Exercises — Hydrogen atom — solving in spherical coordinates

Hum kya check karte hain, order mein: pehle $0\le \ell \le n-1$, phir $-\ell \le m_{\ell }\le \ell$.

• (i) $n=2\Rightarrow \ell \le 1$. Yahan $\ell =1$ ✓. Phir $|m_{\ell }|\le 1$, aur $m_{\ell }=-1$ ✓. Allowed (ye ek $2p$ state hai).
• (ii) $n=2\Rightarrow \ell \le 1$. Yahan $\ell =2>1$. Illegal — $\ell$, $n-1$ se zyada hai.
• (iii) $n=3,\ell =1\Rightarrow |m_{\ell }|\le 1$. Yahan $m_{\ell }=2>1$. Illegal — $m_{\ell }$, $\ell$ se zyada hai.
• (iv) $n=1\Rightarrow \ell =0$, $m_{\ell }=0$ ✓. Allowed (ground state $1s$).

$\ell =0\to s$, $\ell =1\to p$, $\ell =2\to d$, $\ell =3\to f$ (mnemonic Silly People Dance Funny). To $n=3,\ell =2$ ko 3d likha jata hai.

Ye formula kahan se aata hai (WHAT hai $R$?): full wavefunction $\psi =R(r)Y(\theta ,\varphi )$ ki tarah factorize hoti hai, jahan $R(r)$ radial function hai — ye batata hai ki electron ki probability amplitude nucleus se $r$ distance par kaise vary karti hai, direction ignore karke. Ek standard trick $u(r)=rR(r)$ set karta hai; $r$ se multiply karne par messy radial equation ek clean 1-D Schrödinger equation ban jaati hai. $u(r)$ ko $r\to \infty$ par zero tak decay karne ki requirement (electron infinitely door nahi ho sakta) hi energy ko sirf discrete values lene par majboor karti hai: $E_1=-\frac{13.6}{1}=-13.6\text{ eV},E_2=-\frac{13.6}{4}=-3.40\text{ eV},E_3=-\frac{13.6}{9}=-1.51\text{ eV}.$ Gaps $n$ badhne ke saath shrink hote hain — ladder $0$ (ionization threshold) ki taraf crowd hoti hai.

Step 1 (WHAT): released energy $=|E_4-E_2|$. $E_4-E_2=-13.6\left(\frac{1}{16}-\frac{1}{4}\right)=-13.6\times (-0.1875)=2.55\text{ eV.}$ Sign kyun: atom ek lower (zyada negative) level par girta hai, to wo $2.55$ eV lose karta hai, jo photon le jaata hai. Step 2: $\lambda =1240/2.55=486\text{ nm}$ — Balmer series ki blue-green H-$\beta$ line.

$m_{\ell }$ run karta hai $-\ell ,\dots ,+\ell$, yaani $-2,-1,0,+1,+2$: ye $2\ell +1=2(2)+1=$ 5 states hain.

Ye sum kyun karte hain: har energy level $E_n$ apne saare $\ell$ aur $m_{\ell }$ ke saath shared hoti hai (Coulomb accident), to degeneracy $=$ $(\ell ,m_{\ell })$ pairs ki total count. $g(n)={\sum }_{\ell =0}^{n-1}(2\ell +1).$ Terms odd numbers $1,3,5,\dots ,(2n-1)$ hain ($\ell =0,1,\dots ,n-1$ rakhne par). Algebraic proof ki ye $n^2$ tak sum hote hain (koi picture nahi chahiye). Sum do baar likho, ek baar aage aur ek baar peeche, aur term by term add karo: $S=1+3+5+\cdots +(2n-1),$ $S=(2n-1)+\cdots +5+3+1.$ Dono lines add karne par, $n$ column-pairs mein se har ek same total $1+(2n-1)=2n$ deta hai: $2S=\underset{n\text{ terms}}{\underbrace{2n+2n+\cdots +2n}}=n\cdot 2n=2n^2\Rightarrow S=n^2.$ To $g(n)=n^2$, aur $g(4)=4^2=16$. Same fact, pictured: har naya odd number next L-shaped border fill karta hai jo ek $(n-1)\times (n-1)$ square ko $n\times n$ square mein badal deta hai (neeche red border last wala add hua hai).

Pehle, yahan $\mu$ kya hai? $\mu$ reduced mass hai $\mu =\frac{m_e{m}_p}{m_e+m_p}\approx m_e$ — wo single effective mass jo electron-plus-proton two-body problem ko ek one-body problem mein badal deta hai. (Agar chahein to ise "electron mass" padh lo, kyunki $\mu$ aur $m_e$ mein sirf $0.05\%$ ka fark hai; is problem mein kuch nahi badlega.) Extra term kya hai: centrifugal barrier, ek repulsive wall jo $\ell$ ke saath badhti hai aur $r\to 0$ par diverge karti hai. Figure mein $V_{\text{eff}}$ $\ell =0$ vs $\ell =2$ ke liye dikhta hai: $\ell =2$ (red) ka barrier origin ke paas curve ko lift karta hai, to well ka minimum outward shift ho jaata hai.

Conclusion: bada $\ell$ ⇒ strong barrier ⇒ electron nucleus se door held hota hai (uska probability cloud bade $r$ par peak karta hai). Physically: zyada angular motion matlab zyada "orbiting," jo andar khinche jaane ko resist karta hai — jaise ek spinning skater jo centre ki taraf drag hone se mana kare.

Nahi. Pure $1/r$ (Coulomb) potential ke liye energy sirf $n$ par depend karti hai: $E_3=-1.51$ eV chahe $\ell$ aur $m_{\ell }$ kuch bhi ho.

• $n$ energy set karta hai (radial boundary condition).
• $\ell$ angular momentum $|\vec{L}|=\sqrt{\ell (\ell +1)}\hslash$ ka size set karta hai.
• $m_{\ell }$ $z$-projection $L_z=m_{\ell }\hslash$ set karta hai. Last do cloud ki orientation/shape describe karte hain, uski energy nahi. (Symmetry todo — magnetic field apply karo, Zeeman effect — aur $m_{\ell }$ tab energy ko shift karta hai.)

WHAT ionization ka matlab hai: $E_2$ se seedha $E_{\infty }=0$ tak chadhna. Maximum wavelength ↔ minimum photon energy ↔ wo sabse chhoti energy jo bilkul top tak pahunche. $E_{\text{needed}}=0-E_2=0-(-3.40)=3.40\text{ eV.}$ Ye minimum kyun hai: isse kam ho to electron escape nahi kar sakta; isse zyada ho to sirf leftover kinetic energy milti hai (phir bhi ionize hota hai, lekin wo threshold nahi hai). Sabse lamba wavelength threshold energy correspond karta hai: ${\lambda }_{\max }=\frac{1240}{3.40}=365\text{ nm.}$

Systematic sweep — koi case mat chodo:

• $\ell =0$: $m_{\ell }=0$ → ek state, $2s$.
• $\ell =1$: $m_{\ell }=-1,0,+1$ → teen states, $2p$.

Total $=1+3=4=2^2=n^2$ ✓. Char states hain $\{2s;2p_{-1},2p_0,2p_{+1}\}$.

$\mu$ kyun: proton nailed down nahi hai; dono shared centre of mass ke around orbit karte hain. $m_e$ ko $\mu$ se replace karna Reduced mass and two-body problem ko ek clean one-body problem mein badal deta hai. $\frac{\mu }{m_e}=\frac{m_p}{m_e+m_p}=\frac{1836}{1837}=\mathrm{0.999456.}$ To true energy naive value ka $0.999456$ hai — fraction se chhoti $1-0.999456=5.44\times 10^{-4}\approx 0.0544\%.$ Tiny, lekin measurable — isliye deuterium ka spectrum ordinary hydrogen se thoda shifted hota hai.

Derivation (har step mein WHAT + WHY):

1. Har stationary state ki energy $E_n=-13.6/n^2$ eV hai — radial quantization se.
2. Ek transition energy conserve karta hai: photon energy $=|E_{n_i}-E_{n_f}|$, kyunki atom ki lost/gained energy kahi na kahi jaani chahiye.
3. Substitute karo: $|\Delta E|=\left|-\frac{13.6}{n_i^2}+\frac{13.6}{n_f^2}\right|=13.6\left|\frac{1}{n_f^2}-\frac{1}{n_i^2}\right|\text{ eV.}$ H-$\alpha$ check ($n_i=3\to n_f=2$): $|\Delta E|=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\times 0.13889=1.889\text{ eV},\lambda =\frac{1240}{1.889}=656\text{ nm.}$ Ye wo deep-red line hai jo aap har hydrogen discharge tube mein dekhte ho. ✓

Reason karo ki har number kahan se paida hua:

• $m_{\ell }\in \{0,\pm 1,\pm 2,\dots \}$ sirf $\Phi (\varphi )=e^{i{m}_{\ell }\varphi }$ ki single-valuedness se aaya — ek full turn ke baad. Isme $V(r)$ ki koi feature use nahi hoti. → Survive karta hai, identical rule.
• $\ell$ Spherical harmonics $\Theta (\theta )$ ko poles par finite require karne se aaya — phir se sphere ki pure geometry. → Survive karta hai, identical rule $|m_{\ell }|\le \ell$.
• Radial part wahan hai jahan force matter karta hai. Iske solutions ek radial quantum number $n_r=0,1,2,\dots$ se label hote hain, jo sirf count karta hai ki radial function kitni baar zero cross karta hai (uske "wiggles" ki sankhya). Coulomb atom ke liye combination $n=n_r+\ell +1$ usual principal number hai, aur isliye exactly $\ell \le n-1$ hai (kyunki $n_r\ge 0$). $V(r)$ badlo aur energy formula aur ye combination dono badal jaate hain: 3D harmonic well evenly spaced levels deta hai $E\propto \left(2n_r+\ell +\frac{3}{2}\right)$ $-1/n^2$ ki jagah. Yahan $n_r$ wohi "number of radial wiggles" count hai — sirf wo alag hai ki ye $\ell$ ke saath energy mein kaise bundle hota hai.

Bada lesson: angular quantization spheres ke baare mein ek fact hai; sirf energy pattern force ko fingerprint karta hai. Ye parent ke [!mistake] box mein L1 note ka point hai.

Step 1: orbital states per shell $=n^2$; 2 spins ke saath, capacity $=2n^2$.

• $n=1$: $2(1)=2$
• $n=2$: $2(4)=8$
• $n=3$: $2(9)=18$

Step 2 (compare): ye Quantum numbers and the periodic table ko structure karne wale widths $2,8,18$ se match karte hain. Real periodic table subshells ko reorder karta hai ($4s$ before $3d$) kyunki multi-electron atoms mein energy akele $n$ par depend nahi karti — Coulomb degeneracy accident break ho jaata hai. Lekin raw capacities $2n^2$ bilkul ye hydrogen counting hai. Sundar: poore periodic table ka rhythm chhupe hue $\sum (2\ell +1)$ hai.

$\ell$ ko $n-1$ par kyun cap kiya jaata hai? ::: Kyunki radial power series tab hi ek normalizable polynomial mein terminate hoti hai jab $\ell \le n-1$; equivalently $n=n_r+\ell +1$ with $n_r\ge 0$. $m_{\ell }$ quantization kahan se aata hai? ::: $\Phi (\varphi +2\pi )=\Phi (\varphi )$ se — ek full loop ke baad single-valuedness. Hydrogen ki energy kis cheez par depend karti hai? ::: Sirf $n$ par (ek Coulomb-specific accident); $\ell ,m_{\ell }$ shape aur orientation set karte hain.