Spectral series — Lyman, Balmer, Paschen: Worked examples

2.3.15 · D3 · Physics › Modern Physics › Spectral series — Lyman, Balmer, Paschen

Yeh page parent topic ke liye ek worked-example workshop hai. Parent ne tumhe master equation di thi. Yahan hum har tarah ke question tackle karenge jo koi exam ya real spectrometer throw kar sakta hai — har series, dono endpoints, degenerate cases, ek word problem, aur ek exam twist.

Sab kuch ek equation par tika hai — Rydberg formula — jise hum dobara likhte hain taaki koi bhi symbol unexplained na rahe:

Scenario matrix

Kuch bhi calculate karne se pehle, aao har case class map out karte hain jo is topic mein hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Kya cheez ise alag banati hai	Example
A	Lyman, first line ( $n_{1} = 1, n_{2} = 2$ )	sabse gehri landing, sabse chhhoti series → UV	Ex 1
B	Balmer, first line ( $n_{1} = 2, n_{2} = 3$ )	visible red H-alpha	Ex 2
C	Paschen, general line ( $n_{1} = 3, n_{2} = 5$ )	infrared, non-adjacent jump	Ex 3
D	Series limit ( $n_{2} \to \infty$ )	degenerate/limiting input	Ex 4
E	Reverse problem: $λ$ diya hua hai, transition nikalo	inverse / identification	Ex 5
F	Absorption vs emission (same numbers, opposite direction)	sign/direction case	Ex 6
G	Hydrogen-like ion ( $Z > 1$ )	scaling case, $Z^{2}$ factor	Ex 7
H	Real-world word problem (spectrometer / astronomy)	applied context	Ex 8
I	Exam twist: bina calculator ke kaun si line chhoti hai	reasoning-only	Ex 9

Poore matrix ko do facts control karte hain, toh inhe abhi yaad kar lo:

Figure — Spectral series — Lyman, Balmer, Paschen

Worked examples

Cell A — Lyman first line (UV)

Worked example Ex 1 — Lyman series ki sabse lambi wavelength

Lyman series ( $n_{1} = 1$ ) mein emit hone wali sabse lambi wavelength nikalo.

Forecast: Sabse lamba $λ$ matlab sabse chhota jump. Sabse chhota jump jo level 1 par land kare, woh kaunsa hai? Aage padhne se pehle pair $(n_{1}, n_{2})$ guess karo.

Transition pick karo. Level 1 par sabse chhota jump level 2 se hoga: $n_{1} = 1, n_{2} = 2$ . Yeh step kyun? Sabse lambi wavelength ⇔ sabse kam energy ⇔ sabse paas wala higher level, jo hai $n_{1} + 1 = 2$ .
Plug in karo. $\frac{1}{λ} = 1.097 \times 1 0^{7} (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = 1.097 \times 1 0^{7} \cdot \frac{3}{4} = 8.23 \times 1 0^{6} m^{- 1} .$ Yeh step kyun? Rydberg formula level pair ko seedha wavenumber mein convert kar deti hai.
$λ$ pane ke liye invert karo. $λ = \frac{1}{8.23 \times 1 0 ^{6}} = 1.215 \times 1 0^{- 7} m = 121.5 nm .$ Yeh step kyun? $λ$ wavenumber ka reciprocal hota hai.

Verify karo: 121.5 nm visible light ki 400 nm edge se kaafi neeche hai ⇒ ultraviolet, bilkul waise jaisa "sabse gehri landing = sabse chhota λ" rule predict karta hai. ✓

Cell B — Balmer first line (visible)

Worked example Ex 2 — Red H-alpha line (

$3 \to 2$ ) $n_{2} = 3$ se $n_{1} = 2$ wali Balmer line ki wavelength nikalo.

Forecast: Balmer level 2 par land karta hai (visible). Pehli line ek lamba visible wavelength hona chahiye — kaunse visible colour ki wavelength sabse lambi hoti hai? Colour guess karo.

Pair set karo: $n_{1} = 2, n_{2} = 3$ . Kyun? Pehli (sabse lambi) Balmer line sabse paas wala higher level use karti hai, $n_{1} + 1 = 3$ .
Wavenumber compute karo. $\frac{1}{λ} = 1.097 \times 1 0^{7} (\frac{1}{4} - \frac{1}{9}) = 1.097 \times 1 0^{7} \cdot \frac{5}{36} = 1.524 \times 1 0^{6} m^{- 1} .$ Kyun? $\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$ fractional energy gap hai.
Invert karo: $λ = 6.56 \times 1 0^{- 7} m = 656 nm$ .

Verify karo: 656 nm, 400–700 nm visible band ke red end par hai. Sabse lambi wavelength = red. ✓ Yeh hydrogen lamp ki glowing red line hai. Dekho Emission vs Absorption Spectra.

Cell C — Paschen, non-adjacent jump (IR)

Cell D — Limiting case ( $n_{2} \to \infty$ )

Cell E — Reverse problem (transition identify karo)

Worked example Ex 5 — Wavelength diya hua hai, jump batao

Hydrogen ki ek line $λ = 486 nm$ par measure ki gayi. Kaun sa transition tha?

Forecast: 486 nm blue-green hai — clearly visible — toh yeh kaun si series ki hogi? $n_{1}$ compute karne se pehle guess karo.

Colour se series guess karo: visible ⇒ Balmer ⇒ $n_{1} = 2$ . Kyun? Sirf Balmer hi visible band mein land karta hai.
Wavenumber compute karo: $\frac{1}{λ} = \frac{1}{486 \times 1 0 ^{- 9}} = 2.058 \times 1 0^{6} m^{- 1} .$
$n_{2}$ solve karo. Set karo $\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{n _{2}^{2}})$ : $\frac{1}{n _{2}^{2}} = \frac{1}{4} - \frac{2.058 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.25 - 0.1876 = 0.0624 \Rightarrow n_{2}^{2} = 16.02 \Rightarrow n_{2} = 4.$ Yeh step kyun? Hum unknown level isolate karne ke liye rearrange karte hain, phir reciprocal-square-root lete hain.

Verify karo: $n_{2} = 4, n_{1} = 2$ Balmer $β$ line hai. Forward-check: $R_{H} (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 1 0^{7} \cdot \frac{3}{16} = 2.057 \times 1 0^{6} \Rightarrow λ = 486 nm$ . ✓

Cell F — Absorption vs emission (direction/sign)

Worked example Ex 6 — Same numbers, opposite arrow

Thanda hydrogen gas light absorb karta hai aur electron ko $n = 1$ se $n = 3$ tak lift karta hai. Kaun si wavelength absorb hoti hai, aur yeh emission $3 \to 1$ se kaise compare karti hai?

Forecast: Absorption staircase upar jaata hai; emission neeche jaata hai. Kya "upar 1→3" aur "neeche 3→1" mein wavelength alag hoti hai? Compute karne se pehle guess karo.

Upar jaane ke liye needed energy neeche aane par release hone wali energy ke barabar hoti hai — same gap: $\frac{1}{λ} = R_{H} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = 1.097 \times 1 0^{7} \cdot \frac{8}{9} = 9.75 \times 1 0^{6} m^{- 1} .$ Absorption ke liye bhi lower level pehle kyun likhte hain? Formula tabhi positive wavenumber deta hai jab hum $1/ n_{1}^{2} - 1/ n_{2}^{2}$ likhte hain; photon ki energy dono directions mein same magnitude ki hoti hai.
Invert karo: $λ = 1.026 \times 1 0^{- 7} m = 102.6 nm$ (UV).

Verify karo: Absorption line aur emission line bilkul same wavelength 102.6 nm par hain — dark absorption line aur bright emission line coincide karti hain. Isliye Emission vs Absorption Spectra ek doosre ke negatives lagte hain. ✓ (Yeh Lyman- $β$ line hai.)

Cell G — Hydrogen-like ion ( $Z^{2}$ scaling)

Worked example Ex 7 — Singly ionized helium (

$He^{+}$ , $Z = 2$ ) ki Balmer-analogue line $He^{+}$ mein ek electron hai lekin nuclear charge $Z = 2$ hai. $3 \to 2$ wavelength nikalo.

Forecast: $Z$ double karne par energy gaps $Z^{2} = 4$ se multiply ho jaati hain. Kya wavelength 4 se badhti ya ghatti hai? Guess karo.

Scaled Rydberg formula use karo: $\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ . $Z^{2}$ kyun? Bada nuclear charge electron ko 4 guna zyada tight bind karta hai, toh har gap $Z^{2}$ se scale hoti hai. Dekho Hydrogen-like Ions and Z dependence.
$Z = 2$ , $n_{1} = 2$ , $n_{2} = 3$ plug in karo: $\frac{1}{λ} = 1.097 \times 1 0^{7} \cdot 4 \cdot \frac{5}{36} = 6.094 \times 1 0^{6} m^{- 1} .$
Invert karo: $λ = 1.641 \times 1 0^{- 7} m = 164.1 nm$ (UV).

Verify karo: Hydrogen ki $3 \to 2$ 656 nm thi; $Z^{2} = 4$ se divide karne par $656/4 = 164 nm$ milta hai. ✓ Wavelength exactly 4 se ghati, toh visible red line ultraviolet ban gayi.

Cell H — Real-world word problem

Worked example Ex 8 — Astronomer ka redshift-free ID

Kisi star ke spectrum mein $434 nm$ par ek bright hydrogen line dikhti hai. Astronomer yeh confirm karna chahta hai ki yeh hydrogen hi hai (koi Doppler-shifted impostor nahi). Isse identify karo.

Forecast: 434 nm violet hai — visible — toh phir se Balmer ( $n_{1} = 2$ ). Guess karo ki $n_{2}$ , Ex 5 (jo 486 nm thi) se bada hoga ya chhota.

Wavenumber: $\frac{1}{λ} = \frac{1}{434 \times 1 0 ^{- 9}} = 2.304 \times 1 0^{6} m^{- 1} .$
$n_{2}$ solve karo $n_{1} = 2$ ke saath: $\frac{1}{n _{2}^{2}} = 0.25 - \frac{2.304 \times 1 0 ^{6}}{1.097 \times 1 0 ^{7}} = 0.25 - 0.2100 = 0.0400 \Rightarrow n_{2}^{2} = 25 \Rightarrow n_{2} = 5.$ Kyun? Ex 5 jaisa hi isolate-and-invert method; chhoti wavelength (bada gap) matlab 486 nm line se zyada upar ka starting level.

Verify karo: $n_{2} = 5 \to n_{1} = 2$ Balmer- $γ$ line hai. Forward check: $R_{H} (\frac{1}{4} - \frac{1}{25}) = 1.097 \times 1 0^{7} \cdot \frac{21}{100} = 2.304 \times 1 0^{6} \Rightarrow λ = 434 nm$ . ✓ Yeh hydrogen hai — genuine.

Cell I — Exam twist (bina calculator reasoning)

Worked example Ex 9 — Bina calculate kiye kaun si line chhoti hai?

Sirf reasoning se in teen lines ko wavelength ke hisaab se order karo, sabse chhoti pehle: Lyman-limit ( $\infty \to 1$ ), Balmer $3 \to 2$ , Paschen $4 \to 3$ .

Forecast: Apna ordering guess abhi likh do.

Energy gap se rank karo. Gehra landing = bada $R_{H} / n_{1}^{2}$ scale. Lyman level 1 par land karta hai, Balmer level 2 par, Paschen level 3 par. Plus limit ( $n_{2} = \infty$ ) Lyman ke andar sabse bada jump hai. Yeh step kyun? Chhota $λ$ ⇔ bada energy gap; koi numbers nahi chahiye, bas dominant $1/ n_{1}^{2}$ term aur jump size compare karo.
Order: Lyman-limit (sabse bada gap, $1/ λ = R_{H}$ ) $>$ Balmer $3 \to 2$ $>$ Paschen $4 \to 3$ energy mein, toh wavelength mein (reverse): Lyman-limit sabse chhota, phir Balmer, phir Paschen sabse lamba.

Numbers se verify karo: Lyman-limit $= 91.2 nm$ , Balmer $3 \to 2 = 656 nm$ , Paschen $4 \to 3 = 1875 nm$ . Shortest→longest ordering: 91.2 < 656 < 1875 nm. ✓ Reasoning se match karta hai.

Recall

Recall Matrix bhar ke fast checks

Kisi bhi series ki first line kaun sa start level use karti hai ::: $n_{2} = n_{1} + 1$ (sabse chhota gap, sabse lamba λ) Series limit kaun sa start level use karta hai ::: $n_{2} \to \infty$ (sabse bada gap, sabse chhota λ) Lyman first line wavelength ::: 121.5 nm (UV) Balmer H-alpha (3→2) wavelength ::: 656 nm (red) Paschen 5→3 wavelength ::: lagbhag 1282 nm (IR) Balmer series limit wavelength ::: 365 nm 486 nm hydrogen line kaun sa transition hai ::: Balmer 4→2 Absorption 1→3 wavelength ::: 102.6 nm, emission 3→1 ke barabar He+ (Z=2) 3→2 wavelength ::: 164 nm, hydrogen ki value Z squared se divide ki gayi Lyman series limit wavelength ::: 91.2 nm

Connections

Parent topic — woh derivation jise yeh examples exercise karte hain.
Bohr Model of the Atom — jahan se $E_{n} = - 13.6/ n^{2}$ aata hai.
Photon Energy and Planck's Relation — $λ = 1240/Δ E$ shortcut jo poore mein use hota hai.
Hydrogen-like Ions and Z dependence — Ex 7 ka $Z^{2}$ scaling.
Ionization Energy — Ex 4 ka $n_{2} \to \infty$ series limit.
Emission vs Absorption Spectra — Ex 6 ki same-wavelength coincidence.