Pauli exclusion principle: Worked examples — Hinglish deep dive

2.3.16 · D3 · Physics › Modern Physics › Pauli exclusion principle

Yeh page ek workout hai. Pauli exclusion principle parent note ne idea banaya tha (koi bhi do identical fermions ek poora quantum-number set share nahi kar sakte) aur reason bhi (antisymmetry ki wajah se shared-state wavefunction vanish ho jaati hai). Yahan hum us rule ko har tarah ki situation mein run karte hain — taaki jab exam ya koi real problem ek scenario de, tum uski shape pehle se dekh chuke ho.

Pehle kuch bhi karne se: ek atomic electron ka quantum "address" char numbers hote hain $(n, ℓ, m_{ℓ}, m_{s})$ . Inhe street–number–floor–door plus ek "which-way-it-spins" tag ki tarah padho. Pauli ka poora kaam yeh hai: every electron in one atom gets a different one of these four-part addresses. Woh picture yaad rakho; hum baar baar usi pe lean karte hain.

Scenario matrix

Yeh us topic ke cases ka poora space hai jo tumhare saath ho sakta hai. Neeche har example ko us cell ke saath tag kiya gaya hai jo wo cover karta hai, taaki tum grid ko bharta hua dekh sako.

Cell	Scenario class	Jis "edge" ko test karta hai
A	Full shell bharo, capacity count karo	the $2 n^{2}$ machinery, no leftovers
B	Same orbital, spins bachate hain	two electrons legal because $m_{s}$ differs
C	The forbidden move	a repeated address → what breaks
D	Partial subshell (Hund territory)	many valid arrangements, pick the ground one
E	Degenerate / zero input	$ℓ = 0$ (single orbital), $n = 1$ (smallest system)
F	Limiting / large behaviour	big $n$ : how fast capacity grows
G	Non-electron fermions	protons & neutrons obey it too (nucleus, neutron star)
H	Boson contrast (the sign flip)	same math, $+$ sign → no exclusion
I	Real-world word problem	degeneracy pressure holding up a star
J	Exam-style twist	"how many electrons have $m_{s} = + \frac{1}{2}$ ?" counting subsets

Hum saare das cells examples mein cover karte hain.

Example 1 — $n = 3$ shell ko completely bharo · Cell A, E

Worked example Teesre shell mein legally har electron count karo.

Forecast: padhne se pehle number guess karo. Shell 1 mein 2 aye, shell 2 mein 8… shell 3 ke liye tumhara bet kya hai?

Step 1 — Allowed $ℓ$ list karo. $n = 3$ ke liye, rule hai $ℓ = 0, 1, \dots, n - 1 = 0, 1, 2$ . Yeh step kyun? $ℓ$ $n$ se cap hoti hai; orbital "wiggliness" shell se zyada nahi ho sakti. Yahi woh jagah hai jahan degenerate low case $ℓ = 0$ (Cell E) already appear hota hai — yeh ek bilkul valid single orbital hai.

Step 2 — Har $ℓ$ mein orbitals count karo. Har $ℓ$ mein $m_{ℓ} = - ℓ, \dots, + ℓ$ hota hai, jo $(2 ℓ + 1)$ values hai.

$ℓ = 0$ : $m_{ℓ} = 0$ → $1$ orbital.
$ℓ = 1$ : $m_{ℓ} = - 1, 0, + 1$ → $3$ orbitals.
$ℓ = 2$ : $m_{ℓ} = - 2, - 1, 0, + 1, + 2$ → $5$ orbitals. Yeh step kyun? Har alag $(n, ℓ, m_{ℓ})$ ek "parking spot" hai. Pehle spots count karo, phir spin ke liye double karo.

Step 3 — Spin ke liye double karo, phir add karo. Har orbital $2$ electrons hold karta hai ( $m_{s} = \pm \frac{1}{2}$ ). $N = 2 (1 + 3 + 5) = 2 \cdot 9 = 18.$ Yeh step kyun? Factor $2$ exactly woh freedom hai jo Pauli allow karta hai: same orbital, opposite spin.

Verify: formula $N = 2 n^{2}$ deta hai $2 \cdot 3^{2} = 18$ . ✓ Yeh bhi notice karo ki $1 + 3 + 5 = 9$ pehle $3$ odd numbers ka sum $= 3^{2}$ hai — woh identity jo parent note ne use ki thi. Units-check: yeh pure counts hain (dimensionless). ✓

Agla figure addresses ka poora grid bharta hua dikhata hai.

Worked example Ground-state helium (

$Z = 2$ ). Dono electrons sabse low, sabse chhoti state $n = 1$ chahte hain. Kya yeh legal hai? Forecast: kya Pauli doosre electron ko block karta hai, ya andar aane deta hai?

Step 1 — Pehle electron ka full address likho. Lowest energy: $n = 1 \Rightarrow ℓ = 0 \Rightarrow m_{ℓ} = 0$ . Ek spin chuno: $m_{s} = + \frac{1}{2}$ . Address $= (1, 0, 0, + \frac{1}{2})$ . Yeh step kyun? Pauli identical addresses forbid karta hai, isliye poore char numbers likhne zaroori hain, spin include karke — sirf orbital nahi.

Step 2 — Doosre electron ko same orbital mein try karo. Same $n, ℓ, m_{ℓ}$ . Legal rehne ke liye aakhri slot alag honi chahiye, isliye $m_{s} = - \frac{1}{2}$ . Address $= (1, 0, 0, - \frac{1}{2})$ . Yeh step kyun? Bacha hua sirf ek free coordinate spin hai; yeh ek two-value switch hai, isliye exactly ek alternative exist karta hai.

Step 3 — Dono sets compare karo. $(1, 0, 0, + \frac{1}{2}) \neq = (1, 0, 0, - \frac{1}{2})$ — yeh aakhri entry mein differ karte hain. Legal. Yeh step kyun? "Same orbital" matlab "same state" nahi hai. Yeh classic Cell-B rescue hai.

Verify: shell $n = 1$ ki capacity hai $2 n^{2} = 2 \cdot 1^{2} = 2$ . Helium exactly $2$ electrons place karta hai → shell full, koi violation nahi. ✓

Example 3 — Lithium ka teesra electron: forbidden move · Cell C

Worked example Lithium (

$Z = 3$ ). Kya saare teen electrons $n = 1$ mein reh sakte hain? Forecast: electron #3 ko kahan jaana padega?

Step 1 — $n = 1$ ke addresses enumerate karo. Sirf $(1, 0, 0, + \frac{1}{2})$ aur $(1, 0, 0, - \frac{1}{2})$ exist karte hain. Yeh kul do addresses hain. Yeh step kyun? Jab $ℓ$ forced $0$ hai aur $m_{ℓ}$ bhi forced $0$ hai, spin hi akela variable hai — aur uske sirf do values hain.

Step 2 — Electron #3 ko $n = 1$ mein seat karne ki koshish karo. Har $n = 1$ address pehle se use ho chuka hai. Koi bhi choice ek ko repeat karegi → antisymmetric wavefunction collapse ho jaati hai $Ψ = 0$ mein. Yeh step kyun? Yahi Cell C hai: ek duplicate force karna. Parent ka result yaad karo $a = b \Rightarrow Ψ = 0$ — ek zero-probability, yaani impossible, state.

Step 3 — Isse next shell mein promote karo. Electron #3 $n = 2$ mein jaata hai, jisse Li ka configuration $1 s^{2} 2 s^{1}$ banta hai. Yeh step kyun? Kyunki $n = 1$ genuinely full hai, periodic table ki nayi row shuru hoti hai — Pauli shell structure create karta hai.

Verify: $n = 1$ capacity $= 2$ ; Li ke $3$ electrons hain; $3 > 2$ isliye ek zaroor $n = 2$ mein spill hoga. ✓ Lithium sach mein period 2 start karta hai. ✓

Example 4 — Carbon ka $2 p^{2}$ : kaun sa arrangement jeetega? · Cell D

Worked example Carbon (

$Z = 6$ ): $1 s^{2} 2 s^{2} 2 p^{2}$ . Do $2 p$ electrons — same orbital ya alag? Forecast: ek $p$ -orbital mein paired, ya do mein spread?

Step 1 — $2 p$ orbitals list karo. $ℓ = 1 \Rightarrow m_{ℓ} = - 1, 0, + 1$ : teen orbitals, chhah slots. Yeh step kyun? Hume decide karne se pehle legal seats ka menu chahiye.

Step 2 — 2 electrons place karne ke Pauli-legal tarike count karo. Chhah mein se koi bhi do distinct addresses Pauli-allowed hain. Toh $(2 6) = 15$ legal placements hain. Yeh step kyun? Pauli akela saarey pandrah permit karta hai — yeh kisi ek ko single out nahi karta. Yahi Cell D ka subtle point hai.

Step 3 — Hund's rule ground state pick karne do. 15 mein se, lowest energy alag $m_{ℓ}$ orbitals mein electrons rakhti hai parallel spins ke saath, e.g. $(2, 1, - 1, + \frac{1}{2})$ aur $(2, 1, 0, + \frac{1}{2})$ . Yeh step kyun? Pauli legal ki boundary set karta hai; Hund iske andar sabse sasta chunti hai. Note karo Pauli forbid karta hai do parallel-spin electrons ko same $m_{ℓ}$ mein (woh address duplicate karega), isliye "parallel spins" force karta hai "different orbitals" — Pauli aur Hund cooperate karte hain.

Verify: chune gaye do addresses differ karte hain (in $m_{ℓ}$ ), toh legal. ✓ Count check: $(2 6) = 15$ . ✓

Example 5 — Capacity kitni tezi se badhti hai? · Cell F

Worked example Shells

$n = 1$ se $n = 5$ tak ki electron capacity compare karo, aur growth describe karo. Forecast: linear? Doubling? Kuch zyada tez?

Step 1 — $2 n^{2}$ tabulate karo. $2, 8, 18, 32, 50 for n = 1, 2, 3, 4, 5.$ Yeh step kyun? Derived capacity ka direct use — trend dekhne ka sabse saaf tarika.

Step 2 — Differences dekho. Consecutive gaps: $6, 10, 14, 18$ — ek arithmetic sequence jo $4$ se badhti hai. Yeh step kyun? Capacity $n$ mein quadratic hai, isliye first differences linear hain. Yeh batata hai ki door ki shells electrons ko tezi se nigalti jaati hain.

Step 3 — Limiting behaviour state karo. Jaise $n \to \infty$ , capacity $\sim 2 n^{2} \to \infty$ : koi ceiling nahi ki ek (hypothetical) shell kitne electrons hold kar sakti hai. Yeh step kyun? Cell F limits ke baare mein hai — Pauli kabhi addresses khatam nahi karta; woh sirf demand karta hai ki har naaya alag ho.

Verify: $2 \cdot 5^{2} = 50$ ; aur $2 + 8 + 18 + 32 + 50 = 110$ , jo $\sum_{n = 1}^{5} 2 n^{2}$ se match karta hai. ✓ Differences $8 - 2, 18 - 8, 32 - 18, 50 - 32 = 6, 10, 14, 18$ . ✓

Example 6 — Alpha particle vs. deuteron: fermion ya boson? · Cell G, H

Worked example Ek

$α$ -particle (2 protons + 2 neutrons) aur ek akela neutron. Kaun Pauli exclusion follow karta hai, aur kya dono mein se ek state mein pile ho sakte hain? Forecast: "kaun antisocial hai" guess karo padhne se pehle.

Step 1 — Neutron: half-integer spin. Neutron ka spin $\frac{1}{2}$ hota hai. Half-integer ⇒ fermion ⇒ antisymmetric ⇒ Pauli follow karta hai. Do identical neutrons ek state share nahi kar sakte. Yeh step kyun? Spin-statistics theorem spin- $\frac{1}{2}$ ko minus sign se tie karta hai; parent ka $a = b \Rightarrow Ψ = 0$ apply hota hai (Cell G).

Step 2 — $α$ -particle: spins add karo. Char spin- $\frac{1}{2}$ constituents total spin $0$ (ek integer) mein combine ho sakte hain. Integer total spin ⇒ boson. Yeh step kyun? Statistics composite ke total spin ke baad follow karta hai. Even integer matlab symmetric, $+$ sign.

Step 3 — Boson ke liye sharing test karo. Bosons ke liye $Ψ (1, 2) = + Ψ (2, 1)$ ; $a = b$ set karo: $Ψ = \frac{1}{2} [ψ_{a} ψ_{a} + ψ_{a} ψ_{a}] = 2 ψ_{a} ψ_{a} \neq = 0$ . Sharing allowed. Yeh step kyun? Cell H: same algebra, opposite sign — wavefunction survive karti hai, isliye koi exclusion nahi. Isliye $α$ -particles (aur cold atoms) ek state mein crowd kar sakte hain — dekho Bose-Einstein condensate.

Verify: neutron spin $\frac{1}{2}$ (half-integer) ⇒ fermion ✓; $α$ spin $0$ (integer) ⇒ boson ✓. Boson shared amplitude $2 \neq = 0$ . ✓

Example 7 — White dwarf collapse kyun nahi hota · Cell I (word problem)

Worked example Ek white dwarf star nuclei ki ek ball hai jo electrons mein nhayi hui hai, ab fuel nahi jal raha. Gravity andar dabati hai. Pauli ki terms mein, ise kya rok ke rakha hai?

Forecast: kya yeh electric repulsion hai, ya kuch Pauli-flavoured?

Step 1 — Fermions identify karo. Electrons, spin $\frac{1}{2}$ → fermions. Har ek ek distinct address maangta hai. Yeh step kyun? Sirf fermions degeneracy pressure dete hain; pehle players confirm karne zaroori hain.

Step 2 — Gravity unhe same low state mein squeeze karne ki koshish karta hai. Lekin low-energy addresses sab bhar chuke hain. Naye electrons ko higher-momentum (higher-energy) states mein jaana padega. Yeh step kyun? Yahi parent ki key correction hai "Pauli ek force hai" wali galat soch ki: yeh empty states tak chadhne ki energy cost hai, na electromagnetic push. Star compress karna electrons ko ladder pe upar force karta hai — yeh energy cost lagti hai — aur yahi resistance pressure hai.

Step 3 — Balance. Star wahan settle hota hai jahan outward degeneracy pressure inward gravity ke barabar ho. Koi fuel nahi chahiye — Pauli akela support provide karta hai. Yeh step kyun? Cell I microscopic rule ko ek macroscopic object se link karta hai; dekho White dwarf and neutron star. (Neutron stars mein yahi story neutrons ke saath chalti hai — Cell G at stellar scale.)

Verify (qualitative, dimensional): degeneracy pressure filled fermion states se aati hai → exclusion chahiye → fermions chahiye. Bosons koi aisa pressure nahi dete (woh ground state share kar lete aur collapse ho jaate). Consistent. ✓

Example 8 — Exam twist: argon mein spin-up electrons count karo · Cell J

Worked example Argon (

$Z = 18$ ) ek filled-shell atom hai: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ . Iske kitne electrons mein $m_{s} = + \frac{1}{2}$ hai? Forecast: zyaadatar students instinctively "9" bol dete hain — kya yeh sahi hai, aur kyun?

Step 1 — Full subshells samjho. Har orbital jo doubly occupied hai usme exactly ek $+ \frac{1}{2}$ aur ek $- \frac{1}{2}$ electron hota hai (Pauli pair ko spin mein differ karne force karta hai). Yeh step kyun? Cell J test karta hai ki kya tum electron set ko ek single quantum number se slice kar sakte ho.

Step 2 — Kya argon ke saare orbitals full hain? Haan: $2 + 2 + 6 + 2 + 6 = 18$ , aur listed har subshell completely filled hai (koi half-filled Hund case nahi). Yeh step kyun? Spin split exactly even tabhi hoti hai jab orbitals fully paired hon.

Step 3 — Evenly split karo. Jab saare $9$ orbitals doubly occupied hain, exactly aadhe $18$ electrons spin-up hain: $N_{+ 1/2} = \frac{18}{2} = 9.$ Yeh step kyun? Har full orbital ek $+ \frac{1}{2}$ contribute karta hai; $9$ full orbitals hain.

Verify: orbitals $= 1 (1 s) + 1 (2 s) + 3 (2 p) + 1 (3 s) + 3 (3 p) = 9$ ; spin-up count $= 9$ ; spin-down count $= 9$ ; total $= 18 = Z$ . ✓ Instinct "9" yahan sahi hai, lekin sirf isliye kyunki har orbital paired hai — carbon jaisa Hund-partial atom ke liye yeh fail karta.

Active Recall

Recall

$n = 3$ ke liye full-shell capacity Kitne electrons? ::: $2 n^{2} = 18$ .

Recall Lithium ka 3rd electron shell 1 kyun chhod ta hai

Reason? ::: Dono $n = 1$ addresses $(1, 0, 0, \pm \frac{1}{2})$ le liye gaye hain; teesra ek ko duplicate karega, jisse $Ψ = 0$ ho jaayega.

Recall

$2 p$ subshell mein 2 electrons ki legal placements Count (sirf Pauli)? ::: $(2 6) = 15$ ; Hund phir ground state select karta hai.

Recall Kya

$α$ -particle boson hai ya fermion? Answer? ::: Boson — total spin $0$ (integer); yeh Pauli exclusion follow nahi karta.

Recall Filled argon mein spin-up electrons

Count? ::: $9$ (saare $9$ orbitals doubly occupied, evenly split).

Connections

Pauli exclusion principle — parent rule jise yeh examples exercise karte hain.
Quantum numbers — char-part addresses jo hum enumerate karte hain.
Spin and intrinsic angular momentum — $m_{s} = \pm \frac{1}{2}$ doubling.
Hund's rule — Pauli-legal options ke andar ground state pick karta hai (Example 4, 8).
Aufbau principle and electron configuration — Li, C, Ar ke peeche filling order.
Periodic table structure — full shells nayi rows kyun start karti hain (Example 3).
Spin-statistics theorem — fermion vs boson sign (Example 6).
Bose-Einstein condensate — boson world jahan sharing allowed hai.
White dwarf and neutron star — degeneracy pressure (Example 7).