Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV: Worked examples

Intuition Why a whole page of examples?

The parent note gave you two headline formulas and three tidy examples. But an exam does not hand you a tidy case — it hands you a weird one: a heavy ion, an absorption instead of emission, a "find the wavelength" wrapped inside a "find the speed", or the sneaky " $n \to \infty$ " limit. This page walks every kind of question the Bohr model can produce, so no scenario ever surprises you.

We only ever use the four facts the parent built:

$r_{n} = 0.529 \frac{n ^{2}}{Z} \overset{˚}{A}$ (radius grows with $n^{2}$ , shrinks with $Z$ )
$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$ (energy negative, rises toward $0$ )
$v_{n} = 2.18 \times 1 0^{6} \frac{Z}{n} m/s$ (speed grows with $Z$ , shrinks with $n$ )
$\overset{ν}{ˉ} = \frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , $R_{H} = 1.097 \times 1 0^{7} m^{- 1}$

One conversion we lean on all page: $== 1 \overset{˚}{A} = 1 \times 1 0^{- 10} m ==$ (an ångström is a ten-billionth of a metre — atomic-orbit sized).

The scenario matrix

Every Bohr problem is one (or a combination) of the cells below. Each worked example is tagged with the cell it lands in.

#	Case class	The twist it tests	Example
A	Radius, plain	$n^{2} / Z$ scaling	Ex 1
B	Energy, plain	negative sign, $Z^{2} / n^{2}$	Ex 2
C	Emission (jump down)	$Δ E > 0$ released as light	Ex 3
D	Absorption (jump up)	electron must swallow a photon	Ex 4
E	Ionisation ( $n_{2} = \infty$ )	the $n \to \infty$ limit	Ex 5
F	Speed & period	using $v_{n}$ , geometry of the orbit	Ex 6
G	Hydrogen-like scaling trap	"same energy, different atom" coincidence	Ex 7
H	Degenerate / limiting inputs	$n = 1$ floor, $Z = 1$ , $n \to \infty$	Ex 8
I	Real-world word problem	series limit / colour of light	Ex 9
J	Exam twist (reverse)	given the answer, find $n$ or $Z$	Ex 10

The single picture below is the map of the whole matrix: every spectral cell (C, D, E) is one of the coloured arrows on a hydrogen energy staircase. Keep glancing back at it as you read Ex 3, 4 and 5 — the red arrow is Ex 3's kind of jump, the blue arrow is Ex 4's, the yellow arrow is Ex 5's.

Figure — Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV

Each rung is an allowed energy $E_{n}$ ; the dashed green line at the top ( $E = 0$ ) is the "free electron" ceiling. Down = emit light, up = absorb light, all the way to the top = ionise. That one diagram organises half the examples on this page.

Recall Sign convention we never break

Emitted photon energy ::: $Δ E = E_{high} - E_{low} > 0$ (a positive amount of light leaves the atom). Absorbed photon energy ::: same size, but the electron gains it and climbs up.

Worked examples

Cell A — plain radius

Cell B — plain energy

Cell C — emission (jump down)

Worked example 3. Emission energy for

$He^{+}$ falling $n_{2} = 3 \to n_{1} = 2$ Statement: An electron in $He^{+}$ ( $Z = 2$ ) drops from $n = 3$ to $n = 2$ . How much energy is released, in eV? (This is the red-arrow kind of jump in figure s01.)

Forecast: The electron falls into a deeper well, so it must lose energy — that lost energy leaves as a photon. Answer should be positive.

Start from the raw energies (each carries its own minus sign): $Δ E = E_{n_{2}} - E_{n_{1}} = (- 13.6 \frac{Z ^{2}}{n _{2}^{2}}) - (- 13.6 \frac{Z ^{2}}{n _{1}^{2}}) .$ Why this step? Emission energy = energy of the higher level minus the lower level. Both are negative, so we must track the signs carefully.
Distribute the outer minus onto the second bracket (a minus times a minus becomes plus): $Δ E = - 13.6 \frac{Z ^{2}}{n _{2}^{2}} + 13.6 \frac{Z ^{2}}{n _{1}^{2}} = 13.6 Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) .$ Why this step? This is exactly how the two negative signs cancel into the familiar positive form: the second term flips to $+$ , and factoring out $13.6 Z^{2}$ leaves the $1/ n_{1}^{2} - 1/ n_{2}^{2}$ bracket. Writing small $n$ first keeps the bracket positive for a downward jump.
Bracket: $\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$ . Why this step? Common denominator $36$ keeps it exact before we decimalise.
$Δ E = 13.6 \times 4 \times \frac{5}{36} = 13.6 \times \frac{20}{36} = 7.556$ eV. Why this step? $Z^{2} = 2^{2} = 4$ ; multiply everything through.

Verify: Positive ✓ (energy released, as forecast). For hydrogen the same jump gives only $13.6 \times 5/36 = 1.889$ eV; the $Z^{2} = 4$ boost makes $He^{+}$ emit exactly $4 \times$ more — internally consistent ✓.

Cell D — absorption (jump up)

Worked example 4. Can a

$10.0$ eV photon lift ground-state hydrogen to $n = 2$ ? Statement: Hydrogen ( $Z = 1$ ) sits in $n = 1$ . A photon of $10.0$ eV arrives. Does the electron reach $n = 2$ ? (This is the blue-arrow kind of jump in figure s01.)

Forecast: Absorption is emission run backwards — the electron must swallow exactly the gap $E_{2} - E_{1}$ . Guess: is $10.0$ eV enough, too little, or too much?

$E_{1} = - 13.6$ eV, $E_{2} = - 13.6/4 = - 3.4$ eV. Why this step? We need the two levels' energies to size the gap.
Gap $= E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2$ eV. Why this step? Absorption needs a photon of exactly this energy — atoms are picky, unlike a ramp.
Compare: $10.0 < 10.2$ . The photon is too small; the electron stays in $n = 1$ . Why this step? Bohr allows only exact matches. $10.0$ eV lands between steps, which is forbidden — the photon passes straight through.

Verify: Gap $10.2$ eV is the famous Lyman- $α$ absorption of hydrogen ✓. The photon must match to $\pm 0$ (no partial climbs), so $10.0 \neq = 10.2$ ⇒ no transition ✓.

Cell E — ionisation ( $n_{2} = \infty$ )

Worked example 5. Energy to fully ionise

$He^{+}$ from its ground state Statement: Remove the electron from $He^{+}$ ( $Z = 2$ , starting $n = 1$ ) all the way to freedom. (This is the yellow-arrow all-the-way-to-the-top jump in figure s01.)

Forecast: "Free" means $n \to \infty$ , where $E_{\infty} = 0$ . So ionisation energy $= 0 - E_{1} = ∣ E_{1} ∣$ . It should be bigger than hydrogen's $13.6$ eV because $Z$ is larger.

$E_{1} = - 13.6 \frac{Z ^{2}}{1 ^{2}} = - 13.6 \times 4 = - 54.4$ eV. Why this step? The starting level sets the depth of the well.
As $n \to \infty$ , $\frac{1}{n ^{2}} \to 0$ , so $E_{\infty} = - 13.6 Z^{2} \cdot 0 = 0$ . Why this step? This is the limiting case — the "top of the staircase" where the electron is unbound.
Ionisation energy $= E_{\infty} - E_{1} = 0 - (- 54.4) = 54.4$ eV. Why this step? Energy needed = final minus initial; a positive input because we fight the attraction.

Verify: $54.4 = 4 \times 13.6$ ✓ — the $Z^{2} = 4$ scaling. Positive (we supply energy) ✓. This is why $He^{+}$ is much harder to ionise than H (Ionisation energy).

Cell F — speed & orbital period

Worked example 6. Speed and time-per-revolution of the electron in

$n = 1$ hydrogen Statement: For H ( $Z = 1$ , $n = 1$ ) find the orbital speed $v_{1}$ and the period $T$ (time for one full loop). Use the Bohr radius $r_{1} = 0.529$ Å.

Forecast: The circumference is tiny and the speed is huge ( $\sim 1 0^{6}$ m/s), so the period should be extremely short — femtoseconds?

$v_{1} = 2.18 \times 1 0^{6} \frac{Z}{n} = 2.18 \times 1 0^{6} \frac{1}{1} = 2.18 \times 1 0^{6}$ m/s. Why this step? Straight from the speed formula the parent gave.
Convert the radius to metres first: since $1 \overset{˚}{A} = 1 \times 1 0^{- 10}$ m, $r_{1} = 0.529 \overset{˚}{A} = 0.529 \times 1 0^{- 10}$ m. Why this step? Speed is in metres per second, so the length must be in metres too — otherwise the units won't cancel. This is the ångström conversion stated at the top of the page.
Circumference $C = 2 π r_{1} = 2 π (0.529 \times 1 0^{- 10}) = 3.324 \times 1 0^{- 10}$ m. Why this step? Period is "distance around one loop ÷ speed", and the loop's length is its circumference (see figure s03, where the arrow traces one full lap).
$T = \frac{C}{v _{1}} = \frac{3.324 \times 1 0 ^{- 10}}{2.18 \times 1 0 ^{6}} = 1.525 \times 1 0^{- 16}$ s. Why this step? Uniform circular motion: time $=$ distance $\div$ speed.

Verify: $T \approx 1.5 \times 1 0^{- 16}$ s $= 0.15$ fs — femtosecond scale, matching the forecast that the electron laps the nucleus millions of billions of times per second ✓. Units: m ÷ (m/s) = s ✓.

Cell G — the "same energy" scaling trap

Worked example 7. Which orbit of

$Be^{3 +}$ has the same energy as ground-state hydrogen? Statement: Hydrogen's ground state is $- 13.6$ eV. $Be^{3 +}$ has $Z = 4$ . Find the $n$ for which $Be^{3 +}$ matches $- 13.6$ eV.

Forecast: Energy $\propto Z^{2} / n^{2}$ . To keep $Z^{2} / n^{2} = 1$ we need $n = Z$ . So guess $n = 4$ .

Set $E_{n} (Be^{3 +}) = - 13.6$ : $- 13.6 \frac{Z ^{2}}{n ^{2}} = - 13.6$ . Why this step? Equate the target energy to the formula and solve for $n$ .
Divide out $- 13.6$ : $\frac{Z ^{2}}{n ^{2}} = 1 \Rightarrow n = Z = 4$ . Why this step? Only the ratio matters; $n$ must equal $Z$ for the constant to be $1$ .

Verify: $E_{4} (Be^{3 +}) = - 13.6 \times 16/16 = - 13.6$ eV ✓. This is the general rule: the $n = Z$ orbit of any hydrogen-like ion always sits at $- 13.6$ eV — the trap that fools students into thinking heavy ions have "hydrogen-like" energies everywhere (Quantum mechanical model of atom explains why this exactness breaks for real multi-electron atoms).

Cell H — degenerate / limiting inputs

Worked example 8. The three edge cases:

$n = 1$ floor, $Z = 1$ baseline, $n \to \infty$ ceiling (hydrogen) Statement: For hydrogen ( $Z = 1$ ), state $r_{1}$ , $E_{1}$ , and both limits as $n \to \infty$ : $r_{\infty}$ and $E_{\infty}$ .

Forecast: The staircase has a lowest step ( $n = 1$ , cannot go lower) and an infinitely high top ( $n \to \infty$ ) where the electron floats free. Radius should blow up; energy should climb to $0$ .

Floor: $r_{1} = 0.529 \frac{1}{1} = 0.529$ Å, $E_{1} = - 13.6 \frac{1}{1} = - 13.6$ eV. Why this step? $n = 1$ is the smallest allowed integer — there is no $n = 0$ (that would put the electron on the nucleus, forbidden).
Radius ceiling: as $n \to \infty$ , $r_{n} = 0.529 n^{2} \to \infty$ . Why this step? The $n^{2}$ growth is unbounded — the electron drifts arbitrarily far out.
Energy ceiling: as $n \to \infty$ , $E_{n} = - 13.6/ n^{2} \to 0^{-}$ . Why this step? $1/ n^{2} \to 0$ , so the energy approaches zero from below — the electron becomes just barely free.

Verify: $r_{\infty} = \infty$ and $E_{\infty} = 0$ are consistent: infinitely far apart means no attraction energy left ✓. There is no allowed state below $n = 1$ , which is exactly why hydrogen is stable and doesn't collapse — the point of Bohr's whole model (Rutherford model lacked this floor).

Cell I — real-world word problem

Worked example 9. What colour is the shortest-wavelength line of the hydrogen Lyman series?

Statement: The Lyman series is every jump down to $n_{1} = 1$ . Its series limit (shortest wavelength) comes from $n_{2} \to \infty$ . Find that wavelength (H, $Z = 1$ ) and say which part of the spectrum it lies in.

Forecast: The biggest possible drop into $n = 1$ gives the most energetic photon ⇒ the shortest wavelength. To $n = 1$ the gaps are large (10–13.6 eV), so this is well beyond visible — ultraviolet.

$\overset{ν}{ˉ} = R_{H} Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{\infty ^{2}}) = R_{H} (1 - 0) = R_{H} = 1.097 \times 1 0^{7}$ m⁻¹. Why this step? $n_{2} \to \infty$ kills the second term — the limiting (maximum) wavenumber.
$λ = \frac{1}{ν ˉ} = \frac{1}{1.097 \times 1 0 ^{7}} = 9.116 \times 1 0^{- 8}$ m $= 91.2$ nm. Why this step? Wavelength is the reciprocal of wavenumber.
$91.2$ nm is far below the $400$ – $700$ nm visible band ⇒ ultraviolet. Why this step? Compare to the human-visible window to name the region.

Verify: $91.2$ nm is the accepted Lyman limit of hydrogen ✓. It is shorter than the Balmer- $α$ red line ( $656$ nm from the parent note), as forecast — deeper falls make bluer/UV light (Hydrogen spectrum & Rydberg formula).

Cell J — exam twist (reverse-engineer $n$ )

Worked example 10. A hydrogen electron emits a

$12.09$ eV photon landing in $n = 1$ . Where did it start? Statement: Emission photon $= 12.09$ eV, final level $n_{1} = 1$ , hydrogen ( $Z = 1$ ). Find the initial level $n_{2}$ .

Forecast: $12.09$ eV is close to the $13.6$ eV ionisation energy, so the electron fell from high up — probably $n_{2} = 3$ .

$Δ E = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{n _{2}^{2}}) = 12.09$ . Why this step? Reverse the emission formula: we know the energy, we solve for the unknown level.
$1 - \frac{1}{n _{2}^{2}} = \frac{12.09}{13.6} = 0.889 \Rightarrow \frac{1}{n _{2}^{2}} = 0.111 = \frac{1}{9}$ . Why this step? Isolate $1/ n_{2}^{2}$ ; recognising $0.111 = 1/9$ gives a clean integer.
$n_{2}^{2} = 9 \Rightarrow n_{2} = 3$ . Why this step? Take the positive root ( $n$ is a positive integer).

Verify: Check forward: $13.6 (1 - 1/9) = 13.6 \times 8/9 = 12.09$ eV ✓. Matches the forecast $n_{2} = 3$ ; this is the Lyman- $β$ line ✓.

Recall One-line summary of the matrix

Radius/energy questions ::: plug the dimensionless ratio $n^{2} / Z$ (into $0.529$ Å) or $Z^{2} / n^{2}$ (into $- 13.6$ eV) — cells A, B. Spectral questions ::: emission = drop released as light, absorption = exact gap swallowed, ionisation = jump to $n = \infty$ — cells C, D, E. Reverse questions ::: set the formula equal to the given number and solve for $n$ or $Z$ — cells G, J.

Connections

Hinglish version of the parent
Hydrogen spectrum & Rydberg formula — cells C, D, E, I feed directly into this.
Ionisation energy — cell E is literally its definition for hydrogen-like ions.
Quantum mechanical model of atom — explains why cell G's exactness fails for real atoms.
Rutherford model — cell H's stable floor is the fix Bohr added.

Worked examples — Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV