Spectral series — Lyman, Balmer, Paschen

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What we are explaining

• WHAT: Hydrogen emits/absorbs light only at certain discrete wavelengths, organized into named series.
• WHY: Energy levels are quantized; transitions between them are quantized; therefore wavelengths are quantized.
• HOW: Derive the energy levels, take a difference, convert to wavelength.

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Derivation from scratch (Bohr model)

Where does $13.6$ eV come from? Bohr balanced Coulomb attraction with the centripetal need, then quantized angular momentum $m v r = n\hbar$. The result is $E_n = -\frac{m e^4}{8\varepsilon_0^2 h^2}\cdot\frac{1}{n^2}.$ Plugging constants gives the prefactor $13.6$ eV (the Rydberg energy). You don't need to memorize the constant pile — memorize that it's $-13.6/n^2$.

Now the photon. An electron drops from level $n_2$ (high) to $n_1$ (low), $n_2 > n_1$:

= 13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \text{eV}.$$ This is **positive** (energy is released as a photon). **Why this step?** Lower level is *more negative*, so going down *increases* the magnitude → energy comes out. A photon carries $E_\text{photon} = h\nu = \dfrac{hc}{\lambda}$. Set them equal: $$\frac{hc}{\lambda} = 13.6\,\text{eV}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right).$$ > [!formula] Rydberg formula (the master equation) > $$\boxed{\ \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ }\qquad n_2>n_1$$ > with the **Rydberg constant** > $$R_H = \frac{13.6\ \text{eV}}{hc} \approx 1.097\times10^{7}\ \text{m}^{-1}.$$ > $1/\lambda$ is the **wavenumber** (lines per metre). Bigger gap ⇒ bigger $1/\lambda$ ⇒ **shorter** wavelength. **Why use $1/\lambda$?** Because the energy is *linear* in $1/\lambda$, the formula stays clean. --- ## The named series (grouped by the landing level $n_1$) | Series | $n_1$ (final) | $n_2$ (from) | Region | Why that region | |--------|------|-----|--------|-----------------| | ==Lyman== | 1 | 2,3,4,… | ==Ultraviolet== | Deepest landing ⇒ biggest gaps ⇒ shortest $\lambda$ | | ==Balmer== | 2 | 3,4,5,… | ==Visible== | Medium gaps ⇒ visible light (the ones we *see*) | | ==Paschen== | 3 | 4,5,6,… | Infrared | Shallow gaps ⇒ long $\lambda$ | | Brackett | 4 | 5,6,… | Infrared | Even shallower | | Pfund | 5 | 6,7,… | Far infrared | Shallowest | ![[2.3.15-Spectral-series-—-Lyman,-Balmer,-Paschen.png]] > [!intuition] Series limit & first line > - The **first line** (smallest gap) of a series uses $n_2 = n_1+1$ → **longest** $\lambda$ in that series. > - The **series limit** uses $n_2 \to \infty$ → $\frac{1}{\lambda}=R_H/n_1^2$ → **shortest** $\lambda$ (highest energy) of that series. Beyond it the electron is free (ionized). --- ## Worked examples > [!example] 1 — Longest wavelength of the Lyman series > Lyman ⇒ $n_1=1$. Longest $\lambda$ ⇒ smallest gap ⇒ $n_2=2$. > $$\frac{1}{\lambda}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=1.097\times10^7\left(1-\tfrac14\right) > =1.097\times10^7\cdot\tfrac34=8.23\times10^6\ \text{m}^{-1}.$$ > $$\lambda = \frac{1}{8.23\times10^6}=1.215\times10^{-7}\ \text{m}=121.5\ \text{nm (UV)}.$$ > **Why $n_2=2$?** Longest wavelength = lowest energy = smallest jump = nearest higher level. > [!example] 2 — The red H-alpha line (Balmer, $3\to2$) > $$\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) > =1.097\times10^7\left(\tfrac14-\tfrac19\right)=1.097\times10^7\cdot\tfrac{5}{36}=1.524\times10^6\ \text{m}^{-1}.$$ > $$\lambda = 6.56\times10^{-7}\ \text{m}=656\ \text{nm}\ \text{(red, visible)}.$$ > **Why this matters:** This is the famous deep-red line you actually see in a hydrogen lamp. > [!example] 3 — Series limit of the Paschen series > Paschen ⇒ $n_1=3$, limit ⇒ $n_2\to\infty$ so $1/n_2^2\to0$: > $$\frac{1}{\lambda}=R_H\cdot\frac{1}{9}=1.219\times10^6\ \text{m}^{-1}\Rightarrow \lambda = 820\ \text{nm (infrared)}.$$ > **Why this step?** Series limit = electron arriving from the very edge of being free; gives the highest-energy photon of that series. > [!example] 4 — Energy of a Balmer photon in eV > Use $\Delta E = 13.6(\tfrac1{4}-\tfrac1{9}) = 13.6\cdot\tfrac{5}{36}=1.89\ \text{eV}$. > Check: $\lambda=\frac{1240\ \text{eV·nm}}{1.89\ \text{eV}}=656\ \text{nm}$. ✓ Matches Example 2. > **Why two methods agree:** Both use the same energy gap; eV-route and $R_H$-route are the same physics. --- ## Forecast-then-Verify > [!recall]- Forecast before you compute > Q: Without computing, which has the **shorter** wavelength — Lyman $2\to1$ or Balmer $3\to2$? > > Forecast: Lyman jumps land deeper ⇒ bigger energy gap ⇒ shorter $\lambda$. > Verify: Lyman = 121.5 nm, Balmer = 656 nm. **Lyman is shorter.** ✓ --- ## Common mistakes (Steel-man + fix) > [!mistake] "Swap $n_1$ and $n_2$" — getting a negative $1/\lambda$ > **Why it feels right:** People write the formula by reflex and put the from-level first. > **The trap:** $\frac{1}{n_2^2}-\frac{1}{n_1^2}$ is negative for emission. > **Fix:** For **emission** wavelengths, always **lower level first**: $\frac{1}{n_1^2}-\frac{1}{n_2^2}>0$. > [!mistake] "Balmer is the first/lowest series" > **Why it feels right:** Balmer is taught first historically because it's *visible*. > **Fix:** Lyman ($n_1=1$) is the lowest, most energetic series. Balmer is just the one our eyes catch. > [!mistake] "Longest wavelength uses $n_2=\infty$" > **Why it feels right:** $\infty$ feels like "the biggest" answer. > **Fix:** $n_2=\infty$ gives the **biggest energy** = **shortest** $\lambda$ (the series *limit*). Longest $\lambda$ = smallest jump = $n_2=n_1+1$. > [!mistake] "Rydberg formula works for any atom" > **Fix:** This $R_H$ form is for **hydrogen (one electron)**. For a hydrogen-like ion of charge $Z$, multiply by $Z^2$: $\frac1\lambda=R_H Z^2(\frac1{n_1^2}-\frac1{n_2^2})$. --- ## Active recall #flashcards/physics Hydrogen energy level formula ::: $E_n=-13.6/n^2$ eV Rydberg formula for wavenumber ::: $1/\lambda=R_H(1/n_1^2-1/n_2^2)$, $n_2>n_1$ Value of Rydberg constant $R_H$ ::: $1.097\times10^7\ \text{m}^{-1}$ Lyman series final level and region ::: $n_1=1$, ultraviolet Balmer series final level and region ::: $n_1=2$, visible Paschen series final level and region ::: $n_1=3$, infrared Which transition gives the LONGEST wavelength in a series ::: $n_2=n_1+1$ (smallest gap) Series limit transition ::: $n_2\to\infty$, gives shortest wavelength of the series Wavelength of Lyman first line ::: 121.5 nm (UV) Wavelength of H-alpha (Balmer 3→2) ::: 656 nm (red) Why is the sign positive in emission ::: Lower level is more negative; dropping releases positive energy How to convert eV gap to nm quickly ::: $\lambda(\text{nm})=1240/\Delta E(\text{eV})$ Formula for hydrogen-like ion of charge Z ::: $1/\lambda=R_H Z^2(1/n_1^2-1/n_2^2)$ --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a staircase where the bottom steps are far apart and the top steps are squished together. > A ball (the electron) sits on some step. When it jumps **down**, it gives off a flash of light. > Big jumps near the bottom give **bright, high-energy** flashes (Lyman). Jumps that land on the > 2nd step give **visible** colours we can see (Balmer). Jumps landing on the 3rd step are > **gentle**, giving invisible heat-light (Paschen, infrared). Group all the flashes by **which > step they land on**, and you've got the spectral series. The colour is fixed because the step > heights are fixed — that's why hydrogen always glows the same colours! > [!mnemonic] Series order & landing level > **"L**arry **B**uys **P**ie **B**efore **P**arty**"** → **L**yman(1), **B**almer(2), **P**aschen(3), **B**rackett(4), **P**fund(5). > And: **L**yman→**UV**, **B**almer→**V**isible, **P**aschen→**I**R = "UV-Vis-IR going up the stairs." --- ## Connections - [[Bohr Model of the Atom]] — supplies the quantized energy levels. - [[Photon Energy and Planck's Relation]] — $E=hc/\lambda$ links gap to wavelength. - [[Hydrogen-like Ions and Z dependence]] — generalizes $R_H \to R_H Z^2$. - [[Ionization Energy]] — series limit = the energy to free the electron from level $n_1$. - [[Emission vs Absorption Spectra]] — same lines, opposite direction of jump. - [[Quantization of Angular Momentum]] — the assumption that makes levels discrete. ## 🖼️ Concept Map ```mermaid flowchart TD Q[Quantized energy levels] -->|En equals -13.6 over n squared| E[Energy of level n] B[Bohr model] -->|derived from| Q E -->|take difference n2 to n1| D[Energy gap Delta E] D -->|released as| P[Photon] P -->|E equals hc over lambda| R[Rydberg formula] D -->|set equal to hc over lambda| R R -->|1 over lambda equals RH times gap| W[Wavenumber and wavelength] R -->|group by final level n1| S[Spectral series] S -->|n1 equals 1| L[Lyman ultraviolet] S -->|n1 equals 2| BA[Balmer visible] S -->|n1 equals 3| PA[Paschen infrared] L -->|biggest gaps give| SW[Shorter wavelength] PA -->|shallow gaps give| LW[Longer wavelength] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, hydrogen atom ke andar electron sirf kuch fixed "floors" (energy levels) pe hi reh sakta > hai, jaha har floor ki energy $E_n=-13.6/n^2$ eV hoti hai. Jab electron upar wale floor se neeche > aata hai, to dono floors ki energy ka difference ek photon (light ka packet) banke bahar nikal > jata hai. Photon ki energy fix hai, isliye uska wavelength bhi fix hai — yahi reason hai ki > hydrogen hamesha same colours emit karta hai. > > Ab saari jumps ko unke **landing floor** ke hisaab se group kar do, to series ban jaati hain. > Agar electron floor 1 pe land karta hai → **Lyman** series (UV, sabse zyada energy). Floor 2 pe → > **Balmer** (visible light, jo hum aankhon se dekhte hain). Floor 3 pe → **Paschen** (infrared, > invisible heat). Master formula yaad rakho: $\frac1\lambda=R_H(\frac1{n_1^2}-\frac1{n_2^2})$, > jaha $n_1$ landing floor hai aur $n_2$ upar wala. $R_H=1.097\times10^7$ per metre. > > Do important traps: (1) Emission ke liye hamesha **chhota level pehle** likho warna answer > negative aa jayega. (2) "Longest wavelength" ka matlab **sabse chhoti jump** ($n_2=n_1+1$), > aur "series limit" ka matlab $n_2=\infty$ jo **sabse chhota wavelength** deta hai. Quick trick: > energy gap ko eV me nikaalo aur $\lambda(\text{nm})=1240/\Delta E$ laga do — fatafat answer. ![[audio/2.3.15-Spectral-series-—-Lyman,-Balmer,-Paschen.mp3]]