se ek mein land karta hai. Last column us example ka naam deta hai jo ise cover karta hai.

#	Case class	Usmein kya special hai	Covered by
A	Thin barrier, $κ L$ modest	Approximation risky — exact se check karo	Ex 1
B	Thick barrier, $κ L ≫ 1$	Approximation excellent; exponential rules karta hai	Ex 2
C	Width scaling ( $L \to 2 L$ , etc.)	$L$ ke saath exponential sensitivity	Ex 3
D	$E \to V_{0}^{-}$ (energy near top)	$κ \to 0$ : degenerate/limiting; $T \to$ max	Ex 4
E	$L \to 0$ (no barrier)	Degenerate: $T = 1$ recover karna zaroori	Ex 4
F	Heavy particle ( $m$ large)	$κ \propto m$ : tunneling collapse ho jaata hai	Ex 5
G	Height scaling ( $V_{0}$ up)	$V_{0}$ dependence isolate karo	Ex 6
H	Real-world word problem (STM)	Physics translate karo → tiny $Δ L$ , bada $Δ$ current	Ex 7
I	Exam twist: $E > V_{0}$	Formula toot jaata hai; $κ$ imaginary → resonance	Ex 8

Agar koi bhi symbol naya lage toh prerequisites: Schrödinger Equation, Wavefunction and Boundary Conditions, de Broglie Wavelength. Applications yahan hain: Alpha Decay aur Scanning Tunneling Microscope.

Worked examples

Example 1 — Cell A: thin barrier, kya approximation safe hai?

Worked example Electron, thin wall

Electron ( $m = 9.11 \times 1 0^{- 31}$ kg), $E = 1$ eV, $V_{0} = 2$ eV, $L = 0.5$ nm. $T$ dono tarikhon se nikalo aur compare karo.

Forecast: $κ L$ kuch-ek hoga — bahut bada nahi. Guess karo: kya simple $G e^{- 2 κ L}$ exact answer se zyada deta hai ya kam? Aage padhne se pehle apna guess likho.

Step 1 — Energies ko joules mein convert karo. $V_{0} - E = 1 eV = 1.602 \times 1 0^{- 19}$ J. Yeh step kyun? $κ$ ke formula mein SI units use hoti hain; eV se galat metres aayenge.

Step 2 — $κ$ compute karo. $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 1.602 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} \approx 5.12 \times 1 0^{9} m^{- 1}$ Yeh step kyun? $κ$ master quantity hai — decay rate jo sab kuch set karta hai.

Step 3 — Exponent $2 κ L$ . $2 (5.12 \times 1 0^{9}) (0.5 \times 1 0^{- 9}) = 5.12$ . Yeh step kyun? Dono formulas ko $κ L = 2.56$ chahiye; probability $e^{- 2 κ L}$ par ride karti hai.

Step 4 — Approximate $T$ . $G = \frac{16 ( 1 ) ( 1 )}{2 ^{2}} = 4$ , toh $T \approx 4 e^{- 5.12} \approx 4 (0.00598) \approx 0.0239$ , yaani 2.4%. Yeh step kyun? Yahi number ek fast estimate deta hai.

Step 5 — Exact $T$ . $sinh (2.56) = \frac{1}{2} (e^{2.56} - e^{- 2.56}) \approx 6.44$ , toh $T = [1 + \frac{2 ^{2} ( 6.44 ) ^{2}}{4 ( 1 ) ( 1 )}]^{- 1} = [1 + 41.5]^{- 1} \approx 0.0235.$ Yeh step kyun? $κ L = 2.56$ "medium" hai, toh check karna zaroori hai ki " $+ 1$ " drop karna legal tha ya nahi.

Verify: Approximate $0.0239$ vs exact $0.0235$ — $\sim 2%$ ke andar agree. Toh yahan bhi approximation theek hai, aur yeh slightly overshoot karti hai (kyunki usne " $+ 1$ " throw away kiya jo hamesha sirf $T$ ko reduce karta hai). Units: $κ$ ke paas $m^{- 1}$ hai, $L$ ke paas m hai, product dimensionless ✓. $T$ dimensionless aur $< 1$ ✓.

Example 2 — Cell B: thick barrier, exponential dominate karta hai

Worked example Same wall, width teen guna

Electron wahi wala lekin $L = 1.5$ nm. $T$ nikalo.

Forecast: $L$ Ex 1 se teen guna ho gaya. Kya $T$ 3 ke factor se drop hoga? Ek order of magnitude guess karo.

Step 1 — $κ = 5.12 \times 1 0^{9} m^{- 1}$ reuse karo. Yeh step kyun? $κ$ sirf $m, V_{0}, E$ par depend karta hai — inme se kuch nahi badla, toh recompute karne ki zaroorat nahi.

Step 2 — Exponent. $2 κ L = 2 (5.12 \times 1 0^{9}) (1.5 \times 1 0^{- 9}) = 15.36$ . Yeh step kyun? Ab $κ L = 7.68 ≫ 1$ → approximation territory mein deep.

Step 3 — Approximate $T$ . $G = 4$ abhi bhi, toh $T \approx 4 e^{- 15.36} \approx 4 (2.14 \times 1 0^{- 7}) \approx 8.6 \times 1 0^{- 7}$ . Yeh step kyun? Itne bade $κ L$ ke saath " $+ 1$ " bilkul negligible hai, toh approximation hi exact answer hai kaafi digits tak.

Verify: "3× smaller" nahi balki Ex 1 ke $0.024$ se ~28000× chota. Yahi poora lesson hai: $T$ exponentially girta hai, linearly nahi, $L$ ke saath. Check: $0.024/8.6 \times 1 0^{- 7} \approx 2.8 \times 1 0^{4}$ , aur $e^{15.36 - 5.12} = e^{10.24} \approx 2.8 \times 1 0^{4}$ ✓ (ratio purely exponential difference hai).

Example 3 — Cell C: width double karna

Worked example STM sensitivity number

Ex 1 ka setup lo ( $2 κ L = 5.12$ at $L = 0.5$ nm). $L$ double hokar $1.0$ nm ho jaaye toh $T$ kitne factor se change hoga?

Forecast: Factor guess karo. Zyaadatar log "2×" kehte hain. Dekho.

Step 1 — Naya exponent. $L \to 1.0$ nm se $2 κ L = 10.24$ milta hai. Yeh step kyun? Sirf exponent change hota hai; $G$ unchanged rehta hai kyunki $E, V_{0}$ fixed hain.

Step 2 — Do $T$ values ka ratio. $\frac{T ( 1.0 )}{T ( 0.5 )} = \frac{G e ^{- 10.24}}{G e ^{- 5.12}} = e^{- 5.12} \approx 0.00598.$ Yeh step kyun? Divide karne se $G$ completely cancel ho jaata hai — sensitivity pure exponential hai, prefactor se independent.

Step 3 — Invert karo. $1/0.00598 \approx 167$ . Yeh step kyun? Hum chahte hain "kitne times smaller", toh invert karte hain.

Verify: Width double karne se $T$ ~167× suppress hoti hai, 2× nahi. Exactly isliye ek Scanning Tunneling Microscope single atoms resolve karta hai: $0.1$ nm ka height bump tunneling current ko order of magnitude se change kar deta hai. Units ratio mein cancel ho jaate hain ✓.

Example 4 — Cells D & E: do degenerate limits

Ab woh delicate cases jinhe naive user bhool jaata hai: kya hota hai jab barrier disappear hoti hai? Yeh do tarike hain jismein yeh disappear ho sakti hai — wall thin ho jaati hai ( $L \to 0$ ) ya particle ki energy wall ke top tak pahunch jaati hai ( $E \to V_{0}$ ). Approximation $G e^{- 2 κ L}$ illegal hai dono mein (ise $κ L ≫ 1$ chahiye), toh hum exact $sinh$ form use karni padegi. Yahan $sinh$ apni jagah earn karta hai.

Worked example Cell D — energy top ki taraf creep karti hai,

$E \to V_{0}^{-}$ $L$ fix rakho aur $E$ ko $V_{0}$ ke neeche se approach karne do. $T$ kiske paas jaata hai? ( $V_{0} = 2$ eV, $L = 0.5$ nm use karo pehle ki tarah.)

Forecast: Wall barely bahut unchi hai ab. Guess karo: kya $T \to 1$ , ya kuch number 1 se neeche?

Step 1 — $κ$ ko dekho. $κ = 2 m (V_{0} - E) /ℏ \to 0$ jab $E \to V_{0}$ . Yeh step kyun? Height jisko wave fight karti hai, $V_{0} - E$ , zero ho jaati hai — decay rate vanish ho jaata hai.

Step 2 — Correction ka small- $κ L$ form. $κ L$ tiny hone par, $sinh (κ L) \approx κ L$ , toh $sinh^{2} (κ L) \approx κ^{2} L^{2} = \frac{2 m ( V _{0} - E )}{ℏ ^{2}} L^{2}$ . Yeh step kyun? Hum ek " $0/0$ " resolve karna chahte hain — numerator mein $sinh^{2}$ se $(V_{0} - E)$ hai aur denominator mein bhi $(V_{0} - E)$ hai, toh woh cleanly cancel ho jaate hain.

Step 3 — Cancel karo. $\frac{V _{0}^{2} s i n h ^{2} ( κ L )}{4 E ( V _{0} - E )} \approx \frac{V _{0}^{2} \cdot \frac{2 m ( V _{0} - E )}{ℏ ^{2}} L ^{2}}{4 E ( V _{0} - E )} = \frac{V _{0}^{2} m L ^{2}}{2 E ℏ ^{2}} .$ Jab $E \to V_{0}$ : $\to \frac{V _{0} m L ^{2}}{2 ℏ ^{2}}$ . Yeh step kyun? $(V_{0} - E)$ 's cancel hone ke baad yeh finite leftover hai — $T$ blowup nahi hota ya vanish nahi hota.

Step 4 — Number. $\frac{V _{0} m L ^{2}}{2 ℏ ^{2}}$ with $V_{0} = 2 \times 1.602 \times 1 0^{- 19}$ J, $L = 0.5 \times 1 0^{- 9}$ m: $= \frac{( 3.204 \times 1 0 ^{- 19} ) ( 9.11 \times 1 0 ^{- 31} ) ( 0.5 \times 1 0 ^{- 9} ) ^{2}}{2 ( 1.055 \times 1 0 ^{- 34} ) ^{2}} \approx 3.28.$ Toh $T \to [1 + 3.28]^{- 1} \approx 0.234$ , yaani 23%. Yeh step kyun? Wall barely particle se uunchi hone par bhi, is width ke liye transmission sirf ~23% hai — ek striking, non-obvious limiting value.

Verify: $T$ finite rehta hai aur $< 1$ jab $E \to V_{0}$ , jaisa hona chahiye (finite width ki wall phir bhi kuch wave reflect karti hai even jab particle almost clear kar leta hai). Koi division-by-zero blowup nahi — Step 2 mein cancellation hi poora point tha.

Example 5 — Cell F: heavy particle (mass dependence)

Worked example Electron ki jagah Proton

Ex 1 ke numbers repeat karo ( $E = 1$ eV, $V_{0} = 2$ eV, $L = 0.5$ nm) lekin proton ke saath, $m_{p} = 1836 m_{e}$ .

Forecast: Heavier particle. Guess karo: kya $T$ 1836 ke factor se drop hoga, ya bahut zyada?

Step 1 — $κ$ kaise scale karta hai. $κ \propto m$ , toh $κ_{p} = 1836 κ_{e} \approx 42.85 κ_{e}$ . Yeh step kyun? Mass sirf $m$ ke through enter karta hai; scaling scratch se recompute karne se behtar hai.

Step 2 — Naya exponent. Ex 1 se, electron ka $2 κ_{e} L = 5.12$ . Toh $2 κ_{p} L = 42.85 \times 5.12 \approx 219$ . Yeh step kyun? Exponent mein, prefactor mein nahi, mass detonate karta hai.

Step 3 — $T$ estimate karo. $T \approx 4 e^{- 219} \approx 4 \times (2 \times 1 0^{- 96}) \approx 1 0^{- 95}$ . Yeh step kyun? Yeh effectively zero hai — ek proton simply is wall ko tunnel nahi karta.

Verify: $e^{- 219}$ unimaginably small hai. Isliye macroscopic objects kabhi tunnel nahi karte: mass ek exponential ke andar square root ke neeche baitha hai, toh even 1836× mass (abhi bhi tiny!) $T$ ko 2% se $1 0^{- 95}$ tak le jaata hai. Units: ratio mein $m$ scaling dimensionless hai ✓.

Example 6 — Cell G: barrier height badhana

Worked example Uunchi wall

Electron, $E = 1$ eV, $L = 0.5$ nm, lekin $V_{0}$ ko $2$ eV se badhakar $5$ eV karo. $T$ nikalo aur Ex 1 se compare karo.

Forecast: Wall height se zyada se double ho gayi. $T$ ke order of magnitude ka guess karo.

Step 1 — Naya $κ$ . $V_{0} - E = 4$ eV $= 6.408 \times 1 0^{- 19}$ J. $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 6.408 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} \approx 1.024 \times 1 0^{10} m^{- 1} .$ Yeh step kyun? $G$ aur exponent dono $V_{0}$ par depend karte hain; $κ$ change hota hai kyunki $V_{0} - E$ badh gaya.

Step 2 — Exponent. $2 κ L = 2 (1.024 \times 1 0^{10}) (0.5 \times 1 0^{- 9}) = 10.24$ . Yeh step kyun? Note karo yeh Ex 1 ke $5.12$ se double ho gaya kyunki $V_{0} - E$ quadruple hua aur $κ \propto V_{0} - E$ .

Step 3 — Prefactor. $G = \frac{16 ( 1 ) ( 4 )}{5 ^{2}} = \frac{64}{25} = 2.56$ . Yeh step kyun? $G$ shrink ho gaya (4 se) kyunki denominator mein $V_{0}^{2}$ numerator se tezi se badha.

Step 4 — Combine karo. $T \approx 2.56 e^{- 10.24} \approx 2.56 (3.58 \times 1 0^{- 5}) \approx 9.2 \times 1 0^{- 5}$ , yaani 0.009%. Yeh step kyun? Final number deliver karta hai.

Verify: $2.4%$ (Ex 1) se gir ke $0.009%$ — sirf wall uthane se ~260× drop — confirm karta hai parent-note ki warning ki " $T$ sirf width par depend karta hai" galat hai: height utni hi strongly matter karti hai $κ$ ke through.

Example 7 — Cell H: real-world STM word problem

Worked example Atom ko current se read karna

Ek Scanning Tunneling Microscope mein tunneling current $I \propto T \approx G e^{- 2 κ L}$ hai, jahan $L$ tip–surface gap hai. Ek electron ke liye $V_{0} - E \approx 4$ eV (typical work-function gap) ke saath, tip $0.1$ nm kareeb aane par current kitne factor se change hogi?

Forecast: $0.1$ nm ek atom ke size ka fraction hai. Guess karo: kya current kuch percent change hogi, ya kaafi zyada factor se?

Step 1 — $κ$ compute karo. $V_{0} - E = 4$ eV ke saath, Ex 6 se $κ \approx 1.024 \times 1 0^{10} m^{- 1}$ . Yeh step kyun? $κ$ set karta hai ki current gap changes ke liye kitni sharply respond karti hai.

Step 2 — Gap change $Δ L$ ke liye current ratio. $\frac{I ( L - Δ L )}{I ( L )} = e^{- 2 κ (L - Δ L)} / e^{- 2 κ L} = e^{+ 2 κ Δ L} .$ Yeh step kyun? $Δ L$ se kareeb aane par tunnel path chhoti ho jaati hai; prefactor $G$ (same $E, V_{0}$ ) cancel ho jaata hai.

Step 3 — Number. $2 κ Δ L = 2 (1.024 \times 1 0^{10}) (0.1 \times 1 0^{- 9}) = 2.048$ , toh factor $= e^{2.048} \approx 7.75$ . Yeh step kyun? Physics ko ek measurable multiplier mein turn karta hai.

Verify: Sirf $0.1$ nm approach karne se current ~7.8× jump karti hai. Isliye ek STM single-atom bumps feel karta hai: exponential sub-atomic height differences ko large, easily-measured current swings mein turn karta hai. Units: $κ Δ L$ dimensionless ✓, ratio dimensionless ✓.

Example 8 — Cell I: exam twist, $E > V_{0}$ (formula toot jaata hai)

Worked example "Bas

$E > V_{0}$ plug in karo"? Ek student $E = 3$ eV, $V_{0} = 2$ eV (toh $E > V_{0}$ ) tunneling formula mein set karta hai. Kya galat hota hai, aur actually kya hota hai?

Forecast: Particle ke paas wall se zyada energy hai. Classically woh hamesha paar ho jaata hai. Guess karo: kya $T$ exactly 1 hai?

Step 1 — $κ$ ko dekho. $κ = 2 m (V_{0} - E) /ℏ$ with $V_{0} - E = - 1$ eV $< 0$ . Yeh step kyun? Square root ke andar negative number ka matlab hai $κ$ imaginary ho jaata hai — $E < V_{0}$ formula apne domain se bahar hai.

Step 2 — $κ$ ko real wavenumber mein convert karo. Likho $κ = i k^{'}$ with $k^{'} = 2 m (E - V_{0}) /ℏ$ real. Yeh step kyun? Imaginary decay = real oscillation. Barrier ke andar wave ab decay karne ki jagah oscillate karti hai.

Step 3 — $sinh$ ban jaata hai $sin$ . $sinh (i u) = i sin (u)$ use karte hue, toh $sinh^{2} (κ L) = - sin^{2} (k^{'} L)$ : $T = [1 + \frac{V _{0}^{2} s i n ^{2} ( k ^{'} L )}{4 E ( E - V _{0} )}]^{- 1} .$ Yeh step kyun? Yeh correct over-barrier formula hai — aur $L$ mein oscillatory hai.

Step 4 — Resonance. Jab bhi $k^{'} L = nπ$ ( $n = 1, 2, \dots$ ), $sin = 0$ aur $T = 1$ exactly (perfect transmission). Doosri widths par $T < 1$ even though $E > V_{0}$ ! Yeh step kyun? Yeh Ramsauer–Townsend effect hai: barrier ke upar bhi, quantum reflection tab bhi hoti hai except resonant widths par.

Verify: Toh $E > V_{0}$ ke liye $T$ hamesha 1 nahi hota — quantum mechanics kuch wave reflect karta hai even ek chhoti wall ke upar se, $k^{'} L = nπ$ par perfect-transmission windows ke saath. Sanity: $sin = 0$ par, $T = [1 + 0]^{- 1} = 1$ ✓; $sin^{2} = 1$ aur $E = 3, V_{0} = 2$ par, $T = [1 + \frac{4}{4 ( 3 ) ( 1 )}]^{- 1} = [1 + \frac{1}{3}]^{- 1} = 0.75$ , genuinely 1 se kam ✓. Kabhi bhi blindly $E > V_{0}$ ko $sinh$ form mein plug mat karo.

Recall Scenario-matrix self-test

"Covered by" column cover karo aur naam bolo ki kaunsa example har cell handle karta hai. Thin barrier check ::: Ex 1 (approx vs exact ~2% mein agree) Thick barrier ::: Ex 2 (approximation kaafi digits tak exact) Doubling width ::: Ex 3 (~167× drop) $E \to V_{0}$ limit ::: Ex 4 (finite $T \approx 0.23$ , koi blowup nahi) $L \to 0$ limit ::: Ex 4 ( $T \to 1$ ) Heavy particle ::: Ex 5 ( $T \sim 1 0^{- 95}$ ) Raising $V_{0}$ ::: Ex 6 (~260× drop) STM word problem ::: Ex 7 (~7.8× current per 0.1 nm) $E > V_{0}$ twist ::: Ex 8 ( $sinh \to sin$ , resonances)