Exercises — Spectral series — Lyman, Balmer, Paschen

Hum kya pooch rahe hain: kaunsi family ek visible line ki owner hai. Kyun: har series apne landing floor $n_1$ se define hoti hai, aur sirf ek series visible band mein land karti hai.

• $121.5\text{nm}$ region = ultraviolet = Lyman ($n_1=1$).
• $\approx 400$–$700\text{nm}$ = visible = Balmer ($n_1=2$).
• $>800\text{nm}$ = infrared = Paschen ($n_1=3$). Kyunki $434\text{nm}$ visible hai, yeh Balmer series hai, landing level $n_1=2$. (Yeh actually $5\to 2$ line hai, jise H-$\gamma$ kehte hain.)

Idea: chhoti wavelength ⇔ bada energy gap ⇔ gehri landing floor. Floors $n$ badhne ke saath ek-doosre ke kareeb aati jaati hain (spacing $1/n^2$ ki tarah shrink karti hai), toh floor 1 par land karna sabse bada jump hai, floor 3 par land karna sabse chhota.

• $2\to 1$: floor 1 par land karta hai → sabse bada gap → shortest $\lambda$.
• $3\to 2$: floor 2 par land karta hai → medium.
• $4\to 3$: floor 3 par land karta hai → sabse chhota gap → longest $\lambda$. Order (short → long): $2\to 1,3\to 2,4\to 3$.

Kya: $n_1=2$, $n_2=5$ ko Rydberg mein plug karo. Kyun ye numbers: Balmer ⇒ landing floor $n_1=2$; "5" starting floor hai. $\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{25}\right).$ $\frac{1}{4}-\frac{1}{25}=\frac{25-4}{100}=\frac{21}{100}=\mathrm{0.21.}$ $\frac{1}{\lambda }=1.097\times 10^7\times 0.21=2.304\times 10^6{\text{m}}^{-1}.$ $\lambda =\frac{1}{2.304\times 10^6}=4.34\times 10^{-7}\text{m}=434\text{nm}.$ Exercise 1.1 mein observe ki gayi blue-green line se match karta hai. ✓

Kya: energy-difference form use karo. Kyun: isse hum seedha eV mein answer de sakte hain, phir nm mein bridge kar sakte hain. $\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=13.6\left(1-\frac{1}{9}\right)=13.6\times \frac{8}{9}=12.09\text{eV}.$ Wavelength mein bridge: $\lambda =\frac{1240}{12.09}=102.5\text{nm (ultraviolet)}.$ Gehri landing (floor 1) + oocha start (floor 3) ⇒ bada gap ⇒ chhoti UV wavelength, bilkul expected ki tarah.

Hum kya prove karte hain: sabse lambi Lyman wavelength bhi sabse chhoti Balmer wavelength se chhoti hai. Agar worst-case Lyman line best-case Balmer line ko beat karti hai, toh bands touch nahi kar sakte. Kyun yeh pairing: kisi series mein, sabse lambi $\lambda$ ↔ sabse chhota wavenumber (pehli line, $n_2=n_1+1$); sabse chhoti $\lambda$ ↔ sabse bada wavenumber (series limit, $n_2\to \infty$). Hum sabse chhote Lyman wavenumber ko sabse bade Balmer wavenumber se compare karte hain — agar sabse chhota Lyman bhi sabse bade Balmer se zyada hai, toh ho gaya.

• Sabse chhota Lyman wavenumber ($2\to 1$, uski pehli line): $\frac{1}{\lambda }=R_H\left(1-\frac{1}{4}\right)=\frac{3}{4}{R}_H=0.75R_H.$
• Sabse bada Balmer wavenumber (series limit, $n_2\to \infty$, toh $1/n_2^2\to 0$): $\frac{1}{\lambda }=R_H\left(\frac{1}{4}-0\right)=\frac{1}{4}{R}_H=0.25R_H.$ Kyunki $0.75R_H>0.25R_H$, sabse chhota Lyman wavenumber bhi sabse bade Balmer wavenumber se zyada hai. Bada wavenumber ⇔ chhoti wavelength, toh poora Lyman band poore Balmer band se chhoti wavelengths par hai. Woh kabhi overlap nahi karte. ∎

Kya: Paschen band ke dono edges nikalo — uski sabse lambi aur sabse chhoti wavelength. Kyun hum sabse lambi line ke liye $n_2=4$ chunte hain: sabse lambi wavelength matlab sabse kam photon energy, jo sabse chhote energy gap se aati hai; $n_1=3$ par land karne wali series ke liye sabse chhota gap ek fall bilkul agli floor se hai, $n_2=n_1+1=4$.

Sabse lambi line = sabse chhota jump = $n_2=4$: $\frac{1}{\lambda }=R_H\left(\frac{1}{9}-\frac{1}{16}\right)=1.097\times 10^7\times \frac{16-9}{144}=1.097\times 10^7\times \frac{7}{144}.$ $=1.097\times 10^7\times 0.04861=5.333\times 10^5{\text{m}}^{-1}\Rightarrow \lambda =1875\text{nm}.$

Kya: ab doosra edge. Kyun $n_2\to \infty$: sabse chhoti wavelength = sabse bada energy gap = electron bilkul free hone ki kagar se aa raha hai, yaani $n_2\to \infty$ toh $1/n_2^2\to 0$. $\frac{1}{\lambda }=R_H\times \frac{1}{9}=1.219\times 10^6{\text{m}}^{-1}\Rightarrow \lambda =820\text{nm}.$ Toh Paschen lines roughly 820 nm (limit, sabse chhoti) se 1875 nm (pehli line, sabse lambi) tak hain — sab infrared.

Strategy: $2.55\text{eV}$ itna chhota hai ki visible ho sakta hai, toh Balmer ($n_1=2$) guess karo aur $n_2$ solve karo. $2.55=13.6\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\Rightarrow \frac{2.55}{13.6}=0.25-\frac{1}{n_2^2}.$ $0.1875=0.25-\frac{1}{n_2^2}\Rightarrow \frac{1}{n_2^2}=0.0625=\frac{1}{16}\Rightarrow n_2=4.$ Transition: $4\to 2$ (Balmer H-$\beta$ line). Check: $\lambda =1240/2.55=486\text{nm}$, blue — correct.

Tool: nuclear charge $Z$ wale hydrogen-like ion ke liye, har wavenumber $Z^2$ se scale hota hai (dekho Hydrogen-like Ions and Z dependence): $\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right).$ Kyun $Z^2$: electron $Z$ guna nuclear pull feel karta hai, aur energy levels $Z^2$ se gehre hote hain, toh gaps — hence wavenumbers — $Z^2$ se badhte hain. Same $n$'s ⇒ same bracket, toh wavenumber $Z^2=9$ guna bada hai, matlab wavelength $9$ guna chhoti hai: ${\lambda }_{{\text{Li}}^{2+}}=\frac{121.5\text{nm}}{9}=13.5\text{nm (extreme UV / soft X-ray)}.$

Lines count karna: $\{1,2,3,4\}$ mein levels ke har downward pair se ek line milti hai. 4 mein se 2 floors choose karne ke tarike hain $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{4\times 3}{2}=6\text{lines}.$ List karo: $4\to 3,4\to 2,4\to 1,3\to 2,3\to 1,2\to 1$. Kaun si Balmer (visible) hain? Jo $n_1=2$ par land karti hain: $3\to 2$ aur $4\to 2$. Toh 6 mein se 2 lines visible hain; baaki $2\to 1,3\to 1,4\to 1$ (Lyman, UV) aur $4\to 3$ (Paschen, IR) hain.

$\frac{1}{\lambda }=R_H\left(\frac{1}{4}-\frac{1}{16}\right)=1.097\times 10^7\times \frac{4-1}{16}=1.097\times 10^7\times \frac{3}{16}.$ $=1.097\times 10^7\times 0.1875=2.057\times 10^6{\text{m}}^{-1}\Rightarrow \lambda =486\text{nm (blue, visible)}.$ Haan — yeh exactly H-$\beta$ line hai, $n=4$ cascade ki do Balmer lines mein se ek (doosri $3\to 2$ hai $656\text{nm}$ par, H-$\alpha$).

Kya: Lyman series limit $n_1=1$, $n_2\to \infty$ hai. $\frac{1}{\lambda }=R_H\left(\frac{1}{1}-0\right)=R_H=1.097\times 10^7{\text{m}}^{-1}\Rightarrow \lambda =91.2\text{nm}.$ Energy mein convert karo: $E=\frac{1240\text{eV⋅nm}}{91.2\text{nm}}=13.6\text{eV}.$ Yeh us energy ke barabar hai jo electron ko ground state ($n=1$) se freedom ($n=\infty$) tak uthane ke liye chahiye — yaani ionization energy. Series limit wahi hai jo ionization threshold hai, light ki side se dekha hua. ∎

• 1.1 ::: Balmer, $n_1=2$
• 1.2 ::: order $2\to 1,3\to 2,4\to 3$
• 2.1 ::: $434\text{nm}$
• 2.2 ::: $12.09\text{eV}$, $102.5\text{nm}$
• 3.1 ::: no overlap ($0.75R_H$ vs $0.25R_H$)
• 3.2 ::: $820$–$1875\text{nm}$
• 4.1 ::: $4\to 2$
• 4.2 ::: $13.5\text{nm}$
• 5.1 ::: chhe lines, 2 visible
• 5.2 ::: $486\text{nm}$
• 5.3 ::: $91.2\text{nm}=13.6\text{eV}$