Exercises — Quantum numbers n, l, mₗ, mₛ

KYA: hum $l$ ko rule $l=0,1,\dots ,(n-1)$ se list karte hain. KYU: $l$ ek $n$ se kam par rukta hai kyunki radial wave ko bound state hone ke liye kam se kam ek unit ka "node budget" chahiye. $n=4$ ke saath: $l=0,1,2,3$ — yahi s, p, d, f sub-shells hain. Char values.

$m_l$ integer steps mein $-l$ se $+l$ tak jaata hai: $m_l=-3,-2,-1,0,+1,+2,+3$. Count $=2l+1=2(3)+1=\boxed{7}$ values. Negatives kyun aate hain: $m_l$ ek vector ka $z$-axis par projection $L_z/\hslash$ hai — ek projection neeche (negative) ki taraf utni hi aasaani se point kar sakta hai jitna upar ki taraf.

$m_s=+\frac{1}{2}$ aur $m_s=-\frac{1}{2}$. Yeh Schrödinger equation se nahi aate. Inhe humpar Stern-Gerlach Experiment ne force kiya (silver atoms ki ek beam exactly do spots mein split hui) aur baad mein Dirac ki relativistic equation ne inhe derive kiya. Dekho Angular Momentum in Quantum Mechanics.

KYA: $l=0,1,2$ par $2(2l+1)$ ko sum karo. $l=0$: $2(1)=2$. $l=1$: $2(3)=6$. $l=2$: $2(5)=10$. Total $=2+6+10=18$. Check: $2n^2=2(3{)}^2=18$. ✓ Factor 2 kyun: har $(n,l,m_l)$ orbital do electrons rakhta hai, ek $m_s=+\frac{1}{2}$ aur ek $m_s=-\frac{1}{2}$ (Pauli Exclusion Principle).

$4f\Rightarrow n=4,l=3$. $|\vec{L}|=\sqrt{l(l+1)}\hslash =\sqrt{3\cdot 4}\hslash =\sqrt{12}\hslash =2\sqrt{3}\hslash \approx 3.464\hslash .$ $\sqrt{l(l+1)}$ kyun, $l$ kyun nahi: Bohr ka purana $L=l\hslash$ galat hai (woh parent mein Mistake B hai). Eigenvalue equation solve karne par $|\vec{L}{|}^2=l(l+1){\hslash }^2$ milta hai.

$3d\Rightarrow l=2$, toh $m_l=-2,-1,0,+1,+2$. $L_z=m_l\hslash$, toh $L_z\in \{-2\hslash ,-\hslash ,0,+\hslash ,+2\hslash \}$. Note karo ki sabse bada $L_z=2\hslash$, $|\vec{L}|=\sqrt{6}\hslash \approx 2.449\hslash$ se chota hai — vector kabhi poori tarah $z$ ke saath align nahi hota (space quantization). Neeche wali figure dekho.

(a) Invalid. $l=2$ hai lekin maximum $n-1=1$ hai. Broken rule: $l\le n-1$. (b) Valid. $l=2\le 2$ ✓, $m_l=-2$, $[-2,2]$ mein hai ✓, $m_s=-\frac{1}{2}$ ✓. (c) Invalid. $l=0\Rightarrow m_l$ sirf $0$ ho sakta hai, lekin yahan $m_l=-1$ hai. Broken rule: $|m_l|\le l$. (d) Invalid. $m_s$ $\pm \frac{1}{2}$ hona chahiye, kabhi $0$ nahi. Yeh caps kyun exist karte hain: $l\le n-1$ aur $|m_l|\le l$ dono wavefunction ko blow up hone se rokne ki demand se aate hain (respectively $r\to \infty$ par aur poles $\theta =0,\pi$ par) — Schrödinger's equation solve karne ke boundary conditions.

KYA: $\sqrt{l(l+1)}=\sqrt{6}$ solve karo, toh $l(l+1)=6\Rightarrow l=2$. Sabse chhota $n$ kyun: humein $l\le n-1$ chahiye, yaani $n\ge l+1=3$. Sabse chhota $n=3$ hai. $l=2$ d sub-shell hai, toh jawab $3d$ hai.

Yeh formula kyun: $L_z$ adjacent side hai ($z$ par projection) aur $|\vec{L}|$ hypotenuse hai us right triangle ka jo $\vec{L}$ aur $z$-axis se banta hai — figure dekho. Toh $\cos \theta =\text{adjacent}/\text{hypotenuse}=L_z/|\vec{L}|$. $\cos \theta =\frac{m_l\hslash }{\sqrt{l(l+1)}\hslash }=\frac{2}{\sqrt{6}}\approx \mathrm{0.8165.}$ $\theta =\arccos (0.8165)\approx 35.26^{\circ }.$ Sanity check: $\theta \ne 0$, yeh confirm karta hai ki vector maximum $m_l$ par bhi exactly axis par nahi ho sakta.

KYA: $n=3,l=1$ fix karo ($3p$ sub-shell). Baaki free labels: $m_l$ aur $m_s$. $m_l\in \{-1,0,+1\}$ → 3 orbitals. Har ek 2 spins rakhta hai. $3\times 2=\boxed{6}$ electrons. KYU: Pauli ke anusaar, 6 distinct $(m_l,m_s)$ combinations mein se har ek ek unique seat hai.

$1s\Rightarrow n=1,l=0,m_l=0$ dono electrons ke liye. Electron A: $(1,0,0,+\frac{1}{2})$. Electron B: $(1,0,0,-\frac{1}{2})$. Sirf $m_s$ alag hai. Agar yeh bhi nahi hota, toh saare char numbers match karte — Pauli se forbidden. Isliye ek orbital maximum 2 electrons tak hi cap hota hai.

KYA: $2n^2=32\Rightarrow n^2=16\Rightarrow n=4$ solve karo. Sub-shells: $l=0,1,2,3$ → $4s,4p,4d,4f$. Count check: $2(1)+2(3)+2(5)+2(7)=2+6+10+14=32$ ✓ (dekho Electron Configuration & Periodic Table kyun filling order is raw capacity se alag hoti hai).

KYA: loop-matching condition impose karo. $\Phi (\varphi +2\pi )=e^{i{m}_l(\varphi +2\pi )}=e^{i{m}_l\varphi }{e}^{i{m}_{l}2\pi }=\Phi (\varphi )\cdot e^{i{m}_{l}2\pi }.$ Iske $\Phi (\varphi )$ ke barabar hone ke liye $e^{i{m}_{l}2\pi }=1$ chahiye. KYU: Euler's relation $e^{ix}=\cos x+i\sin x$ se, $e^{i{m}_{l}2\pi }=1$ sirf tab hoga jab $\cos (2\pi m_l)=1$ aur $\sin (2\pi m_l)=0$, yaani jab $2\pi m_l$ full turns ka ek whole number ho: $m_l\in \{\dots ,-1,0,1,\dots \}$. Non-integer $m_l$ ki physical nonsense: electron wave same point in space par alag value ke saath wapas aati — toh cloud ki density ek hi location par two-valued hoti — yeh impossible hai.

KYA: $\sum _{l=0}^{n-1}2(2l+1)=2\sum _{l=0}^{n-1}(2l+1)$ compute karo. Inner sum pehle $n$ odd numbers ka sum hai $1+3+5+\cdots +(2n-1)$, jo ek classic identity hai aur $n^2$ ke barabar hai. $n^2$ kyun: odd numbers stack karne se ek perfect square banta hai, layer by layer (ek $k\times k$ square mein $2k+1$ dots ka L-shaped shell add karne se $(k+1)\times (k+1)$ banta hai). Figure dekho. Isliye total $=2\cdot n^2=2n^2$. $■$

Bohr: $L=n\hslash =1\cdot \hslash =\hslash$ (non-zero). Schrödinger: $|\vec{L}|=\sqrt{l(l+1)}\hslash =\sqrt{0\cdot 1}\hslash =0$. Kaunsa sahi hai: Schrödinger. Ek $s$-electron ka probability cloud perfectly spherical hota hai jisme koi preferred rotation axis nahi hoti, isliye woh orbital angular momentum carry nahi kar sakta. Spectroscopic fine-structure aur hydrogen spectrum confirm karte hain ki $l=0$ states mein zero orbital angular momentum hai. Bohr ka $n\hslash$ rule energies ke liye ek lucky approximation tha lekin angular momentum ke liye galat.

• $n=4$ mein max electrons? ::: $2n^2=32$.
• $3d$ electron ke liye $|\vec{L}|$? ::: $\sqrt{6}\hslash$.
• $z$ se $m_l=+2$ state ($l=2$) ka angle? ::: $\arccos (2/\sqrt{6})\approx 35.26^{\circ }$.
• $(4,0,-1,+\frac{1}{2})$ invalid kyun hai? ::: $l=0$ force karta hai $m_l=0$, toh $m_l=-1$ forbidden hai.
• Kaunsa boundary condition $m_l$ ko quantise karta hai? ::: $\Phi (\varphi +2\pi )=\Phi (\varphi )$ ⇒ $e^{i{m}_{l}2\pi }=1$ ⇒ integer $m_l$.