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

Here, $Z$ is the nuclear charge number (how many protons), $n$ is the principal quantum number (which "step" the electron sits on, $n = 1, 2, 3, \dots$ ), $\overset{ν}{ˉ}$ ("nu-bar") is the wavenumber = number of wave crests per metre, and $λ$ ("lambda") is the wavelength = distance from one crest to the next.

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

Level 1 — Recognition

Recall Solution 1

WHAT: plug into $r_{n} = 0.529 (n^{2} / Z)$ . WHY: L1 is just "spot the formula and substitute." $r_{4} = 0.529 \times \frac{4 ^{2}}{1} = 0.529 \times 16 = 8.464 \overset{˚}{A}$ The orbit is 16× the Bohr radius — because $r \propto n^{2}$ and $4^{2} = 16$ .

Recall Solution 2

WHAT: use $E_{n} = - 13.6 (Z^{2} / n^{2})$ . $E_{1} = - 13.6 \times \frac{2 ^{2}}{1 ^{2}} = - 13.6 \times 4 = - 54.4 eV$ He⁺ sits 4× deeper than ground-state hydrogen because the doubled nuclear charge appears squared.

Level 2 — Application

Recall Solution 3

WHAT: evaluate $n^{2} / Z$ first, then multiply. $\frac{n ^{2}}{Z} = \frac{9}{3} = 3, r_{3} = 0.529 \times 3 = 1.587 \overset{˚}{A}$ Even though Li²⁺ has a heavier nucleus, this orbit is exactly 3× the Bohr radius. The strong $Z = 3$ pull is partly undone by the large $n = 3$ spreading.

Recall Solution 4

WHAT: use $v_{n} = 2.18 \times 1 0^{6} (Z / n)$ with $Z = 1, n = 1$ . $v_{1} = 2.18 \times 1 0^{6} m/s$ As a fraction of $c$ : $\frac{v _{1}}{c} = \frac{2.18 \times 1 0 ^{6}}{3.00 \times 1 0 ^{8}} = 7.27 \times 1 0^{- 3} \approx \frac{1}{137}$ WHY it matters: the electron moves at less than 1% of light speed, so ignoring relativity (as Bohr does) is reasonable for hydrogen. That $1/137$ is the famous fine-structure constant $α$ .

Recall Solution 5

WHAT: ionisation = raise the electron to $n = \infty$ , where $E_{\infty} = 0$ . $E_{2} = - 13.6 \times \frac{1}{2 ^{2}} = - 3.4 eV$ $Δ E = E_{\infty} - E_{2} = 0 - (- 3.4) = + 3.4 eV$ WHY positive: you must supply energy to free a bound (negative-energy) electron. It costs less than the 13.6 eV from $n = 1$ because the electron already sits higher up the well.

Level 3 — Analysis

Recall Solution 6

(a) WHAT: use the Rydberg formula with $n_{1} = 2$ (lower), $n_{2} = 3$ (upper). $\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} = 0.1389$ $\overset{ν}{ˉ} = 1.097 \times 1 0^{7} \times 0.1389 = 1.524 \times 1 0^{6} m^{- 1}$ (b) $λ = 1/ \overset{ν}{ˉ} = 1/ (1.524 \times 1 0^{6}) = 6.56 \times 1 0^{- 7} m = 656 nm$ (red). (c) Photon energy via level difference: $Δ E = 13.6 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times 0.1389 = 1.89 eV$ Consistency check: using $E (eV) = 1240/ λ (nm)$ gives $1240/656 = 1.89$ eV. ✓ The three routes agree — evidence the model hangs together.

Recall Solution 7

WHAT: two equations, two unknowns. From radius: $0.529 (n^{2} / Z) = 0.1764 \Rightarrow n^{2} / Z = 0.3333 = 1/3$ . From energy: $- 13.6 (Z^{2} / n^{2}) = - 122.4 \Rightarrow Z^{2} / n^{2} = 9$ . WHY combine them: notice $Z^{2} / n^{2} = 9$ means $Z / n = 3$ , and $n^{2} / Z = 1/3$ means $Z = 3 n^{2}$ . Substitute $Z = 3 n^{2}$ into $Z = 3 n$ : $3 n^{2} = 3 n \Rightarrow n = 1$ , then $Z = 3$ . So it is Li²⁺ in its ground state ( $n = 1$ , $Z = 3$ ). Check: $r = 0.529/3 = 0.176$ Å ✓, $E = - 13.6 \times 9 = - 122.4$ eV ✓.

Recall Solution 8

WHAT: Rydberg with the $Z^{2}$ factor, $Z = 2$ so $Z^{2} = 4$ . $\frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} = 0.1875$ $\overset{ν}{ˉ} = 1.097 \times 1 0^{7} \times 4 \times 0.1875 = 8.228 \times 1 0^{6} m^{- 1}$ $λ = 1/ \overset{ν}{ˉ} = 1.215 \times 1 0^{- 7} m = 121.5 nm$ WHY interesting: this is ultraviolet, and it coincides with the H Lyman-α line — because the He⁺ $4 \to 2$ energies mirror H $2 \to 1$ (the $Z^{2}$ boost shifts everything by exactly a factor of 4).

Level 4 — Synthesis

Recall Solution 9

WHAT: write the ratio symbolically. $\frac{r _{2}}{r _{1}} = \frac{0.529 ( 2 ^{2} / Z )}{0.529 ( 1 ^{2} / Z )} = \frac{4/ Z}{1/ Z} = 4$ The $0.529$ and the $Z$ both cancel top-and-bottom, leaving $2^{2} / 1^{2} = 4$ . WHY $Z$ cancels: because $Z$ multiplies both orbits by the same factor $1/ Z$ , so any ratio of orbits of the same ion forgets $Z$ entirely — spacing depends only on the $n$ values.

Recall Solution 10

WHAT: find the photon's energy, then locate the level whose gap from $n = 1$ matches. $E_{photon} = \frac{1240}{97.2} = 12.76 eV$ The electron starts at $E_{1} = - 13.6$ eV, so after absorbing it reaches: $E_{n} = - 13.6 + 12.76 = - 0.84 eV$ Solve $- 13.6/ n^{2} = - 0.84 \Rightarrow n^{2} = 13.6/0.84 = 16.2 \Rightarrow n = 4$ . WHY round to 4: only integer $n$ are allowed (postulate 2). The tiny excess over 16 is rounding in the constants. So the electron lands on $n = 4$ .

Level 5 — Mastery

Recall Solution 11

WHAT — the two starting facts: $(1) m v^{2} = \frac{k Z e ^{2}}{r}, (2) m v r = \frac{nh}{2 π}$ WHY these two: (1) is force balance (Coulomb = centripetal), (2) is quantised angular momentum. Two equations, two unknowns $v, r$ . STEP — isolate $r$ from (2): $r = \frac{nh}{2 π m v}$ . STEP — substitute into (1): $m v^{2} = \frac{k Z e ^{2}}{nh / ( 2 π m v )} = \frac{2 π mk Z e ^{2} v}{nh}$ Cancel one $m$ and one $v$ from each side: $v = \frac{2 π k Z e ^{2}}{nh}$ Everything except $Z / n$ is a fixed bundle of constants, so $v_{n} \propto Z / n$ . Evaluating the bundle gives $2.18 \times 1 0^{6}$ m/s — exactly the number the parent note quotes. The ratio: Li²⁺( $Z = 3, n = 1$ ) vs H( $Z = 1, n = 1$ ): $\frac{v _{Li^{2 +}}}{v _{H}} = \frac{3/1}{1/1} = 3$ The Li²⁺ ground-state electron whirls 3× faster because the triple nuclear charge pulls harder.

Recall Solution 12

WHAT — find $Z$ from the given line. Rydberg for $2 \to 1$ : $\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = R_{H} Z^{2} \cdot \frac{3}{4}$ Solve for $Z^{2}$ : $Z^{2} = \frac{1}{λ R _{H} ( 3/4 )} = \frac{1}{( 30.4 \times 1 0 ^{- 9} ) ( 1.097 \times 1 0 ^{7} ) ( 0.75 )}$ $Z^{2} = \frac{1}{0.2501} = 3.998 \approx 4 \Rightarrow Z = 2$ So it is the He⁺ ion. STEP — predict the $3 \to 1$ line for $Z = 2$ : $\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}} = 1 - \frac{1}{9} = \frac{8}{9} = 0.8889$ $\overset{ν}{ˉ} = 1.097 \times 1 0^{7} \times 4 \times 0.8889 = 3.900 \times 1 0^{7} m^{- 1}$ $λ = 1/ \overset{ν}{ˉ} = 2.564 \times 1 0^{- 8} m = 25.6 nm$ WHY shorter: a bigger drop ( $3 \to 1$ vs $2 \to 1$ ) releases more energy, and higher energy means shorter wavelength. Both lines are deep ultraviolet, as expected for a strongly-bound He⁺.

Recall Self-test checklist

Did you get every level? Score yourself. L1 ✅ formula recognition (Ex 1–2) ::: recall the two headline formulas cold L2 ✅ single substitution (Ex 3–5) ::: handle $Z^{2}$ , ionisation-to- $\infty$ , speed L3 ✅ multi-step analysis (Ex 6–8) ::: Rydberg, back-solve for $n, Z$ , keep the $Z^{2}$ L4 ✅ synthesis (Ex 9–10) ::: prove ratios, climb the well with a photon L5 ✅ mastery (Ex 11–12) ::: derive from postulates, identify an ion, predict a new line

Connections

Parent topic — all formulas derived
Hydrogen spectrum & Rydberg formula — Exercises 6, 8, 12 are pure applications.
Ionisation energy — Exercise 5 computes it directly.
de Broglie wavelength — the standing-wave view behind the quantisation used in Ex 11.