Spectral series — Lyman, Balmer, Paschen: Visual walkthrough

We are going to answer one question:

Why does hydrogen only glow at a few fixed colours, and what fixes them?

Keep that question in mind. Every step below is a small move toward the answer.

Step 1 — Draw the energy staircase

WHAT. The electron inside a hydrogen atom cannot sit anywhere. It is only allowed to rest on certain "floors", labelled by a whole number $n = 1, 2, 3, \dots$ We call $n$ the principal quantum number — think of it simply as the floor number.

WHY. This "only-certain-floors" rule comes from Quantization of Angular Momentum inside the Bohr Model of the Atom: the electron's angular momentum can only be whole-number multiples of a tiny unit, and that forces the floors to be discrete. Nature does not offer a floor $n = 2.5$ .

PICTURE. Look at the figure. The floors are not evenly spaced — they crowd together as you climb. The bottom floor $n = 1$ is far below everything; higher floors squeeze toward a ceiling.

Figure — Spectral series — Lyman, Balmer, Paschen

Step 2 — Give each floor a number (its energy)

WHAT. We attach an energy value to every floor: $E_{n} = - \frac{13.6 eV}{n ^{2}}$

Let us read this symbol by symbol, right where each piece sits:

$E_{n}$ — the energy of floor number $n$ . Measured in electron-volts ( $eV$ ), a tiny energy unit.
the minus sign — says the electron is trapped (bound). It would need energy added to escape.
$13.6 eV$ — a fixed number, the Rydberg energy, the depth of the ground floor. (Its origin is the Ionization Energy of hydrogen.)
$n^{2}$ — the floor number squared. Because it sits in the denominator, bigger $n$ makes the fraction smaller in size, so higher floors are less negative — closer to the top.

WHY squared, and why negative? The $n^{2}$ is what makes the floors crowd together near the top (Step 1's picture): $E_{1} = - 13.6$ , $E_{2} = - 3.4$ , $E_{3} = - 1.51$ , $E_{4} = - 0.85$ — the gaps shrink. The negative sign is the "trapped below ground level" bookkeeping: level $n \to \infty$ gives $E = 0$ , the "free" ceiling.

PICTURE. Same staircase, now with a value written on every floor. The height of a floor above the bottom is its energy value.

Step 3 — The electron falls, and light is born

WHAT. An electron on a high floor $n_{2}$ can drop to a lower floor $n_{1}$ (with $n_{2} > n_{1}$ ). When it drops, the atom spits out one particle of light — a photon.

WHY a photon? Energy must be conserved. The electron loses energy by falling; that lost energy cannot vanish, so it leaves as light. This is exactly the Emission vs Absorption Spectra story: emission = falling and giving out light.

PICTURE. An arrow from the upper floor $n_{2}$ down to the lower floor $n_{1}$ , with a wavy photon shooting out sideways. The length of the drop is the energy of the photon.

Step 4 — Measure the drop (subtract the two floor-values)

WHAT. The energy released is the difference between the two floors: $Δ E = E_{n_{2}} - E_{n_{1}} = - 13.6 (\frac{1}{n _{2}^{2}} - \frac{1}{n _{1}^{2}}) eV .$

Reading it in place:

$Δ E$ — the "change in energy", the size of the drop (the arrow's length in Step 3).
$E_{n_{2}} - E_{n_{1}}$ — (energy at top) minus (energy at bottom). Literally "how much height did we lose".
pulling the common $- 13.6$ out front leaves the two fractions inside.

Now flip the order inside the bracket to kill the awkward minus (this is just algebra: $- (a - b) = b - a$ ): $Δ E = 13.6 (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) eV$

WHY flip it? Because $n_{1} < n_{2}$ means $\frac{1}{n _{1}^{2}} > \frac{1}{n _{2}^{2}}$ , so the bracket is positive. Energy comes out — a positive amount. If you ever get a negative number here, you subtracted in the wrong order (see the mistake box).

PICTURE. The two floor-heights drawn as two horizontal levels, and a bracket measuring the vertical gap between them — that gap is $Δ E$ .

Step 5 — Turn energy into a colour (wavelength)

WHAT. A photon's energy and its wavelength are locked together by Photon Energy and Planck's Relation: $E_{photon} = \frac{h c}{λ} .$

$λ$ — the wavelength, the length of one ripple of the light wave (in vacuum, as agreed at the top). Short $λ$ = blue/UV, long $λ$ = red/IR.
$h$ — Planck's constant, a fixed number of nature.
$c$ — the speed of light in vacuum, also fixed.
$h c$ — a fixed bundle; only $λ$ changes from photon to photon.

Notice: energy sits on top, $λ$ on the bottom. So more energy ⇒ shorter wavelength. That one fact explains the whole colour ordering.

WHY this tool and not another? We need a bridge from an energy (what we computed in Step 4) to a colour (what an eye or a spectrometer measures). Planck's relation is exactly that bridge — it is the only equation that converts energy directly into wavelength.

PICTURE. Two wavy lines: a tightly-packed high-energy wave (short $λ$ ) and a stretched-out low-energy wave (long $λ$ ), with the arrow " $E$ up ⇒ $λ$ down".

Step 6 — Glue the two halves together

WHAT. The energy released (Step 4) is the photon's energy (Step 5). Set them equal: $\frac{h c}{λ} = 13.6 eV (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) .$

Now divide both sides by $h c$ to get $1/ λ$ alone:

R_H \;=\; \frac{13.6\ \text{eV}}{hc}.$$ **Mind the units — turn eV into joules first.** The bundle $R_H = 13.6\ \text{eV}/(hc)$ only comes out in "per metre" if the top and bottom use the **same** energy unit. Planck's constant and $c$ live in SI ($h$ in $\text{J·s}$, $c$ in $\text{m/s}$, so $hc$ is in $\text{J·m}$), while $13.6$ is in eV. So we first convert using $1\ \text{eV} = 1.602\times10^{-19}\ \text{J}$: $$R_H \;=\; \frac{(13.6)(1.602\times10^{-19}\ \text{J})}{(6.626\times10^{-34}\ \text{J·s})(2.998\times10^{8}\ \text{m/s})} \;\approx\; 1.097\times10^{7}\ \text{m}^{-1}.$$ Watch the units cancel: $\dfrac{\text{J}}{\text{J·s}\cdot\text{m/s}}=\dfrac{\text{J}}{\text{J·m}}=\dfrac{1}{\text{m}}$. Clean "per metre", exactly a wavenumber. (This $1.097\times10^7$ already carries the reduced-mass correction from Step 2's footnote.) Reading the master equation in place: - $\dfrac{1}{\lambda}$ — the ==wavenumber== (ripples per metre, in vacuum). We keep $1/\lambda$ rather than $\lambda$ because energy is **linear** in $1/\lambda$, which keeps the formula clean. - $R_H$ — the **Rydberg constant**, the whole pile of fixed numbers ($13.6\,\text{eV}$ converted to joules, $h$, $c$) squashed into one. - $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ — the only part that changes: it depends purely on *which two floors* the electron jumps between. **WHY this is the answer to our question.** Because $n_1,n_2$ are whole numbers, the bracket can only take **certain discrete values**. So $1/\lambda$ — and therefore the colour — can only be certain discrete values. **That is why hydrogen glows at fixed colours.** For a heavier one-electron ion, multiply by $Z^2$ — see [[Hydrogen-like Ions and Z dependence]]. **PICTURE.** The two floors on the left feeding an arrow into the master formula, and the formula firing out a single sharp spectral line on a wavelength axis. ![[deepdives/dd-physics-2.3.15-d2-s06.png]] --- ## Step 7 — Group the jumps: the named series **WHAT.** Fix the **landing floor** $n_1$ and let *all* higher floors drop onto it. Each choice of $n_1$ gives one **family** of lines — a spectral series. - $n_1 = 1$: ==Lyman== (lands on the ground floor — deepest drops — **ultraviolet**). - $n_1 = 2$: ==Balmer== (lands on floor 2 — medium drops — **visible**, the ones we see). - $n_1 = 3$: ==Paschen== (lands on floor 3 — gentle drops — **infrared**). **WHY grouped this way?** Every line in one series shares the *same* $\dfrac{1}{n_1^2}$ term (same landing strip); only $n_2$ varies. That common landing energy is what makes them look like a family. **PICTURE.** The staircase with three coloured "landing strips" and arrows raining down onto each one, each strip labelled with its series name and light region. ![[deepdives/dd-physics-2.3.15-d2-s07.png]] --- ## Step 8 — The two edge cases of a series **WHAT.** Within one series, two special jumps bound everything: 1. **First line** — the *smallest* jump, $n_2 = n_1 + 1$. Smallest energy ⇒ **longest** $\lambda$. 2. **Series limit** — the *biggest* jump, $n_2 \to \infty$. Then $\dfrac{1}{n_2^2}\to 0$, so $$\frac{1}{\lambda} \;=\; \frac{R_H}{n_1^{2}} \quad\Rightarrow\quad \text{shortest } \lambda \text{ of the series.}$$ **WHY $n_2\to\infty$ gives the *shortest* wavelength (the sneaky one).** As $n_2$ climbs, the term $1/n_2^2$ shrinks toward $0$, so the bracket *grows* to its maximum value $1/n_1^2$. Bigger bracket ⇒ bigger $1/\lambda$ ⇒ **shortest** $\lambda$. Beyond the series limit the electron arrives from the "free" ceiling — it was essentially unbound (this ceiling is the [[Ionization Energy]] edge). **PICTURE.** One landing strip with two arrows marked: a tiny hop from just above (longest $\lambda$) and a giant plunge from the ceiling $\infty$ (shortest $\lambda$, the limit). ![[deepdives/dd-physics-2.3.15-d2-s08.png]] > [!mistake] "$n_2=\infty$ gives the longest wavelength" > **Why it feels right:** infinity feels like "the biggest λ". > **Fix:** infinity = **biggest energy gap** = **shortest** λ. Longest λ is the *tiny* hop $n_2=n_1+1$. --- ## The one-picture summary The whole derivation in a single frame: staircase → drop → photon → glued formula → a real spectrum. ![[deepdives/dd-physics-2.3.15-d2-s09.png]] > [!formula] Everything on one line > $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\quad n_2>n_1,\quad R_H\approx1.097\times10^7\ \text{m}^{-1}\ (\lambda \text{ in vacuum}).$$ > [!recall]- Feynman: tell the whole story in plain words > There is a staircase inside the atom, and the steps get squished together near the top. A tiny ball > (the electron) can only stand on the steps, never between them — that is the quantum rule. Every step > has a "depth" number, most negative at the bottom. When the ball hops **down**, the depth it loses > flies off as a flash of light. A big drop makes a high-energy flash (short, bluish/UV wavelength); a > small drop makes a gentle flash (long, reddish/IR wavelength). Because the steps are at *fixed* > heights, the drops are of *fixed* sizes, so the flashes are of *fixed* colours — that is why hydrogen > always glows the same colours (measured in vacuum). If we sort all the flashes by **which step they > land on**, we get the families: land on step 1 → Lyman (UV), step 2 → Balmer (the visible ones), > step 3 → Paschen (IR). In each family, the smallest hop makes the longest wavelength, and a plunge > from the very top (the ceiling) makes the shortest — the series limit, right at the edge of the ball > breaking free. One tiny honesty: the proton is not truly nailed down, so the exact staircase numbers > use a "reduced mass" — a $0.05\%$ nudge already inside the measured $13.6\ \text{eV}$ and $R_H$. --- ## Active recall Which quantum number labels the floors ::: $n$, the principal quantum number ($n=1,2,3,\dots$) Why is $E_n$ negative ::: the electron is bound (trapped); energy must be added to free it Why put the landing level $n_1$ first in the bracket ::: so $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ is positive (emission releases energy) Which relation converts energy to wavelength ::: $E=hc/\lambda$ (Planck) Why keep $1/\lambda$ instead of $\lambda$ ::: energy is linear in $1/\lambda$, keeping the formula clean What unit conversion is needed to compute $R_H$ ::: convert $13.6\ \text{eV}$ to joules ($1\ \text{eV}=1.602\times10^{-19}\ \text{J}$) so J cancels in $13.6\,\text{eV}/(hc)$ What is $R_H$ physically ::: the bundle $13.6\,\text{eV}/(hc)\approx1.097\times10^7\ \text{m}^{-1}$ Do the wavelengths refer to air or vacuum ::: vacuum wavelengths Why is the value $13.6$ eV slightly below the infinite-mass Bohr value ::: reduced-mass correction (finite proton mass), about $0.05\%$ Which jump gives the longest wavelength in a series ::: the smallest hop, $n_2=n_1+1$ Which jump gives the shortest wavelength (series limit) ::: $n_2\to\infty$, giving $1/\lambda=R_H/n_1^2$ --- ## Connections - [[Bohr Model of the Atom]] — where the quantized floors come from. - [[Quantization of Angular Momentum]] — the rule that forces discrete floors. - [[Photon Energy and Planck's Relation]] — the energy↔wavelength bridge (Step 5). - [[Emission vs Absorption Spectra]] — falling = emission (Step 3). - [[Ionization Energy]] — the $E=0$ ceiling and the series limit. - [[Hydrogen-like Ions and Z dependence]] — generalises $R_H\to R_H Z^2$. - [[2.3.15 Spectral series — Lyman, Balmer, Paschen (Hinglish)|Yeh note Hinglish mein padho →]]