Hydrogen energy levels Eₙ = −13.6 - n² eV: Worked examples

2.3.14 · D3 · Physics › Modern Physics › Hydrogen energy levels Eₙ = −13.6 - n² eV

Scenario matrix

Kuch bhi solve karne se pehle, case-classes ka poora set list karte hain. Har exam question inhi cells mein se ek hota hai (ya inhi ka combination). Neeche har worked example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Case class	Tricky kyun hai	Example
A	Single level, "shelf $n$ kitni gehri hai?"	bas $n^{2}$ se divide karo	Ex 1
B	Excited state se ionization (ground nahi)	$E = 0$ tak jaana hai, $n = 1$ tak nahi	Ex 2
C	Emission (electron girta hai) → photon bahar	do levels ka difference, wavelength	Ex 3
D	Absorption (electron chadhta hai) → photon andar	same $Δ E$ , lekin sign/direction ulta	Ex 4
E	Limiting case $n \to \infty$ (series limit)	$1/ n^{2} \to 0$ boundary	Ex 5
F	Degenerate / "impossible" input	$n = 0$ , non-integer, ya downward "emission" jo ho nahi sakti	Ex 6
G	One-electron ion, hidden $Z^{2}$	formula pure hydrogen wala nahi hai	Ex 7
H	Real-world word problem	ek story se physics translate karo	Ex 8
I	Exam twist: wavelength di hai, transition dhundho	machine ko ulta chalao	Ex 9

Do constants jo hum har jagah reuse karte hain:

Neeche diya figure woh map hai jis par hum baar baar point karte rahenge. Ise apne dimag mein pin kar lo.

Figure — Hydrogen energy levels Eₙ = −13.6 - n² eV

Worked examples

Cell A — ek level

Worked example Ex 1 ·

$n = 4$ shelf kitni gehri hai? Hydrogen ke liye $E_{4}$ nikalo, aur batao yeh ground state se kitna upar hai.

Forecast: Pehle guess karo — kya $E_{4}$ , $- 13.6$ eV ke zyada kareeb hai ya $0$ ke? (Upar wali shelves zero ke paas bheed lagati hain, toh… $0$ ke zyada kareeb.)

$n^{2}$ se divide karo. $E_{4} = - \frac{13.6}{4 ^{2}} = - \frac{13.6}{16}$ . Yeh step kyun? Formula kehta hai har shelf ki gehraai basement depth $13.6$ eV hai jo $1/ n^{2}$ factor se chhoti ho jaati hai. Bada $n$ ⇒ chhota fraction ⇒ shallower shelf.
Compute karo. $- \frac{13.6}{16} = - 0.85 eV$ .
Ground se upar kitna. $E_{4} - E_{1} = - 0.85 - (- 13.6) = 12.75 eV$ . Yeh step kyun? "Ground state se upar" matlab sabse gehri shelf $E_{1} = - 13.6$ eV se difference.

Verify: $- 0.85$ , $- 13.6$ se kahin zyada $0$ ke kareeb hai — hamara forecast match karta hai. Units: poora eV mein. ✓

Cell B — excited state se ionization

Worked example Ex 2 ·

$n = 2$ mein baitha hydrogen ionize karna Ek atom pehle excited state ( $n = 2$ ) mein hai. Electron ko poori tarah free karne ke liye photon ko kitni energy supply karni padegi? Ground state se ionize karne se compare karo (woh famous $13.6$ eV — dekho Ionization energy).

Forecast: Kya ise $13.6$ eV se zyada chahiye ya kam? Electron pehle se well mein upar hai, toh kam chahiye hoga.

Free matlab $E = 0$ . Energy chahiye $= 0 - E_{2}$ . Yeh step kyun? Ionization = well ke bilkul top tak chadhna, jahan electron rest mein hai aur infinitely door ( $E = 0$ ). Yeh reference humne parent note mein build kiya tha.
$E_{2}$ nikalo. $E_{2} = - \frac{13.6}{2 ^{2}} = - 3.40 eV$ .
Subtract karo. $Δ E = 0 - (- 3.40) = 3.40 eV$ .

Verify: $3.40 < 13.6$ — ground state se kam chahiye, jaise forecast kiya tha. Yeh exactly $n = 2$ shelf ki gehraai hai, aur yahi baat hai. ✓

Cell C — emission, drop, photon bahar

Worked example Ex 3 ·

$n = 4 \to n = 2$ Balmer line Electron $n = 4$ se $n = 2$ par girta hai. Emitted photon ki energy aur wavelength nikalo. Kya yeh visible hai?

Forecast: $n = 2$ par khatam hona ek Balmer transition hai, aur Balmer lines visible hoti hain. Guess: kuch eV, kuch hundred nm.

Transition energy. $Δ E = 13.6 (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}) = 13.6 (\frac{1}{4} - \frac{1}{16})$ . Yeh step kyun? Photon do shelves ka difference carry karta hai, ek shelf ki value nahi. Yahan $n_{f} = 2$ (neeche, toh uska $1/ n^{2}$ bada term hai ⇒ $Δ E > 0$ , energy release hoti hai).
Arithmetic. $\frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16}$ , toh $Δ E = 13.6 \times \frac{3}{16} = 2.55 eV$ .
Wavelength. $λ = \frac{h c}{Δ E} = \frac{1240}{2.55} \approx 486 nm$ . Yeh step kyun? Release hui energy light ban jaati hai; $h c = 1240$ eV·nm ek division mein eV ko nm mein convert karta hai.

Verify: $486$ nm blue-green hai — visible band (400–700 nm) ke andar, ek real Balmer line ( $H β$ ). Forecast match karta hai. ✓

Cell D — absorption, chadhna, photon andar

Worked example Ex 4 · Absorb karke upar jaana:

$n = 1 \to n = 3$ Ek ground-state atom ek photon absorb karta hai aur $n = 3$ par jump karta hai. Ise kaunsi photon energy aur wavelength chahiye? Kaunsi spectral series?

Forecast: $n = 1$ wale jumps sabse bade gaps hote hain → high energy → short wavelength → ultraviolet (Lyman).

Same $Δ E$ machine. Atom ko exactly shelf difference gain karna padega: $Δ E = E_{3} - E_{1} = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) .$ Yeh step kyun? Absorption, emission ka ulta hai. Photon ko precisely gap $E_{3} - E_{1}$ deliver karna padega; kuch bhi aur aur atom simply absorb nahi karega. Discrete shelves ⇒ colour ke baare mein choosy hota hai.
Arithmetic. $1 - \frac{1}{9} = \frac{8}{9}$ , toh $Δ E = 13.6 \times \frac{8}{9} = 12.09 eV$ .
Wavelength. $λ = \frac{1240}{12.09} \approx 102.6 nm$ .

Verify: $102.6$ nm deep UV hai — ek Lyman line, $n = 1$ par khatam/shuru hoti hai, bahut badi energy. Forecast match karta hai. Note karo $12.09 < 13.6$ : ionize karne ke liye kaafi nahi, exactly $n = 3$ tak pahunchne ke liye kaafi. ✓

Cell E — limiting case $n \to \infty$

Worked example Ex 5 · Series limit (jahan lines pile up karti hain)

Jaise $n_{f} = 2$ par khatam hone wale transitions ke liye $n_{i} \to \infty$ hota hai, Balmer lines kaunsi wavelength approach karti hain? Yeh "series limit" woh jagah hai jahan discrete lines ek continuum mein crowd ho jaati hain.

Forecast: Aur aur upar ki shelves $E = 0$ ke aur kareeb hoti jaati hain. Toh $Δ E$ ek fixed ceiling approach karta hai — $n_{f} = 2$ shelf ki poori gehraai — jo ek shortest Balmer wavelength deta hai.

Limit lo. Jaise $n_{i} \to \infty$ , $\frac{1}{n _{i}^{2}} \to 0$ , toh $Δ E \to 13.6 (\frac{1}{2 ^{2}} - 0) = \frac{13.6}{4} = 3.40 eV .$ Yeh step kyun? Starting shelf well ke top ( $E = 0$ ) mein pighal jaati hai, toh photon simply $n = 2$ shelf ki poori gehraai le jaata hai — jo $3.40$ eV hai (Ex 2 se compare karo!).
Wavelength. $λ_{min} = \frac{1240}{3.40} \approx 365 nm$ .

Verify: $365$ nm jaana-maana Balmer series limit hai (violet ke theek baad, near-UV mein). Kyunki yeh is series ke liye sabse bada possible $Δ E$ use karta hai, yeh sabse chhoti wavelength hai — line pile-up ka edge. Consistency check: same $3.40$ eV jaisa $n = 2$ ionization energy, jaise hona chahiye. ✓

Cell F — degenerate / impossible inputs

Worked example Ex 6 · Jab formula answer dene se mana kar de

(a) $E_{0}$ kya hai? (b) Kya electron $n = 2$ se $n = 3$ jaate hue photon emit kar sakta hai?

Forecast: Dono "off" lagte hain. $n = 0$ ek real shelf nahi hai; aur $2 \to 3$ chadhna hai, girna nahi — upar jaate hue emit nahi kar sakte.

(a) $n = 0$ try karo. $E_{0} = - \frac{13.6}{0 ^{2}}$ — zero se division, undefined. Yeh step kyun? Bohr quantization tha $m v r = n ℏ$ with $n = 1, 2, 3, \dots$ . Koi $n = 0$ orbit nahi hoti (zero angular momentum matlab electron seedha proton mein girna). Formula ka domain simply $n = 1$ par shuru hota hai.
(b) $2 \to 3$ ke liye $Δ E$ compute karo. $Δ E = E_{2} - E_{3} = - 3.40 - (- 1.51) = - 1.89 eV$ . Yeh step kyun? Negative $Δ E$ matlab atom ko energy release nahi karni padegi, gain karni padegi. Toh $2 \to 3$ absorption hai, spontaneous emission ke roop mein impossible hai. Emit karne ke liye, tumhe $n_{i} > n_{f}$ chahiye (girna), jo $Δ E > 0$ deta hai.

Verify: (a) $1/0$ genuinely undefined ⇒ $n = 0$ excluded, sahi. (b) $∣ Δ E ∣ = 1.89$ eV same size hai jitna $3 \to 2$ emission (sirf opposite sign) — ladder energy mein symmetric hai, sirf direction flip hota hai. ✓

Cell G — one-electron ion, hidden $Z^{2}$

Worked example Ex 7 · He

$^{+}$ ground state Singly-ionized helium (He $^{+}$ ) mein ek electron hai lekin nuclear charge $Z = 2$ hai. Iska ground-state energy aur ionization energy nikalo.

Forecast: Do guna nuclear charge electron ko bahut zyada kheenchti hai. Guess: hydrogen se kahin zyada tightly bound — $Z^{2} = 4$ ke factor se.

$Z^{2}$ form use karo. One-electron ion ke liye, $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ eV. Yeh step kyun? Coulomb force ko $e$ ki jagah $Z e$ charge ke saath redo karo: derivation mein $e^{2}$ ban jaata hai $Z e^{2}$ , aur yeh energy mein squared enter hua, toh $Z^{2}$ factor aata hai. (Dekho Rydberg constant and formula.)
Ground state. $n = 1, Z = 2$ : $E_{1} = - 13.6 \times \frac{4}{1} = - 54.4 eV$ .
Ionization energy. $0 - (- 54.4) = 54.4 eV$ .

Verify: $54.4 = 4 \times 13.6$ — exactly $Z^{2} = 4$ boost jaise forecast kiya tha. He $^{+}$ ko H se kahin zyada mushkil se ionize kiya ja sakta hai. ✓

Cell H — real-world word problem

Worked example Ex 8 · Ek hydrogen lamp aur ek solar cell

Ek hydrogen discharge lamp apni $H α$ red line $656$ nm par emit karta hai. Ek student check karna chahta hai ki kya ye photons kisi metal se electron free kar sakti hain jiska "work function" (electron eject karne ke liye minimum energy) $2.5$ eV hai. Kya red photons kaam karenge?

Forecast: Red light kaafi low-energy hoti hai. Work function $2.5$ eV ek decent barrier hai. Guess: probably kaafi nahi.

Wavelength se photon energy. $E_{photon} = \frac{h c}{λ} = \frac{1240}{656} \approx 1.89 eV$ . Yeh step kyun? Humhe ek colour ( $λ$ ) diya gaya hai; colour aur energy ke beech physics lever $E = h c / λ$ hai. Yahi $1.89$ eV hai jo pehli jagah line ( $3 \to 2$ ) banata hai.
Barrier se compare karo. $1.89 eV < 2.5 eV$ . Yeh step kyun? Ek electron tab hi eject hota hai jab ek single photon kam se kam work function ke barabar energy carry kare (photoelectric threshold — ek photon, ek electron).

Verify: $1.89 < 2.5$ , toh koi electron eject nahi hoga — forecast match karta hai. Zyada bright lamp (zyada photons) bhi help nahi karegi; tumhe bluer (higher-energy) photons chahiye honge, jaise Ex 3 ka $n = 4 \to 2$ line $2.55$ eV par, jo ise just clear kar deta. ✓

Cell I — exam twist: $λ$ diya hai, transition dhundho

Worked example Ex 9 · Ek spectral line reverse-engineer karna

Hydrogen ki ek line $λ = 1216 \overset{˚}{A} = 121.6 nm$ par measure ki gayi. Transition identify karo ( $n_{i} \to n_{f}$ ).

Forecast: $121.6$ nm deep UV hai — bahut high energy — toh yeh zaroor $n_{f} = 1$ (Lyman) involve karta hai. Probably sabse chhota aisa jump, $2 \to 1$ .

Wavelength → photon energy. $E_{photon} = \frac{1240}{121.6} \approx 10.2 eV$ . Yeh step kyun? Photon rule ulta chalao: colour se hum woh energy recover karte hain jo transition ne release ki.
Level gap se match karo. Humhe chahiye $13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) = 10.2$ . Try karo $n_{f} = 1, n_{i} = 2$ : $13.6 (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} = 10.2 eV . ✓$ Yeh step kyun? High energy ⇒ $n = 1$ par khatam hota hai; pehle sabse chhota Lyman jump test karo. Exactly land karta hai.
Conclusion. Transition hai $n_{i} = 2 \to n_{f} = 1$ — Lyman- $α$ line.

Verify: $10.2$ eV se waapas wavelength: $1240/10.2 \approx 121.6$ nm — perfectly round-trip karta hai. UV, Lyman, jaise forecast kiya tha. ✓

Recall Matrix ko memory se rebuild karo

Table cover karo. Har cell ke liye, ek idea batao: A single level ::: $13.6$ ko $n^{2}$ se divide karo. B level $n$ se ionization ::: cost hai $13.6/ n^{2}$ , yani $E = 0$ tak chadhna, $n = 1$ tak nahi. C emission ::: girna, $n_{i} > n_{f}$ , $Δ E > 0$ , photon bahar. D absorption ::: chadhna, $n_{i} < n_{f}$ , atom exactly gap gain karta hai. E limit $n_{i} \to \infty$ ::: $1/ n_{i}^{2} \to 0$ , series-limit (shortest) wavelength deta hai. F degenerate ::: $n = 0$ undefined; upward "emission" impossible ( $Δ E < 0$ ). G one-electron ion ::: $Z^{2}$ se multiply karo. I reverse ::: $λ$ se $E = 1240/ λ$ nikalo, phir level gap se match karo.

Connections

2.3.14 Hydrogen energy levels Eₙ = −13.6 - n² eV (Hinglish) — parent topic
Photon energy E = hf = hc/λ — $λ = h c /Δ E$ tool jo Ex 3, 4, 5, 8, 9 mein use hua
Hydrogen spectral series — Lyman/Balmer naming poore mein use hue
Ionization energy — Cells B aur G
Rydberg constant and formula — Ex 7 mein $Z^{2}$ generalisation
Bohr model of the atom — $n$ kyun $1$ se shuru hota hai (Ex 6)
Coulomb's law — $Z^{2}$ factor ka origin
Quantum numbers — $n$ ka matlab