Quantum tunneling — concept, transmission coefficient: Exercises

2.3.11 · D4 · Physics › Modern Physics › Quantum tunneling — concept, transmission coefficient

Toolkit se pehle ye chaar symbols samjho (har problem mein exactly yahi use hote hain):

Poore chapter mein use hone wale constants (inhe ek sticky note par likh lo):

Intuition Prefactor

$G$ aata kahan se hai — aur kya iske upar trust kiya ja sakta hai? $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}}$ koi fudge factor nahi hai. Ye dono walls par wavefunction aur uski slope ko match karne se nikalta hai ( $x = 0$ aur $x = L$ par) — wo Wavefunction and Boundary Conditions jo Region I ko Region II se aur Region II ko Region III se jodte hain. Physically ye do steps par impedance mismatch measure karta hai: kitna amplitude walls se bounce karke andar/bahar jaane ki bajaye reflect ho jaata hai. Jab $E$ , $V_{0}$ ke close hota hai toh wave aasaani se andar jaati hai aur $G$ apne maximum $4$ ke paas pahunchta hai; jab $E ≪ V_{0}$ hota hai toh walls hard reflect karti hain aur $G$ $0$ ki taraf shrink karta hai. Validity ka domain: compact form $T \approx G e^{- 2 κ L}$ thick-barrier limit hai ( $κ L ≳ 2$ ), jo $sinh (κ L)$ ke growing part ko hi rakhne se milta hai. Kyunki $G$ itna bada $4$ tak ho sakta hai, ek thin barrier ke liye ( $κ L \to 0$ , jahan $e^{- 2 κ L} \to 1$ ) blindly $G e^{- 2 κ L}$ evaluate karne se bakwaas answer $T > 1$ aa sakta hai. Ye signal hai ki aap approximation ke domain se bahar nikal gaye ho — exact $sinh^{2}$ formula par wapas aao, jo hamesha $T \leq 1$ obey karta hai. Neeche ke sabhi problems mein $κ L$ comfortably bada hai, isliye approximation safe hai.

Yeh picture har problem ke peeche hai — kaam karte waqt ise khula rakho. Region I (blue, $x < 0$ ) incoming + reflected wave hai; Region II (red, $0 < x < L$ ) barrier ke andar width $L$ ki exponential decay $e^{- κ x}$ hai (bottom axis ke saath double-headed arrow se marked); Region III (green, $x > L$ ) woh chhoti surviving transmitted wave hai jiska squared amplitude $T$ hai. White dashed line energy $E$ ko mark karti hai jo yellow barrier top $V_{0}$ se neeche baith rahi hai — woh vertical gap, figure mein labelled, $(V_{0} - E)$ hai, aur red curve ki steepness exactly $κ$ hai. Neeche har problem isi figure ke baare mein ek numerical question hai:

Figure — Quantum tunneling — concept, transmission coefficient

Figure 1 — Height $V_{0}$ aur width $L$ ki rectangular barrier jahan $E < V_{0}$ . Blue: Region I mein incoming/reflected wave ( $x < 0$ ). Red: barrier ke across exponential decay $e^{- κ x}$ , Region II ( $0 < x < L$ ). Green: Region III mein transmitted wave ( $x > L$ ), jiska squared amplitude $T$ deta hai. Bottom double arrow barrier width $L$ hai; plateau ke andar vertical arrow energy deficit $V_{0} - E$ hai.

L1 — Recognition

Problem 1.1

Barrier ke andar wavefunction $e^{- κ x}$ ki tarah behave karti hai (Figure 1 mein red curve). Apne words mein batao, $κ$ (kappa) ka physically kya matlab hai, aur iske formula mein kaun si teen quantities ise bada banati hain?

Recall Solution 1.1

$κ$ classically forbidden region ke andar wavefunction ka decay rate hai — Figure 1 mein red curve kitni steeply girti hai. $1/ κ$ distance ke andar $ψ$ ka amplitude $e \approx 2.718$ factor se drop ho jaata hai. Teen formal factors seedhe $κ = \frac{2 m ( V _{0} - E )}{ℏ}$ se padho:

mass $m$ — bhaari particle → bada $κ$ → faster decay → kam tunneling.
energy deficit $(V_{0} - E)$ — particle ki energy se upar taller/steeper barrier → bada $κ$ .
$1/ℏ$ — $κ$ Planck's constant $ℏ$ ke inversely proportional hai. Chhota $ℏ$ bada $κ$ deta hai; hypothetical classical limit $ℏ \to 0$ mein, $κ \to \infty$ aur tunneling bilkul khatam ho jaata hai. Yahi teesra factor hai jis wajah se tunneling purely quantum effect hai.

Bada $κ$ matlab mushkil tunneling, kyunki $T \propto e^{- 2 κ L}$ shrink ho jaata hai.

Problem 1.2

Formula $T \approx G e^{- 2 κ L}$ mein exponent $- 2 κ L$ kyun hai, $- κ L$ kyun nahi?

Recall Solution 1.2

Wavefunction ka amplitude $e^{- κ x}$ ki tarah decay karta hai, isliye poori width $L$ ke across ye $e^{- κ L}$ tak gir jaata hai. Lekin transmission ek probability hai, aur probability $\propto ∣ ψ ∣^{2}$ — tum amplitude ko square karte ho: $∣ e^{- κ L} ∣^{2} = e^{- 2 κ L} .$ 2 ka factor literally $∣ ψ ∣^{2}$ mein square hai.

L2 — Application

Problem 2.1

Ek electron $E = 2 eV$ ke saath height $V_{0} = 5 eV$ aur width $L = 0.4 nm$ ki barrier se takraata hai. $κ$ compute karo.

Recall Solution 2.1

Step 1 (WHAT): $V_{0} - E = 5 - 2 = 3 eV = 3 \times 1.602 \times 1 0^{- 19} = 4.806 \times 1 0^{- 19}$ J nikalo. Step 2 (WHY): $κ$ ko energy deficit $(V_{0} - E)$ chahiye, forbidden region mein kitni gehraai hai. Step 3 (compute): $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 4.806 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}}$ Numerator: $2 \cdot 9.11 \times 1 0^{- 31} \cdot 4.806 \times 1 0^{- 19} = 8.756 \times 1 0^{- 49} = 9.357 \times 1 0^{- 25}$ . $κ = \frac{9.357 \times 1 0 ^{- 25}}{1.055 \times 1 0 ^{- 34}} \approx 8.87 \times 1 0^{9} m^{- 1} .$

Problem 2.2

Problem 2.1 ke $κ$ ka use karke, exponent $2 κ L$ aur prefactor $G$ compute karo, phir $T$ estimate karo.

Recall Solution 2.2

Exponent: $2 κ L = 2 (8.87 \times 1 0^{9}) (0.4 \times 1 0^{- 9}) = 7.10$ (yahan $L = 0.4$ nm barrier width hai). Prefactor: $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}} = \frac{16 ( 2 ) ( 3 )}{5 ^{2}} = \frac{96}{25} = 3.84$ . Ye $4$ ki cap se neeche hai, jaisa expected tha. Kyunki $κ L = 3.55 ≫ 1$ hai, hum safely thick-barrier domain mein hain, isliye $G e^{- 2 κ L} ≪ 1$ hai aur koi $T > 1$ wali pathology nahi aati. Transmission: $T \approx G e^{- 2 κ L} = 3.84 e^{- 7.10} = 3.84 (8.23 \times 1 0^{- 4}) \approx 3.16 \times 1 0^{- 3}$ . Toh $T \approx 3.2 \times 1 0^{- 3} \approx 0.32%$ — chhota lekin nonzero, tunneling ka fingerprint.

L3 — Analysis

Problem 3.1

Problem 2.1 ke electron ke liye ( $κ = 8.87 \times 1 0^{9} m^{- 1}$ , $G = 3.84$ ), barrier width $L = 0.4$ nm se badhaakar $L = 0.8$ nm kar di jaati hai. $T$ kis factor se drop karti hai?

Recall Solution 3.1

$G$ unchanged rehta hai (ye width $L$ par depend nahi karta). Sirf exponential badalta hai: $\frac{T ( 0.8 )}{T ( 0.4 )} = \frac{e ^{- 2 κ (0.8 nm)}}{e ^{- 2 κ (0.4 nm)}} = e^{- 2 κ (0.4 nm)} = e^{- 7.10} .$ $e^{- 7.10} \approx 8.23 \times 1 0^{- 4} \Rightarrow T lagbhag 1215 ke factor se girta hai.$ Samajhna: width double karne se $T$ aadha nahi hua — balki ~1200× kat gaya. Yahi exponential sensitivity hai jis wajah se Scanning Tunneling Microscope single atoms resolve kar sakta hai: sub-nanometre tip-height change current ko orders of magnitude swing kar deta hai.

Problem 3.2

Problem 2.1 ke electron ke liye exact aur approximate formulas compare karo (yaad karo $κ L = 3.55$ ). Kya thick-barrier approximation yahan trustworthy hai?

Recall Solution 3.2

Approximation $sinh (κ L)$ ko $\frac{1}{2} e^{κ L}$ se replace karti hai. $κ L$ ka size check karo: $κ L = 8.87 \times 1 0^{9} \times 0.4 \times 1 0^{- 9} = 3.55.$ Kyunki $κ L = 3.55 ≫ 1$ hai, $sinh (3.55) = 17.31$ jabki $\frac{1}{2} e^{3.55} = 17.34$ — ye $0.2%$ mein agree karte hain. Exact: $T = [1 + \frac{V _{0}^{2} sinh ^{2} ( κ L )}{4 E ( V _{0} - E )}]^{- 1} = [1 + \frac{25 ( 17.31 ) ^{2}}{4 ( 2 ) ( 3 )}]^{- 1} = [1 + 312.1]^{- 1} = 3.19 \times 1 0^{- 3} .$ Approx: $3.16 \times 1 0^{- 3}$ (Problem 2.2). Difference $< 1%$ — approximation excellent hai jab bhi $κ L ≳ 2$ ho. (Agar $κ L$ chhota hota, toh exact formula $T \leq 1$ maintain karta jabki approximation overshoot kar sakti — toolkit note on $G$ dekho.)

L4 — Synthesis

Problem 4.1

Problem 2.1 se sab kuch same rakho (same $E$ , $V_{0}$ , $L$ ) lekin electron ko ek deuteron se replace karo (mass $\approx 3670 m_{e}$ ). $κ$ kis factor se badalta hai, aur roughly naya exponent $2 κ L$ kya hoga? $T$ par comment karo.

Recall Solution 4.1

$κ \propto m$ hai, isliye $\frac{κ _{deut}}{κ _{e}} = 3670 = 60.6.$ Naya exponent: $2 κ L \to 60.6 \times 7.10 \approx 430$ . $T \sim e^{- 430} \approx 1 0^{- 187} .$ Samajhna: deuteron $\sim 3670 \times$ bhaari hai, uska $κ$ $\sim 61 \times$ bada hai, aur transmission ek unimaginably tiny number tak collapse kar jaati hai. Yahi wajah hai ki macroscopic aur nuclear-mass objects bhi ordinary barriers ke across effectively kabhi tunnel nahi karte — seedha Alpha Decay se connect hota hai, jahan tunneling sirf isliye survive karti hai kyunki barrier patla hai aur alpha (bhaari hone ke bawajood) enormous collision rates face karta hai.

Problem 4.2

Do barriers ek hi same transmission $T$ deti hain. Barrier A: $V_{0} - E = 3$ eV, width $L_{A} = 0.4$ nm. Barrier B: $V_{0} - E = 12$ eV. Order-1 prefactor ignore karte hue, wo width $L_{B}$ nikalo jo barrier A ka exponent reproduce kare.

Recall Solution 4.2

Equal $T$ (prefactor aside) ka matlab equal exponents: $κ_{A} L_{A} = κ_{B} L_{B}$ . Kyunki $κ \propto V_{0} - E$ : $\frac{κ _{B}}{κ _{A}} = \frac{12}{3} = 4 = 2.$ Isliye $L_{B} = \frac{κ _{A}}{κ _{B}} L_{A} = \frac{1}{2} (0.4 nm) = 0.2 nm .$ Samajhna: 4× taller barrier ko utna hi leak karne ke liye sirf aadhi width chahiye — kyunki height square root ke neeche aata hai jabki width exponent mein linearly aata hai. Height aur width $V_{0} - E \cdot L = const$ ke through trade off karte hain.

L5 — Mastery

Problem 5.1 (design)

Tum ek Scanning Tunneling Microscope tip bana rahe ho. Tunnel gap ek barrier ki tarah kaam karta hai jahan $V_{0} - E = 4.0$ eV (ek typical metal work function electron energy se upar). Tum chahte ho ki jab tip move kare toh tunnel current $e$ (lagbhag $2.718 \times$ ) ke factor se change ho. Kitna bada vertical tip motion $Δ L$ current mein ek factor of $e$ deta hai? (Current $\propto T \propto e^{- 2 κ L}$ , jahan $L$ tip-to-surface gap width hai.)

Recall Solution 5.1

Current mein $e$ ka factor matlab exponent 1 se change ho: $2 κ Δ L = 1 \Rightarrow Δ L = \frac{1}{2 κ} .$ $κ$ compute karo $V_{0} - E = 4.0 eV = 6.408 \times 1 0^{- 19}$ J ke saath: $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 6.408 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} = \frac{1.0806 \times 1 0 ^{- 24}}{1.055 \times 1 0 ^{- 34}} \approx 1.024 \times 1 0^{10} m^{- 1} .$ $Δ L = \frac{1}{2 ( 1.024 \times 1 0 ^{10} )} = 4.88 \times 1 0^{- 11} m \approx 0.049 nm .$ Samajhna: tip ko sirf ~0.05 nm — ek atom ke diameter ke fraction — hilane se current poore $e$ ke factor se change ho jaata hai. Yahi extreme sensitivity STM ka poora operating principle hai: ye ek aisa ruler hai jo atoms read kar sakta hai kyunki tunneling current gap width mein exponential hai.

Problem 5.2 (reason backwards)

Ek experiment $T = 1.0 \times 1 0^{- 5}$ measure karta hai electrons ke liye jo width $L = 1.0$ nm ki barrier cross karte hain. Prefactor $G \approx 4$ lete hue, energy deficit $V_{0} - E$ (eV mein) estimate karo.

Recall Solution 5.2

Step 1 — exponent isolate karo. $T \approx G e^{- 2 κ L}$ se: $e^{- 2 κ L} = \frac{T}{G} = \frac{1 0 ^{- 5}}{4} = 2.5 \times 1 0^{- 6} .$ $- 2 κ L = ln (2.5 \times 1 0^{- 6}) = - 12.90 \Rightarrow 2 κ L = 12.90.$ Step 2 — $κ$ solve karo. Width $L = 1.0 \times 1 0^{- 9}$ m ke saath: $κ = \frac{12.90}{2 ( 1.0 \times 1 0 ^{- 9} )} = 6.45 \times 1 0^{9} m^{- 1} .$ Step 3 — energy deficit ke liye $κ$ invert karo. $κ = 2 m (V_{0} - E) /ℏ$ se: $V_{0} - E = \frac{( ℏ κ ) ^{2}}{2 m} = \frac{( 1.055 \times 1 0 ^{- 34} \cdot 6.45 \times 1 0 ^{9} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} .$ Numerator: $(6.805 \times 1 0^{- 25})^{2} = 4.63 \times 1 0^{- 49}$ . $V_{0} - E = \frac{4.63 \times 1 0 ^{- 49}}{1.822 \times 1 0 ^{- 30}} = 2.54 \times 1 0^{- 19} J = 1.59 eV .$ Toh barrier electron energy se lagbhag 1.6 eV upar hai — ek plausible tunnel-junction value.

Recall Master checklist (memory se fill karo)

$κ$ ki units? ::: inverse length, m $^{- 1}$
$L$ kise represent karta hai? ::: the barrier width — forbidden region ki horizontal thickness jiske across wave decay karti hai
$κ$ compute karne se pehle energies ___ mein convert karo? ::: joules
$T$ , $L$ ke saath kaise scale karta hai? ::: exponentially, $e^{- 2 κ L}$ ki tarah
$κ$ , mass ke saath kaise scale karta hai? ::: $m$ ki tarah
$T$ fixed rakhne ke liye height–width trade-off? ::: $V_{0} - E \cdot L = const$
STM current mein ek factor of $e$ ke liye $Δ L =$ kya chahiye? ::: $1/ (2 κ)$
Prefactor $G$ kahan se aata hai, aur ise kya cap karta hai? ::: dono walls par boundary-condition matching se; ye $4$ par cap hai (jo tab reach hota hai jab $E \to V_{0}$ )

Related build-up notes: Wavefunction and Boundary Conditions, Potential Barrier and Reflection, Schrödinger Equation, de Broglie Wavelength.