Hydrogen atom — solving in spherical coordinates: Worked examples

2.3.12 · D3 · Physics › Modern Physics › Hydrogen atom — solving in spherical coordinates

Yeh page ek drill hai. Parent note ne machinery banayi — quantum numbers $n, ℓ, m_{ℓ}$ , energy ladder $E_{n} = - 13.6/ n^{2}$ eV, degeneracy $n^{2}$ . Yahan hum har tarah ke questions handle karenge jo yeh machinery throw kar sakti hai: sabse chhota case, sabse bada allowed case, illegal case (taaki tum use reject karna seekho), limit $n \to \infty$ , lab mein ek real photon, aur ek exam-style trap.

Shuru karne se pehle, chaar chhoti reminders taaki koi symbol unearned na rahe:

Definition Teen counters aur do constants (parent se)

$== n ==$ = principal quantum number = $1, 2, 3, \dots$ Yeh energy aur atom ki size set karta hai.
$== ℓ ==$ = orbital angular-momentum quantum number = $0, 1, \dots, n - 1$ . Yeh shape set karta hai aur electron kitna "orbit" karta hai.
$== m_{ℓ} ==$ = magnetic quantum number = $- ℓ, - ℓ + 1, \dots, + ℓ$ . Yeh tilt set karta hai (orbit kis direction mein point karti hai).
$ℏ$ = reduced Planck constant = $h /2 π$ , angular momentum ka fundamental "chunk size" ( $\approx 1.055 \times 1 0^{- 34}$ J·s).
$μ$ = electron–proton pair ki reduced mass, $μ = \frac{m _{e} m _{p}}{m _{e} + m _{p}} \approx m_{e}$ . Yeh effective mass hai jo do-body atom ko ek one-body problem mein convert karti hai (dekho Reduced mass and two-body problem).

Arrows ko ek waterfall ki tarah padhna: $n$ decide karta hai $ℓ$ kitna upar chadh sakta hai; $ℓ$ decide karta hai $m_{ℓ}$ kitna spread ho sakta hai. Koi level skip nahi hoti.

Scenario matrix

Har hydrogen-atom problem in cells mein se ek hai. Neeche ke worked examples mein cell tag diya gaya hai.

Cell	Case class	Tricky kyon hai	Example
A	Sabse chhota input ( $n = 1$ )	Sirf ek state exist karti hai — "degenerate/trivial" end	Ex 1
B	Ek beech wala $n$ , saari states count karo	Har $ℓ$ ke liye $2 ℓ + 1$ sum karna padega	Ex 2
C	Illegal quantum numbers	Tum reject karo, compute nahi	Ex 3
D	Energy difference (emission)	$Δ E$ ka sign, photon kaun sa hai	Ex 4
E	Energy difference (absorption)	Ulta sign, atom upar chadhta hai	Ex 5
F	Limit $n \to \infty$ (ionisation)	Ladder ki top rung; series limit	Ex 6
G	Real-world word problem (photon → wavelength)	eV ko nanometres mein convert karo	Ex 7
H	Exam twist: max $ℓ$ , phir angular momentum magnitude	$ℓ_{m a x} = n - 1$ ko $ℓ (ℓ + 1) ℏ$ se combine karo	Ex 8
I	Centrifugal barrier: degenerate $ℓ = 0$ (zero barrier) vs large $ℓ$	Division-by-zero edge case	Ex 9

Hum do figures reference karte hain: energy ladder (Ex 4–6) aur centrifugal barrier (Ex 9).

Figure — Hydrogen atom — solving in spherical coordinates

Cell A — Sabse chhota input

Worked example Ex 1 — Ground state,

har allowed state list karo $n = 1$ ke liye, saare allowed $(ℓ, m_{ℓ})$ pairs aur energy likhna.

Forecast: Padhne se pehle guess karo — $n = 1$ par kitne distinct states expect karte ho? (Hint: $n^{2}$ socho.)

Allowed $ℓ$ dhundho. Rule: $ℓ = 0, 1, \dots, n - 1 = 0, \dots, 0$ . Toh $ℓ = 0$ sirf. Yeh step kyon? $n$ , $ℓ$ ko $n - 1$ par cap karta hai. $n = 1$ ke saath cap $0$ hai, toh chadhne ki koi jagah nahi.
Allowed $m_{ℓ}$ dhundho. Rule: $m_{ℓ} = - ℓ, \dots, + ℓ = 0$ sirf. Yeh step kyon? $ℓ = 0$ $m_{ℓ}$ ko spread hone ki koi jagah nahi deta.
Energy batao. $E_{1} = - \frac{13.6}{1 ^{2}} = - 13.6$ eV. Yeh step kyon? Energy sirf $n$ par depend karti hai, aur yahan $n = 1$ .

Verify: States ki count $n^{2} = 1^{2} = 1$ honi chahiye. Humne exactly ek pair $(ℓ, m_{ℓ}) = (0, 0)$ paya. ✓ Energy negative hai (bound state, free-electron zero se neeche). ✓ Yeh $1 s$ orbital hai.

Cell B — Ek beech wale $n$ par saari states count karo

Worked example Ex 2 —

$n = 4$ ki Degeneracy Kitne orbital states (electron spin ignore karo) energy level $n = 4$ share karte hain?

Forecast: Odd numbers aane wale hain. Sum karne se pehle total guess karo.

Allowed $ℓ$ list karo: $0, 1, 2, 3$ . Yeh step kyon? $ℓ$ , $0$ se $n - 1 = 3$ tak chalta hai.
Har $ℓ$ ke liye $m_{ℓ}$ count karo: har $ℓ$ $2 ℓ + 1$ values deta hai.
- $ℓ = 0 : 1$ , $ℓ = 1 : 3$ , $ℓ = 2 : 5$ , $ℓ = 3 : 7$ . Yeh step kyon? $m_{ℓ}$ , $- ℓ$ se $+ ℓ$ tak span karta hai, jo $2 ℓ + 1$ integers hain.
Add karo: $1 + 3 + 5 + 7 = 16$ . Yeh step kyon? Is $n$ par har distinct $(ℓ, m_{ℓ})$ ek alag state hai jiski same energy hai.

Verify: Pehle $n$ odd numbers ka sum $n^{2}$ hota hai. Yahan $n = 4 \Rightarrow n^{2} = 16$ . ✓ Yeh tidy match confirm karta hai ki energy $ℓ$ aur $m_{ℓ}$ ko ignore karti hai — pure $1/ r$ potential ki special "accidental" degeneracy (dekho Zeeman effect, jahan magnetic field break karta hai ise aur 16 ko split karta hai).

Cell C — Illegal case ko reject karo

Worked example Ex 3 — In mein se kaun se impossible hain?

Har proposed triple ke liye, batao ki yeh ek legal hydrogen state hai ya nahi, aur agar nahi, toh kaun si rule tooti. (i) $(n, ℓ, m_{ℓ}) = (2, 2, 0)$ (ii) $(3, 2, - 2)$ (iii) $(2, 1, - 2)$ (iv) $(3, 0, 0)$

Forecast: Compute karne se pehle, expect karo ki kam se kam do forbidden honge — sawaal yeh hai ki kaun si rule.

(i) $ℓ \leq n - 1$ check karo: $ℓ \leq 1$ chahiye, par $ℓ = 2$ . Illegal. Yeh step kyon? Radial solution sirf $ℓ \leq n - 1$ ke liye terminate hota hai; $ℓ = n$ blow up kar deta hai.
(ii) Dono rules check karo: $ℓ = 2 \leq n - 1 = 2$ ✓, aur $∣ m_{ℓ} ∣ = 2 \leq ℓ = 2$ ✓. Legal. Yeh step kyon? Dono waterfall steps satisfy hue.
(iii) $∣ m_{ℓ} ∣ \leq ℓ$ check karo: yahan $∣ - 2 ∣ = 2 > ℓ = 1$ . Illegal. Yeh step kyon? $m_{ℓ}$ magnitude mein $ℓ$ se zyada nahi ho sakta — full turn ki single-valued condition ise cap karti hai.
(iv) Dono check karo: $ℓ = 0 \leq 2$ ✓, $m_{ℓ} = 0$ (single choice) ✓. Legal.

Verify: Legal ones hain (ii) aur (iv). Sanity check: (ii) ke liye state $3 d$ hai; $d$ matlab $ℓ = 2$ , aur $n = 3$ $ℓ = 2$ tak allow karta hai, toh $3 d$ orbital exist karta hai — reality se match karta hai. ✓ (iii) ke liye, $2 p$ shell mein sirf $m_{ℓ} \in {- 1, 0, + 1}$ hain, toh $m_{ℓ} = - 2$ genuinely exist nahi kar sakta. ✓

Cell D — Emission (energy bahar aati hai)

Worked example Ex 4 —

$H α$ line, $n = 3 \to n = 2$ Jab electron $n = 3$ se $n = 2$ par girti hai toh kitni energy release hoti hai. Energy kis direction flow karti hai?

Forecast: Neeche lower rung par girna — kya atom energy gain ya lose karta hai? Sign guess karo.

Dono levels compute karo. $E_{3} = - 13.6/9 = - 1.511$ eV, $E_{2} = - 13.6/4 = - 3.400$ eV. Yeh step kyon? Har rung ki height $- 13.6/ n^{2}$ hai; difference lene ke liye dono chahiye.
$Δ E = E_{final} - E_{initial} = E_{2} - E_{3}$ lo. $Δ E = - 3.400 - (- 1.511) = - 1.889 eV .$ Yeh step kyon? Final minus initial batata hai ki atom ki energy kaise badi. Negative ⇒ atom ne energy khoyi.
Photon energy = $∣Δ E ∣ = 1.889$ eV. Yeh step kyon? Energy conserved hoti hai: atom jo khoye, photon le jaaye.

Verify: Ladder figure (s01) par, $n = 3$ se $n = 2$ ka arrow neeche point karta hai aur chhota hai. $∣Δ E ∣ = 1.89$ eV red visible band mein land karta hai — yeh famous red $H α$ line hai. ✓ Positive photon energy, atom ke liye negative $Δ E$ — consistent. ✓

Cell E — Absorption (energy andar jaati hai)

Worked example Ex 5 —

$n = 1 \to n = 4$ chadhna Ek photon absorb hoti hai, electron ko ground state se $n = 4$ par kick karti hai. Kitni photon energy chahiye?

Forecast: Ladder par upar chadhna — is baar atom ke liye $Δ E$ ka sign kya hoga?

Dono levels compute karo. $E_{1} = - 13.6$ eV, $E_{4} = - 13.6/16 = - 0.850$ eV. Yeh step kyon? Same ladder formula, alag rungs.
$Δ E = E_{4} - E_{1} = - 0.850 - (- 13.6) = 12.750$ eV. Yeh step kyon? Final minus initial. Ab Positive — atom ne energy gain ki.
Photon exactly $12.75$ eV supply karti hai. Yeh step kyon? Absorption sirf tab hoti hai jab incoming photon exactly gap se match kare (isliye absorption spectra sharp lines hoti hain).

Verify: $4 \to 1$ se emission same $12.75$ eV release karegi — arrow reverse karna sign flip karta hai par magnitude nahi. ✓ Ex 4 ke $1.89$ eV se bada jump, jaisa ki lambe chadhne ke liye expected hai. ✓

Cell F — Limit $n \to \infty$ (ionisation)

Worked example Ex 6 — Ground state se ionisation energy

$n = 1$ se start karke electron ko completely free karne mein kitni energy lagti hai?

Forecast: "Completely free" matlab kaun si final rung? Guess karo ladder kaunse number approach karta hai.

Ladder ki top identify karo. Jab $n \to \infty$ , $E_{n} = - 13.6/ n^{2} \to 0$ . Yeh step kyon? Ek free (unbound) electron ki zero binding energy hoti hai — yeh ceiling hai jiske toward rungs crowd karte hain.
Us limit tak difference lo. $Δ E = E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6$ eV. Yeh step kyon? Ionisation = electron ko apni rung se $E = 0$ tak lift karo.
Result batao: ground-state hydrogen ki ionisation energy $13.6$ eV hai. Yeh step kyon? Yeh single number exactly $∣ E_{1} ∣$ hai — sabse gehra well ki depth.

Verify: Ladder (s01) par, rungs bunch up hote hain aur $E = 0$ line ke against pile karte hain — gaps $1/ n^{2}$ ki tarah shrink karte hain, confirming convergence. ✓ Hydrogen ki measured ionisation energy $13.6$ eV hai. ✓ Note karo yeh Bohr model of hydrogen prediction se match karta hai — full Schrödinger solution Bohr ka number reproduce karta hai.

Cell G — Real-world word problem (photon → wavelength)

Worked example Ex 7 —

$H α$ line ka colour kya hai? $n = 3 \to 2$ transition ek $1.889$ eV photon emit karta hai (Ex 4). Nanometres mein wavelength dhundho.

Forecast: Visible light roughly $400$ – $700$ nm run karta hai. Compute karne se pehle guess karo $H α$ blue hai ya red.

Photon relation yaad karo. Photon ki energy aur wavelength $E = \frac{h c}{λ}$ follow karte hain, toh $λ = \frac{h c}{E}$ . Yeh step kyon? Energy aur wavelength inversely linked hain — zyada energy, chhoti wave. $λ$ paane ke liye invert karo.
Ek handy constant use karo. $h c = 1240$ eV·nm. Yeh step kyon? Yeh packaged value seedha eV divide karne deta hai aur nm milta hai — koi unit gymnastics nahi.
Divide karo: $λ = \frac{1240 eV\cdotnm}{1.889 eV} = 656.4 nm$ . Yeh step kyon? Ex 4 ki photon energy seedha daalo.

Verify: Units: $\frac{eV\cdotnm}{eV} = nm$ . ✓ $656$ nm visible spectrum ke red part mein hai — observed deep-red hydrogen line se match karta hai jo astronomers glowing gas map karne mein use karte hain. ✓

Cell H — Exam twist: max $ℓ$ aur angular momentum magnitude

Worked example Ex 8 —

$n = 3$ par sabse bada orbital angular momentum $n = 3$ ke andar, sabse bada allowed $ℓ$ kya hai, aur us state ke liye angular momentum ki actual magnitude $∣ L ∣$ kya hai?

Forecast: Ek common trap kehta hai $∣ L ∣ = ℓ ℏ$ . Step 3 se pehle guess karo kya yeh sahi hai.

Max $ℓ$ . $ℓ_{m a x} = n - 1 = 2$ . Yeh step kyon? Waterfall rule $ℓ$ ko $n$ se ek neeche cap karta hai.
Magnitude formula yaad karo (parent ke $\hat{L}^{2}$ eigenvalue se). $\hat{L}^{2} Y = ℓ (ℓ + 1) ℏ^{2} Y$ , toh length hai $∣ L ∣ = ℓ (ℓ + 1) ℏ$ . (Yahan $ℏ$ reduced Planck constant hai, angular momentum ka chunk size.) Yeh step kyon? Measurable magnitude $\hat{L}^{2}$ eigenvalue ka square root hai — $ℓ ℏ$ nahi.
$ℓ = 2$ daalo: $∣ L ∣ = 2 \cdot 3 ℏ = 6 ℏ \approx 2.449 ℏ$ . Yeh step kyon? Earned formula mein direct substitution.

Verify: $6 ℏ \approx 2.449ℏ$ , naive $ℓ ℏ = 2ℏ$ se bada hai — trap answer under-count karta hai. ✓ Dekho Angular momentum operators aur Spherical harmonics yeh samajhne ke liye ki $ℓ (ℓ + 1)$ , $ℓ$ nahi, true length kyun hai. Aur $ℓ = 2$ ek $d$ -orbital hai, Ex 3 ke "3d exists" se match karta hai. ✓

Cell I — Centrifugal barrier: $ℓ = 0$ zero aur division-by-zero edge case

Yahan effective potential hai jo parent note ne banaya. Coulomb attraction ke aage extra piece centrifugal barrier hai: $V_{cf} (r) = \frac{ℏ ^{2} ℓ ( ℓ + 1 )}{2 μ r ^{2}},$ jahan $ℏ$ reduced Planck constant hai aur $μ$ electron–proton pair ki reduced mass hai (dono opening definition box mein recall kiye). Notice karo ki barrier ki state par sirf $ℓ (ℓ + 1)$ factor ke through dependence hai — constants $ℏ, μ$ aur radius $r$ har state mein common hain, toh kisi bhi ratio mein cancel ho jaate hain.

Worked example Ex 9 —

$ℓ = 0$ zero, aur kyun ek naive ratio fail karta hai Ek fixed radius $r$ par, $ℓ = 0$ , $ℓ = 1$ , aur $ℓ = 2$ ke centrifugal barrier compare karo. Task " $ℓ = 0$ se $ℓ = 2$ jaane" ka ratio maangta hai — ise carefully handle karo.

Forecast: $ℓ = 0$ ke liye barrier... kuch special hai. Step 1 se pehle guess karo, phir guess karo ki requested ratio kya karta hai.

$ℓ = 0$ par state-factor evaluate karo. $ℓ (ℓ + 1) = 0 \cdot 1 = 0$ , toh $V_{cf} (ℓ = 0) = 0$ exactly, har radius par. Yeh step kyon? Ek $s$ -state ( $ℓ = 0$ ) mein koi angular motion nahi hai, isliye koi centrifugal barrier nahi — electron nucleus ke paas baith sakta hai. Yeh degenerate/zero case hai.
$ℓ = 1$ aur $ℓ = 2$ ke liye evaluate karo. $ℓ (ℓ + 1) = 1 \cdot 2 = 2$ aur $2 \cdot 3 = 6$ . Yeh step kyon? Higher $ℓ$ mein zyada angular motion hai, toh bada barrier factor.
Requested ratio $\frac{V _{cf} ( ℓ = 2 )}{V _{cf} ( ℓ = 0 )}$ banao — aur ruko. Denominator $0$ hai, toh yeh ratio undefined (yeh $+ \infty$ tak diverge karta hai) hai. Yahan koi finite number report nahi kar sakte. Yeh step kyon? Division by zero "ek bada number" nahi hai — yeh ek genuine edge case hai. Kyunki $ℓ = 0$ literally zero barrier deta hai, "kitne times bada" ka koi jawab nahi: koi bhi positive barrier none se infinitely bada hai. Honest statement yeh hai: $ℓ = 2$ mein ek barrier hai jahan $ℓ = 0$ mein exactly kuch nahi.
Ek well-defined comparison do. Do nonzero barriers compare karne ke liye, $ℓ = 2$ ko $ℓ = 1$ se ratio karo: $\frac{V _{cf} ( ℓ = 2 )}{V _{cf} ( ℓ = 1 )} = \frac{6}{2} = 3$ . Yeh step kyon? Dono denominators nonzero hain, toh constants aur $r$ cleanly cancel ho jaate hain aur hume ek finite, meaningful factor milta hai: $ℓ = 2$ wall har radius par $ℓ = 1$ wall se $3 \times$ taller hai.

Verify: Barrier figure (s02) par, $ℓ = 0$ curve pure attraction hai (koi bump nahi — " $0$ " case), jabki $ℓ = 1$ aur $ℓ = 2$ origin ke paas steeply rise karte hain, $ℓ = 2$ har $r$ par $ℓ = 1$ se teen times zyada. ✓ $ℓ = 2$ -to- $ℓ = 0$ ratio ka koi finite value nahi hai kyunki tum us flat-zero curve se divide karte — exactly jo step 3 ne warn kiya. ✓ Bada $ℓ$ ⇒ stronger barrier ⇒ wavefunction outward push hoti hai, jaisa parent ke skater analogy ne kaha. ✓

Recall Self-test (pehle khud jawab do, phir reveal karo)

Q — $n = 4$ par states ki number? A — $n^{2} = 16$ .

Q — $3 \to 2$ ke liye photon energy? A — $1.89$ eV (red $H α$ , $656$ nm).

Q — Ground state se ionisation energy? A — $13.6$ eV.

Q — $ℓ = 2$ ke liye $L$ ki magnitude? A — $6 ℏ \approx 2.45ℏ$ .

Q — $s$ -state ke liye centrifugal barrier, aur $ℓ = 2$ -to- $ℓ = 0$ ratio? A — Zero ( $ℓ = 0$ ); ratio undefined hai (zero se divide hota hai).