Spectral series — Lyman, Balmer, Paschen: Question bank

2.3.15 · D5 · Physics › Modern Physics › Spectral series — Lyman, Balmer, Paschen

Sach ya jhooth — justify karo

Har jawab sirf verdict nahi, kyun bhi batata hai. Reveal karne se pehle reasoning cover karo.

Balmer series hydrogen mein sabse kam energy wali series hai.

Jhooth — Balmer

n_{1} = 2

par land karta hai, lekin Paschen, Brackett, Pfund aur upar land karte hain aur unke gaps chhote hote hain, isliye wo kam-energy (lambi-wavelength) ke photons emit karte hain. Balmer sirf isliye famous hai kyunki yeh visible range mein hota hai.

Lyman series ki har line ki wavelength Balmer series ki har line se chhoti hoti hai.

Sach — sabse chhota Lyman gap (

n_{2} = 2 \to n_{1} = 1

, 122 nm) bhi sabse bade Balmer gap (

3 \to 2

, 656 nm) se bada hai, kyunki

n_{1} = 1

par landing karna har jump ko kisi bhi Balmer jump se zyaada gehra bana deta hai.

Kisi series ki series limit uski sabse lambi-wavelength wali line hoti hai.

Jhooth — limit

n_{2} \to \infty

hai, jo sabse bada energy gap hai aur isliye sabse chhoti wavelength deta hai. Sabse lambi wavelength pehli line hoti hai,

n_{2} = n_{1} + 1

n = 5

n = 3

par girne wala electron ek Paschen line produce karta hai.

Sach — series ka naam final (landing) level se hota hai, yahan

n_{1} = 3

, jo Paschen landing strip hai; starting level

n_{2} = 5

sirf decide karta hai ki kaunsi Paschen line hogi.

Rydberg formula jo

R_{H}

use karta hai, woh singly-ionized helium (He

^{+}

) ke liye kaam karta hai.

Jaisa likha hai galat hai — He

^{+}

hydrogen-like hai jisme

Z = 2

hai, isliye tumhe

\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

use karna hoga; plain

R_{H}

form sirf

Z = 1

ke liye hai. Dekho Hydrogen-like Ions and Z dependence.

Thande hydrogen dwara light ka absorption usi wavelength par hota hai jis par emission hoti hai.

Sach — energy gaps same hote hain chahe electron upar jump kare (absorb kare) ya neeche aaye (emit kare), isliye wavelengths ek jaisi hoti hain; sirf transition ki direction alag hoti hai. Dekho Emission vs Absorption Spectra.

Hydrogen energy levels ke beech ke gaps

n

badhne ke saath bade hote jaate hain.

Jhooth — kyunki

E_{n} = - 13.6/ n^{2}

hai, levels upar jaake paas-paas aate jaate hain; gaps ionization edge ki taraf chhote hote jaate hain, isliye high-

n

transitions lambi-wavelength (kam-energy) wali light dete hain.

13.6

eV energy wala photon ek ground-state hydrogen atom ko

n = 2

par excite kar sakta hai.

Jhooth —

13.6

eV ionization energy hai (

n = 1

se free tak).

1 \to 2

gap

13.6 (1 - \frac{1}{4}) = 10.2

eV hai;

13.6

eV wala photon ionize kar dega, neatly

n = 2

par land nahi karega. Dekho Ionization Energy.

Galti dhundho

Har prompt mein ek mistake hai; reveal usse naam deta hai aur fix karta hai.

"Emission ke liye,

\frac{1}{λ} = R_{H} (\frac{1}{n _{2}^{2}} - \frac{1}{n _{1}^{2}})

jahan

n_{2} > n_{1}

Bracket ulta hai — upar wale level

n_{2} > n_{1}

ke saath ye negative hai, jo ek bakwaas negative

1/ λ

deta hai. Neeche wala (landing) level pehle rakhna chahiye:

\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} > 0

"Lyman series infrared mein hoti hai kyunki

n_{1} = 1

sabse chhote jumps deta hai."

Do galtiyan hain:

n_{1} = 1

par landing karna sabse bade jumps deta hai (sabse gehri landing), aur ye bade gaps ultraviolet dete hain, infrared nahi.

R_{H} \approx 1.097 \times 1 0^{7}

ki units metres hain."

Units galat hain —

R_{H}

ek wavenumber hai, isliye uski units

m^{- 1}

(lines per metre) hain. Ye

1/ λ

scale ke barabar hai, yaani ek length ka reciprocal.

"Kyunki

n = \infty

sabse bada quantum number hai, series limit sabse lambi wavelength deta hai."

Confusion

n

ki size aur gap ki size ko lekar hai.

n_{2} \to \infty

energy gap ko maximize karta hai, isliye ye series ki sabse chhoti wavelength deta hai, sabse lambi nahi.

"H-alpha (656 nm) ek Lyman line hai kyunki ye sabse bright visible red hai."

H-alpha ek Balmer line hai (

3 \to 2

n_{1} = 2

par landing). Lyman lines

n_{1} = 1

par land karti hain aur saari ultraviolet hain — aankhon ko dikhti nahi.

1.89

eV gap ko wavelength mein convert karne ke liye

λ (nm) = 1240 \times 1.89

use karo."

Relation ek division hai:

λ (nm) = 1240/Δ E (eV)

, kyunki

λ

aur

E

inversely related hain (

E = h c / λ

). Multiply karne se answer bahut zyaada badh jaata hai.

"Ground state

n = 1

mein baitha electron photon emit kar sakta hai."

Koi neeche wala level hi nahi hai jahan girega, isliye ground-state electron emit nahi kar sakta — woh sirf absorb karke upar ja sakta hai. Ground state staircase ka sabse neecha floor hai. Dekho Bohr Model of the Atom.

Why wale sawaal

Reveal mein causal chain hai, sirf restatement nahi.

Jab electron neeche girta hai toh released energy ka sign positive kyun hota hai?

n_{2}

initial (upar wala) level ke saath aur

n_{1}

final (neeche wala) level ke saath, neeche wala level zyaada negative hai (

E_{n} = - 13.6/ n^{2}

); emit hone wali energy

E_{n_{2}} - E_{n_{1}}

hai, aur kyunki

E_{n_{2}} > E_{n_{1}}

hai ye difference positive hai aur photon ke roop mein le jaaya jaata hai.

Rydberg formula

λ

directly ki jagah

1/ λ

kyun use karta hai?

Photon energy

E = h c / λ

hai, isliye energy

1/ λ

mein linear hai; kyunki energy

1/ n^{2}

terms ka ek saaf difference hai, ise

1/ λ

mein likhne se formula ek simple subtraction bana rehta hai. Dekho Photon Energy and Planck's Relation.

Ek series ki saari lines ek hi landing level share kyun karti hain, ek hi starting level nahi?

Final level

n_{1}

se group karne par woh "floor" fix ho jaata hai jis par photons build up hote hain; har allowed upar wala floor

n_{2}

ek line contribute karta hai, isliye series ek fan hai jo ek hi destination share karti transitions ka.

Hydrogen sirf discrete wavelengths par hi emit kyun kar sakta hai, ek continuous rainbow ki jagah?

Kyunki energy levels quantized hain (angular momentum quantized hai,

m v r = n ℏ

), sirf fixed energy gaps exist karte hain, isliye sirf fixed photon energies — hence fixed wavelengths — possible hain. Dekho Quantization of Angular Momentum.

Hydrogen ki ionization energy exactly

13.6

eV kyun hoti hai?

Ye energy hai electron ko

n = 1

(

E_{1} = - 13.6

eV) se uthaakar free (

E_{\infty} = 0

) tak le jaane ki; gap

0 - (- 13.6) = 13.6

eV hai. Dekho Ionization Energy.

Ek series ke spectral lines series limit ki taraf kyun bunch up ho jaati hain?

Kyunki

1/ n_{2}^{2}

starting level

n_{2}

badhne ke saath tezi se chhota hota hai, successive lines ke wavenumbers paas-paas aate jaate hain, limit value

R_{H} / n_{1}^{2}

par pile up hote hue jahan electron barely bound hota hai.

Edge cases

Is topic se jude boundary aur degenerate scenarios.

n_{2} = n_{1}

physically kis transition se correspond karta hai?

Kisi se nahi — "gap" zero hai, isliye

1/ λ = 0

aur koi photon emit nahi hota; electron levels ke beech move nahi kar raha, isliye koi line nahi hai.

Series limit (

n_{2} \to \infty

) ke liye, us energy ke thoda upar electron ka kya hota hai?

Woh ab bound nahi rehta —

0

se upar ki energies ek continuum form karti hain, isliye sharp line ki jagah continuous emission/absorption hoti hai; limit bound spectrum ke edge ko mark karti hai.

Kya "sabse lambi wavelength = pehli line" ka rule tab bhi kaama karta hai agar hum

n_{1} + 1

skip kar dein?

Actual sabse lambi line hamesha sabse chhote possible gap,

n_{2} = n_{1} + 1

, se aati hai; koi bhi bada starting level

n_{2}

bada gap aur chhoti wavelength deta hai, isliye rule "sabse chhota jump" ki definition se hi valid rehta hai.

Z = 2

wale hydrogen-like ion ke liye, kya series structure (Lyman/Balmer/Paschen landing levels) badal jaata hai?

Nahi — landing levels $n_{1}$ same rehte hain; sirf wavelengths

1/ Z^{2}

se scale hoti hain (chhoti ho jaati hain), kyunki

\frac{1}{λ}

Z^{2}

se multiply kiya jaata hai. Names/structure wahi rehte hain. Dekho Hydrogen-like Ions and Z dependence.

Kya ek single series mein kitni lines ho sakti hain uski koi upper limit hai?

Nahi — starting level

n_{2}

landing level

n_{1}

se upar koi bhi integer ho sakta hai, isliye infinitely many lines exist karti hain, series limit ke paas bheed jaati hain lekin kabhi actually pahunchti nahi.

Kya hydrogen mein do alag transitions kabhi exactly same wavelength de sakti hain?

Pure Rydberg (Bohr) spectrum mein nahi — har pair

(n_{1}, n_{2})

ek distinct

\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}

deta hai. (Real atoms mein, jab fine-structure/relativistic corrections add karo ya multi-electron atoms mein jaao, accidental near-overlaps ho sakti hain; "kabhi nahi" sirf idealized formula par apply hota hai.)

Connections

Bohr Model of the Atom — in traps mein use hone wale quantized energy levels ka source.
Photon Energy and Planck's Relation — kyun energy $1/ λ$ mein linear hai.
Emission vs Absorption Spectra — same wavelengths, opposite direction.
Ionization Energy — $13.6$ eV aur series-limit edge cases.
Hydrogen-like Ions and Z dependence — $Z^{2}$ scaling traps.
Quantization of Angular Momentum — spectrum discrete kyun hai.