Bohr atomic model and electron shells: Worked examples

Before we start, three things we reuse everywhere:

The scenario matrix

Every Bohr-model problem falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What varies	Watch out for
A Radius, small $n$	plug into $r_{n} = a_{0} n^{2}$	quadratic growth
B Radius, large $n$ / limit	$n \to \infty$	orbit $\to \infty$ , atom "unbound"
C Energy of one level	$E_{n} = - 13.6/ n^{2}$	sign is negative
D Emission jump (down)	$n_{i} > n_{f}$	photon energy positive, atom loses
E Absorption jump (up)	$n_{i} < n_{f}$	atom gains, needs incoming photon
F Ionization	electron to $E = 0$	energy to fully free it
G Hydrogen-like ion ( $Z > 1$ )	He $^{+}$ , Li $^{2 +}$	extra factor $Z^{2}$
H Wavelength / colour	$λ = h c / E$	convert eV $\to$ nm
I Shell filling (Hardware)	shell capacity	it's a max, count valence
J Word / exam twist	reasoning	null-transition & trap cases

We now walk them.

Cell A + B — Radii, from smallest to the limit

Worked example Example 1 (Cell A, B) — Radius at

$n = 1, 2, 4$ , and what happens as $n \to \infty$ Find $r_{1}, r_{2}, r_{4}$ for hydrogen. Then say what $r_{n}$ does as $n$ grows without bound.

Forecast: Guess the ratio $r_{4} / r_{1}$ before reading. Is it 4? (It is not.)

$n = 1$ : $r_{1} = a_{0} \cdot 1^{2} = 0.529$ Å. Why this step? $n = 1$ is the definition of the Bohr radius — the anchor everything scales from.
$n = 2$ : $r_{2} = a_{0} \cdot 2^{2} = 0.529 \times 4 = 2.116$ Å. Why this step? The $n^{2}$ in $r_{n} = a_{0} n^{2}$ means we multiply by $4$ , not $2$ .
$n = 4$ : $r_{4} = a_{0} \cdot 4^{2} = 0.529 \times 16 = 8.464$ Å. Why this step? Same rule, $4^{2} = 16$ . So $r_{4} / r_{1} = 16$ , not $4$ — that's the surprise.
Limit $n \to \infty$ : $r_{n} = a_{0} n^{2} \to \infty$ . Why this step? As $n$ grows the orbit balloons quadratically; there is no largest orbit. Physically the electron is drifting arbitrarily far from the nucleus — the edge of being free.

Verify: $r_{4} / r_{1} = 16/1 = 16 = 4^{2}$ . Units: Å times a pure number stays Å. ✔

Figure 1 (below) — nested Bohr orbits. Four circles centred on the pink nucleus, coloured yellow, blue, pink, off-white for $n = 1, 2, 3, 4$ , each labelled with its radius in ångströms. Notice how each circle jumps out much farther than the last — the gaps widen, because the radius grows as $n^{2}$ , not evenly.

Figure — Bohr atomic model and electron shells

Cell C — Energy of a single level (mind the sign)

Worked example Example 2 (Cell C) — Energies of

$n = 1, 2, 3$ and the sign trap Compute $E_{1}, E_{2}, E_{3}$ for hydrogen and state which is "lowest".

Forecast: Which level is lowest in energy — $n = 1$ or $n = 3$ ? (Careful: lowest means most negative.)

$E_{1} = - 13.6/ 1^{2} = - 13.6$ eV. Why this step? Direct plug into $E_{n} = - 13.6/ n^{2}$ ; the $n = 1$ value is just $- 13.6$ .
$E_{2} = - 13.6/ 2^{2} = - 13.6/4 = - 3.40$ eV. Why this step? $2^{2} = 4$ , divide.
$E_{3} = - 13.6/ 3^{2} = - 13.6/9 = - 1.51$ eV. Why this step? $3^{2} = 9$ , divide (this rounds to $- 1.51$ eV).
Order them: $- 13.6 < - 3.40 < - 1.51$ , so $E_{1}$ is the lowest (most negative), $E_{3}$ the highest of the three. Why this step? On the number line, "more negative" is farther left = lower. The ground state is the deepest well.

Verify: All negative (bound ✔). As $n$ rises the values climb toward $0$ : $- 13.6 \to - 3.4 \to - 1.51$ , a rising sequence. ✔

Figure 2 (below) — the energy staircase. Horizontal blue rungs at $E_{1}, E_{2}, \dots$ with a dashed yellow line at $E = 0$ marked "free / ionized". The rungs are drawn to scale: they crowd together as they approach $0$ , showing exactly why higher levels sit closer and closer to freedom.

Cell D — Emission (electron falls, light comes out)

Worked example Example 3 (Cell D) — The

$3 \to 2$ jump (H-alpha, red) An electron falls from $n_{i} = 3$ to $n_{f} = 2$ . Find the photon energy.

Forecast: Will the answer be bigger or smaller than the whole $13.6$ eV binding? (Much smaller — it's a small hop, not a full escape.)

Both levels: $E_{3} = - 1.51$ eV, $E_{2} = - 3.40$ eV (from Example 2). Why this step? A jump needs the two endpoints.
Photon energy = energy lost: $Δ E = E_{i} - E_{f} = E_{3} - E_{2} = - 1.51 - (- 3.40) = 1.889$ eV. Why this step? The electron drops ( $n_{i} > n_{f}$ ), so — per the sign convention above — it sheds energy and emits a photon. Subtracting a more-negative number gives a positive photon energy, exactly as the convention predicts.
Interpret: $1.89$ eV is red visible light (the famous H-alpha line). Why this step? Ties the number to something you can see, which is the whole point of Bohr's model.

Verify: Using the jump formula $13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) = 13.6 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} = 1.889$ eV. Matches ✔. Positive ✔.

Figure 3 (below) — the emission jump. The same energy rungs for $n = 1, 2, 3$ , with a yellow downward arrow from the $n = 3$ rung to the $n = 2$ rung and a pink arrow labelled "photon out, 1.89 eV (red)" leaving the atom. The arrow's length is the energy gap you just computed.

Cell E — Absorption (electron climbs, light goes in)

Worked example Example 4 (Cell E) — The

$1 \to 2$ jump (must absorb) A ground-state electron is kicked up from $n = 1$ to $n = 2$ . What energy must arrive, and in what form?

Forecast: Compared to the $3 \to 2$ emission, is this energy bigger or smaller?

Endpoints: $E_{1} = - 13.6$ eV, $E_{2} = - 3.40$ eV. Why this step? Same two-endpoint recipe.
Energy needed: $Δ E = E_{f} - E_{i} = E_{2} - E_{1} = - 3.40 - (- 13.6) = 10.2$ eV. Why this step? The electron climbs ( $n_{i} < n_{f}$ ), so by the sign convention the atom must swallow energy. We take (higher $-$ lower) to get a positive amount absorbed — the reverse subtraction from emission; direction matters.
Form: an incoming photon of exactly $10.2$ eV (ultraviolet, Lyman-alpha) must be absorbed. Why this step? Bohr's rule allows only exact matches; a $9$ eV or $11$ eV photon is ignored.

Verify: Jump formula $13.6 (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = 13.6 \times \frac{3}{4} = 10.2$ eV ✔. It's bigger than the $3 \to 2$ gap ( $1.89$ eV) because the lowest rungs are the widest ✔.

Figure 4 (below) — the absorption jump. The same energy rungs for $n = 1, 2, 3$ , with a blue upward arrow from the $n = 1$ rung to the $n = 2$ rung and a yellow arrow labelled "photon in, 10.2 eV (UV)" striking the atom. Compare with Figure 3: the arrow points the other way and is longer — climbing from the deepest rung costs the most.

Cell F — Ionization (all the way to freedom)

Before this example, one symbol to earn:

Worked example Example 5 (Cell F) — Ionize hydrogen from

$n = 1$ and from $n = 2$ How much energy frees the electron completely (up to $E_{\infty} = 0$ ) starting from $n = 1$ ? And starting from $n = 2$ ?

Forecast: Ionizing from a higher shell — easier or harder than from the ground state?

Target is $E_{\infty} = 0$ : the electron must be lifted from its level all the way up to the free mark. Why this step? "Fully free" means reaching $E_{\infty} = 0$ , the top of the well we just defined.
From $n = 1$ : ionization energy $= E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6$ eV. Why this step? You must pay back the full depth of the ground state.
From $n = 2$ : $= E_{\infty} - E_{2} = 0 - (- 3.40) = 3.40$ eV. Why this step? The electron already sits shallower, so less to pay.

Verify: Ground-state ionization energy of hydrogen is the textbook $13.6$ eV ✔. Higher shell is easier ( $3.40 < 13.6$ ) — matches intuition ✔.

Cell G — Hydrogen-like ion (one electron, charge $Z$ )

Definition The five derivation symbols (from the parent note)

The intuition box below re-uses the exact symbols the parent note built in its "HOW to derive" section. So there is no undefined notation, here is what each one means, tied to the physical picture:

$e$ = the magnitude of the electron's charge (the proton carries $+ e$ , the electron $- e$ ).
$m$ = the mass of the electron (the light thing doing the orbiting).
$v$ = the electron's orbital speed (how fast it goes round the circle).
$ε_{0}$ = the permittivity of free space, the constant in Coulomb's law that sets how strong the electric pull is.
$ℏ$ = the reduced Planck constant ( $h /2 π$ ), the tiny unit of angular momentum that Bohr's quantization rule $m v r = n ℏ$ counts in.

These are exactly the quantities the parent note used in its Step 1 (force balance) and Step 2 (quantization). We only reason with them below — every final formula on this page still reduces to the two anchor numbers $a_{0}$ and $13.6$ eV.

Intuition Why the radius shrinks as

$1/ Z$ (earn the formula) Recall the two rules from the parent derivation, now written with the symbols just defined:

Force balance: Coulomb pull = centripetal need, $\frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r ^{2}} = \frac{m v ^{2}}{r}$ . With nuclear charge $Z e$ , the pull is $Z$ times stronger.
Quantization: $m v r = n ℏ$ (unchanged — this rule is about angular momentum, not charge).

Solve exactly as the parent note did but carry the extra $Z$ : eliminating $v$ gives $r_{n} = \frac{4 π ε _{0} ℏ ^{2}}{m Z e ^{2}} n^{2} = \frac{a _{0} n ^{2}}{Z}$ . The single factor of $Z$ lands in the denominator because the stronger pull reels the orbit in — and it's linear ( $1/ Z$ , not $1/ Z^{2}$ ) because only one power of the charge survives after the quantization rule fixes the other. The energy then picks up $Z^{2}$ : one $Z$ from the deeper Coulomb well, another from the tighter orbit.

Cell H — From energy to colour (wavelength)

Cell I — Shell filling (the Hardware payoff)

First, the counting rule we deferred in the intro:

Worked example Example 8 (Cell I) — Fill silicon and germanium; find valence electrons

Silicon has $Z = 14$ , germanium $Z = 32$ . Fill their shells (capacity $2 n^{2}$ ) and read off valence electrons.

Forecast: Both are semiconductors — guess how many valence electrons each ends with.

Capacities: K $(n = 1) = 2$ , L $(n = 2) = 8$ , M $(n = 3) = 18$ , N $(n = 4) = 32$ . Why this step? $2 n^{2}$ sets the maximum per shell — fill from inside out.
Silicon (14): $2 + 8 + 4 = 14$ . Outermost occupied shell M holds $4$ → 4 valence electrons. Why this step? After K $(2)$ and L $(8)$ take $10$ , the remaining $14 - 2 - 8 = 4$ land in M — and M is the outermost occupied shell, so those $4$ are the valence electrons.
Germanium (32): $2 + 8 + 18 + 4 = 32$ . Outermost shell N holds $4$ → 4 valence electrons. Why this step? K $(2)$ , L $(8)$ , M $(18)$ take $28$ ; the remaining $32 - 28 = 4$ go into N, the outermost occupied shell, so again $4$ valence electrons. (For basic Hardware we fill from inside using $2 n^{2}$ as capacity — see the mistake box below for the real-atom caveat.)
Read the payoff: both Si and Ge end with 4 valence electrons. Why this step? That shared "4" is exactly what lets each form a covalent crystal that conducts a little — the definition of a semiconductor. This is the bridge to Semiconductors and the band gap.

Verify: Si: $2 + 8 + 4 = 14$ ✔. Ge: $2 + 8 + 18 + 4 = 32$ ✔. Both end with $4$ valence electrons — that shared "4" is why both are semiconductors ✔.

Common mistake "M shell must fill to 18 before N starts, so germanium's M holds 18 and N holds 4 by simple counting."

Why it feels right: the $2 n^{2}$ rule says M can hold $18$ , and filling strictly inside-out gives Ge exactly M $(18)$ , N $(4)$ — which even lands the correct valence count of $4$ . The fix: $2 n^{2}$ is a ceiling, not a quota, and real atoms fill by sub-shell energy order, not strictly shell-by-shell. In germanium the 4th shell actually begins filling before the 3rd is completely done (the $3 d$ sub-shell fills after $4 s$ ). The simple inside-out method still gives the right valence answer here, but do not believe every shell is packed to $2 n^{2}$ before the next starts. For introductory Hardware, use $2 n^{2}$ as a capacity limit and count valence by inside-out filling — just know it is an approximation, exactly as the parent note warned.

Cell J — Word problem + null-transition / trap cases

Worked example Example 9 (Cell J) — Real-world: a sign glows; which jump, and the edge cases

A gas-discharge sign emits a photon of $2.11$ eV. A student claims it's a hydrogen $3 \to 2$ transition ( $1.89$ eV). Is that consistent? Then handle three edge cases: a claimed $0$ eV transition, a $13.6$ eV photon absorbed from $n = 1$ , and a $14$ eV photon.

Forecast: Bohr only allows exact level gaps. So a mismatch means...?

Compare the claim: $2.11 \neq = 1.889$ eV, so it is not the hydrogen $3 \to 2$ line. The claim fails. Why this step? Emission energies are quantized to fixed gaps; a value between allowed hydrogen gaps cannot come from hydrogen — it must be a different gas.
Null transition $Δ E = 0$ (the trap case $n_{i} = n_{f}$ ): this would mean start and end level are the same — no jump, so no photon at all. Zero is not a real transition; the electron never moved. Why this step? You cannot emit or absorb light without moving between two distinct levels. Plug $n_{i} = n_{f}$ into the jump formula and you get exactly $0$ — the formula agrees there is nothing.
Boundary case $Δ E = 13.6$ eV absorbed from $n = 1$ : this is the ionization edge — the electron reaches $E_{\infty} = 0$ with zero leftover speed. Just barely free. Why this step? $13.6$ eV is the $1 \to \infty$ gap, the largest single-photon jump possible from the ground state.
Over-energy case $Δ E = 14$ eV: exceeds $13.6$ eV, so the electron is freed and keeps the extra $14 - 13.6 = 0.4$ eV as kinetic energy (a free, moving electron). Why this step? Above ionization the energy no longer has to match a discrete gap — free-electron energies form a continuum, so any excess simply becomes motion.

Verify: $2.11 \neq = 1.889$ ✔ (claim rejected). Null case: jump formula at $n_{i} = n_{f}$ gives $0$ ✔ (no photon). Excess energy $14 - 13.6 = 0.4$ eV ✔.

Active recall

Recall Which subtraction for emission vs absorption?

Emission (falls): photon $= E_{i} - E_{f} > 0$ . Absorption (climbs): energy needed $= E_{f} - E_{i} > 0$ . ::: Both give a positive number; you always end with the electron losing (emit) or gaining (absorb) exactly the gap.

Recall Why does He

$^{+}$ bind at $- 54.4$ eV, not $- 13.6$ ? Because energy scales as $Z^{2}$ and $Z = 2$ , so $- 13.6 \times 4 = - 54.4$ eV. ::: The nucleus pulls twice as hard, the electron sits four times deeper.

Recall What is the ionization energy of hydrogen from the ground state?

$13.6$ eV. ::: It is $E_{\infty} - E_{1} = 0 - (- 13.6)$ .

Recall Germanium (

$Z = 32$ ) valence electron count? 4. ::: Fill $2 + 8 + 18 + 4$ ; the last 4 land in shell N.

Connections

Bohr atomic model and electron shells — the parent note with the derivations these examples use.
Hinglish version — same ideas, informal language.
Valence electrons and bonding — where Example 8's "4 valence electrons" leads next.
Semiconductors and the band gap — Si and Ge's shared valence-4 becomes the band gap story.
Conductors insulators and doping — the Hardware endgame.
Quantum mechanical model of the atom — fixes what Bohr approximates.