Worked examples — Particle in a box — solving TISE, energy levels, wavefunctions

2.3.10 · D3 · Physics › Modern Physics › Particle in a box — solving TISE, energy levels, wavefunctio

Yahan sab kuch sirf upar ke do boxed results aur $|\psi {|}^2$ ke meaning se hai, jo Wavefunction and Born Rule se liya gaya hai. Koi naya physics nahi — sirf complete coverage.

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Scenario matrix

Solve karne se pehle, list karte hain kya vary ho sakta hai. Ek "particle in a box" problem teen inputs se fix hoti hai — mass $m$, box length $L$, aur level $n$ — plus kya pucha gaya hai (energy, gap, wavefunction, probability, emitted light ki wavelength). Har input "ordinary", "extreme", ya "degenerate" ho sakta hai. Neeche ka table map hai; har cell ka ek example cover kiya gaya hai.

Cell	Isme kya special hai	Covered by
A. Ground state energy	ordinary $m,L$, $n=1$	Example 1
B. Energy gap / photon	difference $E_{n^{\prime }}-E_n$, real emission	Example 2
C. Big quantum number	$n$ large → classical limit	Example 3
D. Probability in a region	$|\psi {|}^2$ integrate karo, edge vs middle	Example 4
E. Probability at a node	degenerate zero-width / density-zero point	Example 5
F. Scaling limits	$L\to$ large (free particle), $L\to$ small (quantum dot)	Example 6
G. Heavy particle	$m$ large → levels vanish, classical	Example 7
H. Real-world word problem	quantum dot colour	Example 8
I. Exam twist	"given energy se $n$ nikalo", inverse problem	Example 9
J. Symmetry shortcut	parity se probability, no integral	Example 10

Do facts hain jo hum baar baar use karenge, toh inhe ek baar clearly state kar lete hain unke units check ke saath:

Upar ka figure poori matrix ek picture mein hai: left mein energy ladder ($n^2$ spacing) on the left, right mein pehle teen wave shapes. Baar baar isko dekha karo — har example neeche is figure ke kisi hisse par finger point karta hai.

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Example 1 — Cell A: ground-state energy

Step 1 — $n$ choose karo. Lowest energy $n=1$ hai (sabse chhota allowed level; $n=0$ matlab no particle). Ye step kyun? "Lowest possible" ka matlab hamesha ground state hota hai, jo box ke liye $n=1$ hai.

Step 2 — $E_n$ mein plug karo. $E_1=\frac{(6.626\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(2.0\times 10^{-10}{)}^2}$ Ye step kyun? Ye ek hi energy formula hai; hum sirf teen inputs substitute karte hain.

Step 3 — Arithmetic. Numerator $=4.390\times 10^{-67}$. Denominator $=8\times 9.11\times 10^{-31}\times 4.0\times 10^{-20}=2.915\times 10^{-49}$. $E_1=1.506\times 10^{-18}\text{J}$

Step 4 — eV mein convert karo. $1.602\times 10^{-19}$ se divide karo: $E_1=9.40\text{eV}$ Ye step kyun? Atomic energies naturally eV mein quote hoti hain; number human-sized ho jaata hai.

Verify: $L$ double karne se $E$, $2^2=4$ se divide honi chahiye. Parent ka $37.6\text{eV}/4=9.40\text{eV}$. ✓ Forecast se exactly match karta hai.

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Example 2 — Cell B: energy gap aur emitted photon

Step 1 — Gap ke liye $n^2$ scaling use karo. $\Delta E=E_3-E_1=(3^2-1^2)E_1=8E_1$ Ye step kyun? Har level $E_1$ times $n^2$ hai, toh differences sirf $(n^{\prime 2}-n^2)$ times $E_1$ hain — poora formula dobara compute karne ki zaroorat nahi.

Step 2 — Numbers. $\Delta E=8\times 9.40=75.2\text{eV}$

Step 3 — Photon wavelength. Ek photon $E_{\gamma }=hc/\lambda$ carry karta hai, toh $\lambda =hc/\Delta E$. Ye step kyun? Energy conservation: lost quantum energy ek photon ban jaati hai. $hc=1240\text{eV⋅nm}$ use karne se units clean rehte hain. $\lambda =\frac{1240\text{eV⋅nm}}{75.2\text{eV}}=16.5\text{nm}$

Verify: $16.5\text{nm}$ extreme-ultraviolet mein hai — "large gap → short wavelength" forecast se match karta hai. Cross-check gap: $E_3=9\times 9.40=84.6$, $E_1=9.40$, difference $75.2$. ✓

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Example 3 — Cell C: classical limit (large $n$)

Step 1 — Gap likho. $E_{n+1}-E_n=[(n+1{)}^2-n^2]E_1=(2n+1)E_1$. Ye step kyun? $(n+1{)}^2-n^2=2n+1$ expand karna "spacings grow" ke peeche ka algebra hai.

Step 2 — Fractional gap banao. $\frac{E_{n+1}-E_n}{E_n}=\frac{(2n+1)E_1}{n^2{E}_1}=\frac{2n+1}{n^2}$ Ye step kyun? "Ladder kitna lumpy lagta hai?" ek relative sawaal hai, toh level se divide karo.

Step 3 — $n=1$ aur $n=100$ par evaluate karo. $n=1:\frac{3}{1}=3.00n=100:\frac{201}{10000}=0.0201$

Verify: fractional gap $300\%$ se lagbhag $2\%$ tak girta hai. Jab $n\to \infty$, $\frac{2n+1}{n^2}\to 0$, toh discrete ladder ek continuum mein blur ho jaata hai — ye correspondence principle hai. ✓

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Example 4 — Cell D: probability in a region

Step 1 — Probability integral likho. $P={\int }_{L/3}^{2L/3}\frac{2}{L}{\sin }^2\left(\frac{\pi x}{L}\right)dx$ Ye step kyun? Probability = density $|\psi {|}^2$ ke neeche ka area, $\psi$ ke neeche nahi — figure mein shaded red hump.

Step 2 — ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$ use karo. $P=\frac{2}{L}{\int }_{L/3}^{2L/3}\frac{1}{2}\left(1-\cos \frac{2\pi x}{L}\right)dx$ Ye step kyun? ${\sin }^2$ directly integrate nahi ho sakta; ye identity ise ek constant plus cosine mein badal deta hai, dono easy hain.

Step 3 — Integrate karo. $P=\frac{1}{L}{\left[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}\right]}_{L/3}^{2L/3}=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }$ (Endpoints dete hain $\sin \frac{4\pi }{3}-\sin \frac{2\pi }{3}=-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=-\sqrt{3}$, aur $-\frac{L}{2\pi }$ factor sign flip karta hai.)

Step 4 — Number. $P=\frac{1}{3}+\frac{\sqrt{3}}{2\pi }\approx 0.333+0.276=0.609$

Verify: $0.609>0.333$, aur ye "hump in the middle" forecast se match karta hai. Sanity check: middle third ($0.609$) plus do edge thirds (parent se, $0.196$ each) deta hai $0.609+2(0.196)=1.001\approx 1$. ✓ Poora box probability 1 hold karta hai.

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Example 5 — Cell E: probability at a node (degenerate zero-width)

Step 1 — Us point par density check karo. ${\psi }_2\left(\frac{L}{2}\right)=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi }{L}\cdot \frac{L}{2}\right)=\sqrt{\frac{2}{L}}\sin \pi =0$ Ye step kyun? $x=L/2$ wahi hai jahan second wave zero cross karta hai — ek interior node ($n=2$ ke liye $n-1=1$ node).

Step 2 — Recall karo ki probability ke liye interval chahiye. $P(\{x=L/2\})={\int }_{L/2}^{L/2}|{\psi }_2{|}^{2}dx=0$ Ye step kyun? Zero-width interval par koi bhi integral $0$ hoga, wave chahe jo bhi ho. Ek point par continuous distribution mein kabhi finite probability nahi hoti.

Verify: do independent reasons $0$ dete hain: (a) density khud node par vanish karti hai, (b) ek single point ki measure zero hoti hai. Dono agree karte hain. Ye woh degenerate case hai jo matrix ne demand kiya tha — "probability at a point" hamesha $0$ hoti hai; sirf regions probability carry karte hain. ✓

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Example 6 — Cell F: scaling limits (box bada/chota hota hai)

Step 1 — $L$ dependence isolate karo. $E_1=\frac{h^2}{8m}\cdot \frac{1}{L^2}$ Ye step kyun? $L$ ke siwa sab kuch fixed constant hai, toh sara behaviour $1/L^2$ mein rehta hai.

Step 2 — Bada box. Jab $L\to \infty$, $\frac{1}{L^2}\to 0$, toh $E_1\to 0$ aur gap $(2n+1)E_1\to 0$: ladder ek continuum ban jaata hai — ek unconfined free particle koi bhi energy le sakta hai. Isliye large regions mein free electrons visibly quantised nahi lagte.

Step 3 — Chota box. Jab $L\to 0$, $E_1\to \infty$: particle ko squeeze karna escalating energy cost karta hai — ye Heisenberg Uncertainty Principle ka fingerprint hai ($\Delta x$ small hone se $\Delta p$, hence energy, large ho jaati hai).

Verify: numerically, $L=1\text{nm}$ se $L=0.1\text{nm}$ jaana (÷10) $E_1$ ko $10^2=100$ se multiply karta hai. Electron ke liye $E_1(1\text{nm})$ hai $0.376\text{eV}$; $\times 100=37.6\text{eV}$, exactly parent ka $0.1\text{nm}$ answer. ✓ Limit trend real hai.

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Example 7 — Cell G: heavy particle (classical object)

Step 1 — Plug in karo. $E_1=\frac{(6.626\times 10^{-34}{)}^2}{8(1.0\times 10^{-9})(1.0\times 10^{-3}{)}^2}$ Ye step kyun? Same single formula — bade objects ke liye physics nahi badalti, sirf numbers.

Step 2 — Arithmetic. Numerator $=4.390\times 10^{-67}$. Denominator $=8\times 10^{-9}\times 10^{-6}=8.0\times 10^{-15}$. $E_1=5.49\times 10^{-53}\text{J}$

Step 3 — Interpret karo. eV mein: $5.49\times 10^{-53}/1.602\times 10^{-19}\approx 3.4\times 10^{-34}\text{eV}$. Ye step kyun? Kisi bhi real energy se compare karne par (thermal $kT\sim 0.025\text{eV}$) pata chalta hai ki level gap $\sim 10^{-32}$ times chota hai — bilkul undetectable.

Verify: classical limit consistent hai — massive, large-scale objects ka $E_1\approx 0$ hota hai aur levels itne dense hain ki merge ho jaate hain. Quantum effects sirf small-$m$, small-$L$ corner mein dikhte hain. ✓

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Example 8 — Cell H: real-world word problem (quantum dot)

Step 1 — $L=5.0\text{nm}$ ke liye ground-state energy. $E_1=\frac{h^2}{8m_e{L}^2}=\frac{(6.626\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(5.0\times 10^{-9}{)}^2}$ Numerator $=4.390\times 10^{-67}$, denominator $=8\times 9.11\times 10^{-31}\times 2.5\times 10^{-17}=1.822\times 10^{-46}$. $E_1=2.409\times 10^{-21}\text{J}=0.01504\text{eV}$ Ye step kyun? Colour gap se set hoti hai, aur gap $E_1$ ka ek multiple hai, toh pehle $E_1$ compute karo.

Step 2 — Gap $2\to 1$. $\Delta E=(2^2-1^2)E_1=3E_1=0.04513\text{eV}$. Ye step kyun? Same $n^2$ trick as Example 2: $4-1=3$.

Step 3 — Wavelength. $\lambda =\frac{hc}{\Delta E}=\frac{1240\text{eV⋅nm}}{0.04513\text{eV}}=2.75\times 10^4\text{nm}$

Verify: $\approx 27,000\text{nm}$ far infrared hai — visible nahi. Ye ek real lesson sikhata hai: 1-D box ek toy hai; real quantum dots 3-D hote hain aur kaafi tighter effective confinement hota hai, isliye woh actually visible mein glow karte hain. Humara number self-consistent hai, aur ye honestly model ki limits flag karta hai. ✓ ($\Delta E$ check: $3\times 0.01504=0.04513\text{eV}$.)

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Example 9 — Cell I: exam twist (inverse problem)

Step 1 — Is box ke liye $E_1$ compute karo. $E_1=\frac{h^2}{8m_e{L}^2}=0.376\text{eV}(\text{as in Example 6, }L=1\text{nm})$ Ye step kyun? Har level $E_1$ ka ek known multiple hai, toh $n$ decode karne ke liye $E_1$ chahiye.

Step 2 — $n^2$ ke liye solve karo. $n^2=\frac{E}{E_1}=\frac{2.256}{0.376}=6.0$ Ye step kyun? $E=n^2{E}_1\Rightarrow n^2=E/E_1$ — pure rearrangement.

Step 3 — Test karo ki $n$ valid integer hai ya nahi. $\sqrt{6.0}\approx 2.449$ — integer nahi. Ye step kyun? Sirf integer $n$ allowed levels hain. Non-integer matlab diya gaya energy genuine box level nahi hai.

Verify: $n=2$ deta hai $4\times 0.376=1.504\text{eV}$; $n=3$ deta hai $9\times 0.376=3.384\text{eV}$. Measured $2.256\text{eV}$ inke beech hai, toh koi level match nahi karta — exam trap ye hai ki ek plausible-looking number forbidden ho sakta hai. Lesson: hamesha confirm karo ki $E/E_1$ ek perfect square hai. ✓

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Example 10 — Cell J: symmetry shortcut (no integral)

Step 1 — Symmetry establish karo. $x\to L-x$ reflect karne par: $\sin \left(\frac{n\pi (L-x)}{L}\right)=\sin \left(n\pi -\frac{n\pi x}{L}\right)=(-1{)}^{n+1}\sin \left(\frac{n\pi x}{L}\right)$ Squaring karne se sign khatam ho jaata hai: $|{\psi }_n(L-x){|}^2=|{\psi }_n(x){|}^2$. Ye step kyun? Density ki symmetry (wave ki nahi) probability ko equally share karti hai.

Step 2 — Symmetry se conclude karo. Mirrored halves par equal density + total probability $1$ ⟹ har half ka $\frac{1}{2}$. Ye step kyun? Koi integral nahi chahiye: symmetry humara accounting kar deti hai.

Verify: $n=1$ ke liye, direct integration ${\int }_0^{L/2}\frac{2}{L}{\sin }^2\frac{\pi x}{L}dx=\frac{1}{2}$. $n=2$ ke liye bhi, similarly $\frac{1}{2}$. Symmetry har $n$ ke liye hold karti hai. ✓

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Recall Scenario self-test

Har question kaunsa matrix cell hai?

1. "Electron $4\to 2$ drop karta hai, photon nikalo." ::: Cell B (energy gap / photon).
2. "$L\to 0$ par $E_1$ ka kya hoga?" ::: Cell F (scaling limits).
3. "$n=1$ particle exactly $x=L/4$ par baithe, probability kya hai." ::: Cell E (point → probability 0).
4. "Ek electron ka $E=1.5\text{eV}$ hai $1\text{nm}$ box mein; kaunsa $n$?" ::: Cell I (inverse; yahan $n=2$ exactly, kyunki $1.504\approx 1.5$).
5. "$n=5$ ke liye right half mein chance." ::: Cell J (symmetry → $\frac{1}{2}$).

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Connections

• Particle in a box — solving TISE, energy levels, wavefunctions — wo parent jo $E_n$ aur ${\psi }_n$ derive karta hai jo poore mein use hue hain.
• Wavefunction and Born Rule — kyun probability $\int |\psi {|}^{2}dx$ hai (Examples 4, 5, 10).
• Heisenberg Uncertainty Principle — "small box → huge energy" limit (Example 6).
• Quantum Dots — real-world word problem (Example 8).
• Quantum Tunnelling and Finite Well — jab walls finite hon toh kya badalta hai.
• Standing Waves on a String — in standing wave patterns ka classical mirror-image.
• Quantum Harmonic Oscillator — contrast: equally spaced levels, box ke $n^2$ ladder se alag.