Quantum numbers n, l, mₗ, mₛ: Worked examples — Hinglish deep dive

2.3.13 · D3 · Physics › Modern Physics › Quantum numbers n, l, mₗ, mₛ

Full derivation ke liye, parent note dekho.

Rules aur woh ek symbol jo aapko chahiye

Is page par sab kuch chaar rules aur ek physical constant par depend karta hai. Hum unhe yahan state kar rahe hain taaki yeh page apne aap mein complete ho.

Definition Core quantum-number rules

Symbol	Matlab	Allowed values
$n$	principal (size/energy)	$1, 2, 3, \dots$
$l$	azimuthal (shape / angular momentum)	$0, 1, 2, \dots, (n - 1)$ — $n$ se ek pehle ruk jaata hai
$m_{l}$	magnetic (orientation)	$- l, - l + 1, \dots, 0, \dots, + l$ — yaani $2 l + 1$ values, negatives bhi include hain
$m_{s}$	spin projection	$+ \frac{1}{2}$ ya $- \frac{1}{2}$ — sirf yahi do

Rule chain yaad karo: $l < n$ , phir $∣ m_{l} ∣ \leq l$ , phir $m_{s} = \pm \frac{1}{2}$ .

Scenario matrix

Is topic ke har question ka ek case class hota hai. Neeche har worked example mein us cell ka tag diya gaya hai jo woh fill karta hai.

Cell	Case class	Kya tricky hai isme
A	Sabse chhota shell, $n = 1$	degenerate: sirf $l = 0$ exist karta hai, koi orbital angular momentum nahi
B	Ek shell ke saare states list karo	$2 (2 l + 1)$ ko saare $l$ par sum karna hoga, result $2 n^{2}$ aayega
C	Angular momentum magnitude	$l (l + 1) ℏ$ , kabhi $l ℏ$ nahi (Bohr trap)
D	Saari orientations $L_{z} = m_{l} ℏ$	negative projections; $L$ ka tilt angle
E	Ek 4-tuple ki legality check	ek rule violate karo ( $l \geq n$ , $
F	Reverse problem: ek fact diya, $(n, l)$ nikalo	e.g. "is shell mein 18 electrons aate hain"
G	Real-world / spectroscopy word problem	subshell notation ↔ quantum numbers
H	Limiting / degenerate spin & energy	$l = 0$ vs large $l$ ; $n$ ke saath energy scaling

Ab hum A–H ko eight examples se cover karte hain.

Worked example Example 1 — Cell A: sabse chhota, sabse degenerate shell (

$n = 1$ ) $n = 1$ shell mein har quantum state list karo aur electron ka orbital angular momentum batao.

Forecast: pehle guess karo — kitne states honge? Ek 1s electron ke liye $∣ L ∣$ kya hoga?

Step 1 — Allowed $l$ nikalo. $l$ runs $0, 1, \dots, (n - 1)$ . $n = 1$ ke saath, sirf $l = 0$ hai. Yeh step kyun? Rule $l \leq n - 1$ master gate hai; $n = 1$ sirf ek value chodta hai, toh koi 1p, 1d, … nahi hai.

Step 2 — Allowed $m_{l}$ nikalo. $l = 0$ ke liye: $m_{l}$ runs $- l, \dots, + l = 0$ . Sirf ek value. Yeh step kyun? $m_{l}$ orientations count karta hai; ek gol ( $l = 0$ ) cloud mein orient karne ke liye kuch hota hi nahi, toh exactly ek.

Step 3 — Spin attach karo. Har $(n, l, m_{l})$ orbital ko $m_{s} = \pm \frac{1}{2}$ milta hai: 2 states. Yeh step kyun? Spin har spatial orbital ko double kar deta hai, chahe $l$ kuch bhi ho.

Step 4 — Orbital angular momentum. $∣ L ∣ = l (l + 1) ℏ = 0 \cdot 1 ℏ = 0.$ Yeh step kyun? Ek s-electron ( $l = 0$ ) ek spherical, non-rotating cloud hai — yeh degenerate case hai.

Verify: $2 n^{2} = 2 (1)^{2} = 2$ states ✓. Do states = helium ke full 1s shell ke do electrons ✓.

Worked example Example 2 — Cell B: ek poora shell enumerate karo (

$n = 3$ ) $n = 3$ shell mein kitne distinct quantum states hain? Subshell ke hisaab se breakdown karo.

Forecast: compute karne se pehle total guess karo.

Step 1 — Allowed $l$ . $l = 0, 1, 2$ (up to $n - 1 = 2$ ). Yeh step kyun? Same master gate; teen subshells: 3s, 3p, 3d.

Step 2 — Har subshell mein $m_{l}$ count karo. $m_{l}$ values ki sankhya $= 2 l + 1$ .

$l = 0$ : $2 (0) + 1 = 1$ orbital
$l = 1$ : $2 (1) + 1 = 3$ orbitals
$l = 2$ : $2 (2) + 1 = 5$ orbitals

Yeh step kyun? $m_{l}$ , $- l$ se $+ l$ tak inclusive span karta hai, jo $2 l + 1$ integers hote hain.

Step 3 — Har ek ko spin ke liye double karo aur sum karo. $2 (1) + 2 (3) + 2 (5) = 2 + 6 + 10 = 18.$ Yeh step kyun? Har orbital mein 2 electrons aa sakte hain (Pauli), toh 2 se multiply karo.

Verify: $2 n^{2} = 2 (3)^{2} = 18$ ✓. Aur $\sum_{l = 0}^{2} 2 (2 l + 1) = 18$ closed form se match karta hai.

Neeche ki figure is arithmetic ko ek picture mein dikhati hai. Har subshell ke liye do bars dekho: blue bar orbital count $2 l + 1$ hai (1, 3, 5) aur yellow bar electron count $2 (2 l + 1)$ hai (2, 6, 10). Notice karo ki yellow bars hamesha exactly blue se double hote hain — yeh "×2" hi Step 3 ka spin doubling hai. Unka green total, 18, wahi $2 n^{2}$ hai jo humne predict kiya tha.

Worked example Example 3 — Cell C: angular-momentum magnitude, Bohr trap se bachte hue

Ek 4f electron ke liye $∣ L ∣$ nikalo. Ise Bohr ke naive guess se compare karo.

Forecast: letter f kis $l$ ko correspond karta hai? $ℏ$ ki units mein $∣ L ∣$ guess karo.

Step 1 — Letter se $l$ read karo. s,p,d,f $\to l = 0, 1, 2, 3$ . Toh 4f ka matlab $l = 3$ hai (aur $n = 4$ ). Yeh step kyun? Subshell letters $l$ ke liye ek fixed dictionary hain.

Step 2 — Sahi formula apply karo. $∣ L ∣ = l (l + 1) ℏ = 3 \cdot 4 ℏ = 12 ℏ = 23 ℏ \approx 3.464 ℏ.$ Yeh step kyun? Schrödinger ka result $l (l + 1)$ hai, na ki $l ℏ$ . Yeh classic exam trap hai.

Step 3 — Dikhao ki Bohr guess fail karta hai. Naive Bohr-style $l ℏ = 3ℏ$ . Lekin sahi value $\approx 3.464ℏ$ hai, jo zyada hai. Yeh step kyun? Gap $l (l + 1) > l$ exactly wahi cheez hai jo $z$ ke saath full alignment impossible banati hai.

Verify: $(23)^{2} = 12 = l (l + 1) = 3 \cdot 4$ ✓. Aur $3.464 > 3$ , toh quantum magnitude largest projection $l ℏ = 3ℏ$ se zyada hai ✓.

Worked example Example 4 — Cell D: saari orientations aur tilt angle

Ek 3d electron ke liye har allowed $L_{z}$ list karo, aur woh sabse chhota angle nikalo jo $L$ z-axis se bana sakta hai.

Forecast: kitni orientations hongi? Kya angle kabhi $0^{\circ}$ ho sakta hai?

Step 1 — $l$ aur $∣ L ∣$ identify karo. 3d $\Rightarrow n = 3, l = 2$ , toh $∣ L ∣ = 2 \cdot 3 ℏ = 6 ℏ \approx 2.449 ℏ.$ Yeh step kyun? Magnitude $l$ se fix hoti hai, orientation ki baat karne se pehle.

Step 2 — $m_{l}$ list karo aur isse $L_{z} = m_{l} ℏ$ nikalo. $m_{l} = - 2, - 1, 0, + 1, + 2$ , jisse milta hai $L_{z} \in {- 2ℏ, - ℏ, 0, + ℏ, + 2ℏ} .$ Yeh step kyun? $2 l + 1 = 5$ orientations; negatives note karo — projections neeche bhi point karte hain.

Step 3 — Sabse chhota tilt angle. Vector sabse kam tab tilts karta hai jab $L_{z}$ sabse bada ho, $L_{z} = + 2ℏ$ . $z$ -axis se angle $α$ satisfy karta hai: $cos α = \frac{L _{z}}{∣ L ∣} = \frac{2ℏ}{6 ℏ} = \frac{2}{6} \approx 0.8165.$ $α = arccos (0.8165) \approx 35. 3^{\circ} .$ Yeh step kyun? $cos α =$ adjacent/hypotenuse — woh right triangle jisme $L_{z}$ (z ke along) aur $∣ L ∣$ (hypotenuse) hain. Kyunki $2/ 6 < 1$ , $α$ kabhi $0$ nahi ho sakta — vector hamesha tilted rehta hai, yeh "space quantization" ka geometric face hai.

Verify: $cos^{2} α = 4/6 = 2/3$ , toh $sin^{2} α = 1/3$ , jisse $L_{x}^{2} + L_{y}^{2} = ∣ L ∣^{2} sin^{2} α = 6 ℏ^{2} \cdot \frac{1}{3} = 2 ℏ^{2} > 0$ ✓ — perpendicular part kabhi vanish nahi hota, jo uncertainty argument se consistent hai.

Yeh figure "hamesha tilted" fact ko visible banati hai. Har coloured arrow 3d electron ke liye $L$ ki ek allowed orientation hai; uski tip white $z$ -axis par marked height $L_{z} = m_{l} ℏ$ par baithti hai. Panchon arrows ki length same hai $6 ℏ$ (magnitude kabhi nahi badlti) — sirf unka tilt alag hai. Top red arrow dekho ( $m_{l} = + 2$ ): yeh "sabse seedha" wala bhi $\approx 35. 3^{\circ}$ jhuka hua hai aur kabhi $z$ ke along flat nahi hota, kyunki uska dashed horizontal projection kabhi zero nahi ho sakta.

Worked example Example 5 — Cell E: legality checks (ek mein chaar traps)

Inn $(n, l, m_{l}, m_{s})$ sets mein se kaun physically allowed hain? (a) $(2, 1, - 1, + \frac{1}{2})$ (b) $(3, 3, 0, - \frac{1}{2})$ (c) $(4, 2, + 3, + \frac{1}{2})$ (d) $(1, 0, 0, + 1)$

Forecast: guess karo kitne charon mein se bachenge.

Step 1 — Har ek ke liye $l < n$ gate check karo. (a) $l = 1 < n = 2$ ✓. (b) $l = 3$ lekin $n = 3$ → chahiye $l \leq 2$ : fail. (c) $l = 2 < 4$ ✓. (d) $l = 0 < 1$ ✓. Yeh step kyun? Yeh pehla, sabse strict rule hai; (b) ko turant kill kar deta hai.

Step 2 — $∣ m_{l} ∣ \leq l$ check karo. (a) $∣ - 1∣ = 1 \leq 1$ ✓. (c) $∣ + 3∣ = 3$ lekin $l = 2$ → chahiye $∣ m_{l} ∣ \leq 2$ : fail. (d) $∣0∣ \leq 0$ ✓. Yeh step kyun? $m_{l}$ ek projection hai; yeh magnitude cap $l$ se zyada nahi ho sakta.

Step 3 — Check karo ki $m_{s} = \pm \frac{1}{2}$ hi ho. (d) mein $m_{s} = + 1$ hai → fail (spin sirf $\pm \frac{1}{2}$ ho sakta hai). (a) $+ \frac{1}{2}$ ✓. Yeh step kyun? Spin ke exactly do values hain; kuch bhi aur unphysical hai.

Step 4 — Verdict. Sirf (a) allowed hai. (b) $l < n$ todata hai, (c) $∣ m_{l} ∣ \leq l$ todata hai, (d) $m_{s}$ todata hai.

Verify: (a) teenon chained rules $l < n$ , $∣ m_{l} ∣ \leq l$ , $m_{s} \in {\pm \frac{1}{2}}$ simultaneously satisfy karta hai ✓ — chaar mein se exactly ek bachta hai.

Worked example Example 7 — Cell G: spectroscopy word problem

Hydrogen mein ek electron $n = 3$ shell se $n = 1$ shell mein drop karta hai. Emitted photon energy nikalo, aur batao kitne distinct upper quantum states ( $n = 3$ ) se yeh start ho sakta tha.

Forecast: energy eV mein guess karo, aur state count Example 2 se.

Step 1 — Har level ki energy. $E_{n} = - \frac{13.6 eV}{n ^{2}}$ . $E_{3} = - \frac{13.6}{9} = - 1.511 eV, E_{1} = - \frac{13.6}{1} = - 13.6 eV .$ Yeh step kyun? Photon energy levels ke difference se aati hai (energy conservation).

Step 2 — Photon energy. $Δ E = E_{3} - E_{1} = - 1.511 - (- 13.6) = 12.09 eV .$ Yeh step kyun? Electron itna lose karta hai; exactly isi energy ka ek photon emit hota hai.

Step 3 — Possible starting states count karo. Example 2 se, $n = 3$ shell mein 18 distinct $(n, l, m_{l}, m_{s})$ states hain. Yeh step kyun? Hydrogen mein energy sirf $n$ par depend karti hai, toh saare 18 states $E_{3}$ share karte hain — unme se koi bhi origin ho sakta hai.

Verify: $Δ E = 13.6 (1 - \frac{1}{9}) = 13.6 \cdot \frac{8}{9} = 12.09 eV$ ✓; units: eV (energy) ✓; state count $2 n^{2} = 18$ ✓.

Worked example Example 8 — Cell H: energy aur spin ka limiting behaviour

(i) Jab $n \to \infty$ , $E_{n}$ ka kya hota hai? (ii) Kya electron ka spin magnitude $∣ S ∣$ kabhi $n$ , $l$ , ya $m_{l}$ par depend karta hai?

Forecast: kya energy $0$ , $- \infty$ , ya kisi finite non-zero value par jaati hai? Kya spin magnitude kabhi change hoti hai?

Step 1 — Energy ka limit. $E_{n} = - \frac{13.6 eV}{n ^{2}} n \to \infty 0^{-} .$ Yeh step kyun? Bada $n$ = loose binding; electron free (ionised) state ke paas pahunch jaata hai jahan $E = 0$ hota hai. Levels crowd together $0$ ki taraf.

Step 2 — Ionisation reading. Ground state $E_{1} = - 13.6$ eV se, $E_{\infty} = 0$ tak pahunchne mein $+ 13.6$ eV lagta hai — yeh hydrogen ki ionisation energy hai. Yeh step kyun? Yeh limit abstract nahi hai; yeh hi ionisation threshold hai.

Step 3 — Spin magnitude. $∣ S ∣ = s (s + 1) ℏ, s = \frac{1}{2} \Rightarrow ∣ S ∣ = \frac{1}{2} \cdot \frac{3}{2} ℏ = \frac{3}{2} ℏ \approx 0.866 ℏ.$ Yeh step kyun? $s = \frac{1}{2}$ electron ka ek fixed intrinsic constant hai — yeh kabhi $n, l, m_{l}$ par depend nahi karta. Sirf projection $m_{s} = \pm \frac{1}{2}$ flip karta hai.

Verify: $(\frac{3}{2})^{2} = \frac{3}{4} = s (s + 1) = \frac{1}{2} \cdot \frac{3}{2}$ ✓; aur $lim_{n \to \infty} (- 13.6/ n^{2}) = 0$ ✓.

Recall Active recall — answers chhupa lo

$n = 3$ shell mein kitne states hain? ::: $2 n^{2} = 18$ . Ek 4f electron ke liye $∣ L ∣$ ? ::: $12 ℏ = 23 ℏ \approx 3.46ℏ$ . Ek 3d electron ke liye $L$ ka sabse chhota tilt angle? ::: $arccos (2/ 6) \approx 35. 3^{\circ}$ . Hydrogen mein $n = 3 \to n = 1$ ke liye photon energy? ::: $12.09$ eV. Kisi bhi electron ka spin magnitude $∣ S ∣$ ? ::: $\frac{3}{2} ℏ \approx 0.866ℏ$ , $n, l, m_{l}$ se independent. $ℏ$ kya hai? ::: Reduced Planck constant $h /2 π \approx 1.055 \times 1 0^{- 34}$ J·s — angular momentum ka natural unit.

See also: Hydrogen Atom Energy Levels · Angular Momentum in Quantum Mechanics · Pauli Exclusion Principle · Electron Configuration & Periodic Table · Stern-Gerlach Experiment · Bohr Model · Schrödinger Equation.

How many distinct quantum states does the

n = 3

shell hold?

18

2 \cdot 3^{2}

For a 4f electron, what is

∣ L ∣

l (l + 1) ℏ = 12 ℏ = 23 ℏ \approx 3.46ℏ

Why can

L

never lie exactly along the

z

-axis?

Because

max L_{z} = l ℏ < l (l + 1) ℏ = ∣ L ∣

, so the smallest tilt angle is nonzero.

Photon energy emitted for hydrogen

n = 3 \to n = 1

E_{3} - E_{1} = 13.6 (1 - 1/9) = 12.09

eV.

Does spin magnitude depend on

n, l, m_{l}

No —

∣ S ∣ = \frac{3}{2} ℏ

always; only

m_{s} = \pm \frac{1}{2}

changes.

What does

ℏ

mean?

Reduced Planck constant

= h /2 π

, the fundamental packet size of angular momentum.