Derivation of Bohr's radii and energies from electrostatics + quantization: Worked examples

Before anything else, we pin down the four numbers we will re-use so no symbol is ever a mystery:

The scenario matrix

Every question this topic can ask lives in one of these cells. Each row is a "kind of thing that can go wrong or change".

Cell	What changes	The tricky part
C1 · Ground state, $Z = 1$	plain hydrogen, $n = 1$	the base case everything is measured against
C2 · Excited state, $Z = 1$	$n > 1$	radius swells as $n^{2}$ , energy climbs toward 0
C3 · Higher charge, $Z > 1$	He $^{+}$ , Li $^{2 +}$ …	the $Z^{2}$ trap (not $Z$ )
C4 · Transition (two levels)	photon emitted/absorbed	energy is a difference, sign tells emit vs absorb
C5 · Limiting case $n \to \infty$	electron nearly free	radius $\to \infty$ , energy $\to 0^{-}$ = ionization
C6 · Degenerate check $r \to 0$ ?	can the electron sit at the nucleus?	why $n = 0$ is forbidden
C7 · Ratio / scaling question	"how many times bigger?"	constants cancel, only $n^{2} / Z$ survives
C8 · Real-world word problem	speed / period / current	chaining $v_{n}$ into motion quantities
C9 · Exam twist	mixed atom + transition, or reverse-solve for $n$	must invert a formula

The eight worked examples below cover all nine cells (Example 8 hits both C4 and C9).

Worked examples

Worked example Example 1 — Cell C1: the ground state of hydrogen

Find the radius, energy, and speed of the electron in the $n = 1$ orbit of hydrogen ( $Z = 1$ ).

Forecast: guess before reading — will these just equal the three constants above? (Yes — that's the whole point of C1: it is the reference.)

Radius. $r_{1} = a_{0} \frac{n ^{2}}{Z} = 0.529 \times \frac{1 ^{2}}{1} = 0.529 \overset{˚}{A}$ . Why this step? Plug $n = 1, Z = 1$ into $r_{n} = a_{0} n^{2} / Z$ ; it must return the definition of $a_{0}$ itself.
Energy. $E_{1} = - 13.6 \times \frac{Z ^{2}}{n ^{2}} = - 13.6 \times \frac{1}{1} = - 13.6 eV$ . Why this step? Same substitution into $E_{n} = - 13.6 Z^{2} / n^{2}$ ; negative because the electron is bound.
Speed. $v_{1} = 2.18 \times 1 0^{6} \times \frac{Z}{n} = 2.18 \times 1 0^{6} m/s$ . Why this step? Uses $v_{n} \propto Z / n$ ; at $n = 1, Z = 1$ we recover the base speed.

Verify: $v_{1} / c = 2.18 \times 1 0^{6} /3 \times 1 0^{8} \approx 0.0073 \approx 1/137$ — the famous fine-structure ratio. Non-relativistic (well under $c$ ), so Bohr's classical mechanics is justified. ✓

Worked example Example 3 — Cell C3: ground state of Li

$^{2 +}$ (the $Z^{2}$ trap) Lithium has $Z = 3$ ; strip two electrons to get Li $^{2 +}$ (one electron left — hydrogen-like). Find $E_{1}$ and $r_{1}$ .

Forecast: the classic trap. Energy uses $Z^{2} = 9$ , radius uses only $Z = 3$ . Which is which? Forecast both.

Energy. $E_{1} = - 13.6 \times \frac{Z ^{2}}{n ^{2}} = - 13.6 \times \frac{9}{1} = - 122.4 eV$ . Why this step? $E \propto Z^{2}$ : one factor of $Z$ from the Coulomb pull, a second because the orbit itself shrinks as $1/ Z$ . Nine times more bound than hydrogen.
Radius. $r_{1} = a_{0} \frac{n ^{2}}{Z} = 0.529 \times \frac{1}{3} = 0.176 \overset{˚}{A}$ . Why this step? $r \propto 1/ Z$ (single power): triple the charge pulls the electron 3× closer, not 9× closer.

Verify: the Virial Theorem check — $K E = - \frac{1}{2} P E$ , so $∣ E_{1} ∣ = K E = 122.4$ eV, positive kinetic energy, consistent. Units of $E$ in eV, radius in Å — dimensionally clean. ✓

Worked example Example 4 — Cell C4: emission transition

$n = 3 \to n = 2$ in hydrogen An electron drops from $n = 3$ to $n = 2$ . Find the photon energy and wavelength (this is the red Balmer- $α$ line — see Hydrogen Spectrum and Rydberg Formula).

Forecast: the electron falls to a deeper (more negative) level, so it must release energy. The photon energy is positive. Guess: a few eV, visible light.

Level energies. $E_{3} = - 1.51 eV$ (Example 2), $E_{2} = - 13.6/4 = - 3.40 eV$ . Why this step? We need both rungs before we can measure the drop between them.
Photon energy = energy released. $Δ E = E_{2} - E_{3} = - 3.40 - (- 1.51) = - 1.89 eV$ . Why this step? Look at the red arrow in the figure: the electron falls, so the atom's energy decreases by 1.89 eV. That energy leaves as a photon of energy $E_{γ} = + 1.89 eV$ .
Wavelength. $λ = \frac{h c}{E _{γ}} = \frac{1240 eV\cdotnm}{1.89 eV} = 656 nm$ . Why this step? $h c = 1240$ eV·nm is the handy photon shortcut; wavelength is inversely proportional to energy.

Verify: 656 nm is deep red — matches the observed Balmer- $α$ line exactly. Sign of $Δ E$ is negative (energy lost by atom) → emission, consistent with an electron falling. ✓

Worked example Example 5 — Cell C5: the limit

$n \to \infty$ (ionization) What is the ionization energy of hydrogen from the ground state, and what happens to $r_{n}$ as $n \to \infty$ ?

Forecast: at $n = \infty$ the electron is infinitely far and free. Its energy should be exactly $0$ . So ionization energy = depth of the ground level. Guess the number.

Energy at the limit. $E_{\infty} = - 13.6 \times \frac{1}{\infty ^{2}} = 0 eV$ . Why this step? Dividing by $n^{2}$ with $n \to \infty$ sends the level to $0^{-}$ — the boundary between bound (negative) and free (positive).
Ionization energy. $IE = E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6 eV$ . Why this step? Ionization = the work to lift the electron from its rung all the way to freedom (see Ionization Energy).
Radius limit. $r_{n} = a_{0} n^{2} \to \infty$ as $n \to \infty$ . Why this step? $n^{2}$ grows without bound — a free electron has no orbit at all, geometrically consistent.

Verify: experimental hydrogen IE = 13.6 eV. ✓ Energy $\to 0$ from below and radius $\to \infty$ together describe the same physical picture: the electron leaving. ✓

Worked example Example 6 — Cell C6: can the electron sit AT the nucleus (

$r = 0$ )? Is there an orbit with $r = 0$ ? What would $n = 0$ predict, and why is it forbidden?

Forecast: intuitively, $r = 0$ means the electron crashes into the nucleus — Bohr's whole point was to forbid this. Guess: the formula must break.

Try $n = 0$ in the radius. $r_{0} = a_{0} \frac{0 ^{2}}{Z} = 0$ . Why this step? Test the degenerate input directly to see what the formula says.
Try $n = 0$ in the energy. $E_{0} = - \frac{13.6 Z ^{2}}{0 ^{2}} = - \infty$ . Why this step? The energy blows up to negative infinity — an infinitely bound, physically impossible state.
Why it's ruled out. Quantization of Angular Momentum demands $m v r = n ℏ$ with $n \geq 1$ . At $n = 0$ angular momentum is zero, meaning no orbit — a collapsed state Bohr's postulate explicitly excludes. Why this step? The lower cutoff $n = 1$ is exactly what stops the classical spiral-to-death; that is the whole reason quantization was introduced.

Verify: the smallest allowed radius is $r_{1} = a_{0} = 0.529$ Å $\neq = 0$ . The electron never reaches the nucleus. ✓ The de Broglie Wavelength and Standing Waves picture agrees: $n = 0$ would mean zero wavelengths fit the orbit, which is meaningless.

Worked example Example 8 — Cells C8 + C9: word problem + reverse-solve twist

(C8) In hydrogen's ground state, how many times per second does the electron circle the nucleus (its orbital frequency)? (C9 twist) An emission line in hydrogen has photon energy $12.09$ eV from a drop to $n = 1$ . From which level $n$ did the electron fall?

Forecast (C8): speed $\sim 2 \times 1 0^{6}$ m/s, orbit circumference $\sim 3 \times 1 0^{- 10}$ m → frequency should be huge, around $1 0^{16}$ per second. Forecast (C9): a large jump landing on $n = 1$ ; guess $n = 3$ .

(C8) Circumference. $2 π r_{1} = 2 π \times 5.29 \times 1 0^{- 11} = 3.32 \times 1 0^{- 10} m$ . Why this step? Frequency = speed ÷ distance-per-lap, so we need the lap length.
(C8) Frequency. $f = \frac{v _{1}}{2 π r _{1}} = \frac{2.18 \times 1 0 ^{6}}{3.32 \times 1 0 ^{- 10}} = 6.56 \times 1 0^{15} Hz$ . Why this step? Uses Centripetal Force and Circular Motion — the electron goes ~6.6 quadrillion laps per second.
(C9) Set up the transition. $E_{γ} = E_{1} - E_{n} = - 13.6 - (- \frac{13.6}{n ^{2}}) = 13.6 (\frac{1}{n ^{2}} - 1)$ in magnitude $13.6 (1 - \frac{1}{n ^{2}})$ . Why this step? The photon carries the energy dropped; invert the level formula for the unknown $n$ .
(C9) Solve. $12.09 = 13.6 (1 - \frac{1}{n ^{2}}) \Rightarrow 1 - \frac{1}{n ^{2}} = 0.889 \Rightarrow \frac{1}{n ^{2}} = 0.111 \Rightarrow n^{2} = 9 \Rightarrow n = 3$ . Why this step? Algebraically isolate $n^{2}$ , then take the root (positive integer only, since $n$ is an orbit number).

Verify (C8): $f \approx 6.6 \times 1 0^{15}$ Hz is the correct classical orbital frequency of ground-state hydrogen. ✓ Verify (C9): direct check $E_{3} \to E_{1} = - 13.6 - (- 1.51) = - 12.09$ eV → photon of 12.09 eV, matching the Lyman- $β$ line. ✓

Recall

Recall Which power of

$Z$ for radius vs energy? Radius uses $Z^{1}$ (in denominator); energy uses $Z^{2}$ . ::: $r \propto n^{2} / Z$ , but $E \propto Z^{2} / n^{2}$ — energy gets the extra factor of $Z$ from the shrinking orbit.

Recall Why is

$n = 0$ forbidden? Quantization requires $n \geq 1$ ; $n = 0$ gives $r = 0$ , $E = - \infty$ — a collapsed, unphysical state. ::: The lower cutoff is exactly what stops the classical spiral into the nucleus.

Recall Emission vs absorption from the sign of

$Δ E$ ? If the atom's energy decreases (electron falls to lower $n$ ), a photon is emitted; if it increases, a photon is absorbed. ::: $E_{γ} = ∣ E_{final} - E_{initial} ∣$ .

Parent: 1.2.08 Derivation of Bohr's radii and energies from electrostatics + quantization (Hinglish)

Connections

Bohr Model Postulates
Coulomb's Law
Centripetal Force and Circular Motion
Quantization of Angular Momentum
Hydrogen Spectrum and Rydberg Formula
Ionization Energy
Virial Theorem
de Broglie Wavelength and Standing Waves