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

We build on the parent topic and it leans on the Rutherford model (nucleus in the centre) and Planck's quantisation (energy comes in lumps).

Step 1 — Draw the players: nucleus, electron, and the pull between them

WHAT. Put a fat positive dot in the middle — the nucleus, charge $+ Z e$ . A tiny dot circles it at distance $r$ — the electron, charge $- e$ . A red arrow points from the electron back toward the nucleus: the electric attraction.

WHY. Before any equation, we must agree on what each letter is — otherwise every symbol later is a mystery. This is just Rutherford's picture: charge in the middle, electron outside.

WHAT IT LOOKS LIKE.

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

Step 2 — Balance the two forces (this pins down the speed)

WHAT. The electron does not fall in and does not fly away — it circles steadily. For that, the inward electric pull must exactly equal the inward force a circle demands (the centripetal force). We set them equal.

WHY this and not something else? A circling object is always accelerating toward the centre (its direction keeps turning). Newton says a force must cause that, aimed inward. The only inward force here is the electric pull. So: electric pull $=$ centripetal requirement. This is the one equation that ties speed $v$ to radius $r$ .

$k = Coulomb constant \frac{1}{4 π ε _{0}} electric pull \frac{( Z e ) ( e )}{r ^{2}} = centripetal force \frac{m v ^{2}}{r}$

Term by term: $k = \frac{1}{4 π ε _{0}}$ bundles the electric constants into one symbol; $(Z e) (e)$ is "charge times charge" (bigger charges pull harder); $r^{2}$ in the bottom says the pull weakens with the square of distance; $m$ is the electron's mass; $v$ its speed. Cancel one $r$ from each side:

$\boxed{\,\frac{kZe^2}{r} = mv^2\,}\tag{1}$

WHAT IT LOOKS LIKE.

Step 3 — One equation, two unknowns: we need a second rule

WHAT. Equation (1) has two things we don't know: the speed $v$ and the radius $r$ . One equation can't solve for two unknowns. We need a second, independent fact.

WHY. Classically, any $r$ is allowed — pick a radius, equation (1) hands you the matching speed, done. That would give a continuous smear of orbits. But hydrogen glows in sharp lines, so nature forbids most orbits. Bohr's second postulate is exactly that forbidding rule.

$m v r = n ℏ = \frac{nh}{2 π}, n = 1, 2, 3, \dots$

Here $h$ is Planck's constant (the size of nature's smallest action-lump), and $ℏ = h /2 π$ is just " $h$ per radian." $n$ is the quantum number — the step you're standing on.

WHAT IT LOOKS LIKE. The allowed $L$ values are a ladder, not a ramp:

Step 4 — Solve for the radius

WHAT. From the quantisation rule solve for the speed, $v = \frac{nh}{2 π m r}$ , and slot it into equation (1). Now only $r$ is left.

WHY. Two equations, two unknowns → eliminate $v$ to isolate $r$ . This is the moment the allowed radii pop out.

$\frac{k Z e ^{2}}{r} = m v^{2} (\frac{nh}{2 π m r})^{2} = \frac{n ^{2} h ^{2}}{4 π ^{2} m r ^{2}}$

Multiply both sides by $r^{2}$ , divide by $k Z e^{2}$ :

$r_{n} = \frac{n ^{2} h ^{2}}{4 π ^{2} mk Z e ^{2}}$

Term by term of the answer: $n^{2}$ on top → orbits grow fast with the step number; $Z$ on the bottom → a stronger nucleus pulls the electron in, shrinking the orbit; everything else ( $h, m, k, e$ ) is a fixed bundle of constants.

WHAT IT LOOKS LIKE. The allowed circles, spaced by $n^{2}$ ( $1, 4, 9, 16 \dots$ ):

Step 5 — Build the total energy from kinetic + potential

WHAT. The electron's total energy is its energy of motion (kinetic, $K E$ ) plus its energy of position in the electric pull (potential, $P E$ ).

WHY. "Energy of the orbit" is what determines the light emitted — so we need it. $P E$ is negative because the electron is trapped: it sits inside an attractive well, below the "free at infinity" zero line.

$E = K E > 0 \frac{1}{2} m v^{2} + P E < 0 (- \frac{k Z e ^{2}}{r})$

WHAT IT LOOKS LIKE. Picture the well: the nucleus digs a funnel, and the electron sits partway down it.

Step 6 — Use the force balance to simplify (the virial shortcut)

WHAT. From equation (1), $m v^{2} = \frac{k Z e ^{2}}{r}$ , so the kinetic energy is $K E = \frac{1}{2} m v^{2} = \frac{k Z e ^{2}}{2 r}$ . Substitute:

$E = \underbrace{\frac{kZe^2}{2r}}_{KE} \;-\; \underbrace{\frac{kZe^2}{r}}_{|PE|} \;=\; -\frac{kZe^2}{2r}\tag{2}$

WHY. This is why Step 2 was worth keeping around: it lets us replace $m v^{2}$ and get $E$ in terms of $r$ alone. Notice the beautiful result $E = - K E = \frac{1}{2} P E$ — total energy is negative and exactly half the (negative) potential. That's the virial theorem signature of an inverse-square pull.

WHAT IT LOOKS LIKE. Stack the bars: $K E$ up, $P E$ twice as long down, net below zero.

Step 7 — Insert the quantised radius → the energy ladder

WHAT. Put $r_{n}$ from Step 4 into equation (2):

$E_{n} = - \frac{k Z e ^{2}}{2} \cdot \frac{4 π ^{2} mk Z e ^{2}}{n ^{2} h ^{2}} = - \frac{2 π ^{2} m k ^{2} Z ^{2} e ^{4}}{n ^{2} h ^{2}}$

WHY. Only allowed radii exist, so only allowed energies exist. Reading the answer: $Z^{2}$ on top → a stronger nucleus digs a deeper well; $n^{2}$ on the bottom → higher steps are shallower, crowding up toward zero.

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$

WHAT IT LOOKS LIKE. The energy ladder — rungs crammed near $0$ as $n$ climbs:

Step 8 — The limiting case $n \to \infty$ : the electron breaks free

WHAT. Let the step number grow without bound. Then $r_{n} = 0.529 \frac{n ^{2}}{Z} \to \infty$ (orbit huge) and $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} \to 0$ (well vanishes).

WHY. We must show every case, including the edge. At $n = \infty$ the electron is infinitely far and has zero energy — it is free, no longer bound. The energy you must supply to drag it from $n = 1$ out to here is the ionisation energy:

$E_{\infty} - E_{1} = 0 - (- 13.6 Z^{2}) = + 13.6 Z^{2} eV .$

For hydrogen that's exactly $13.6$ eV — the measured ionisation energy. A triumph.

WHAT IT LOOKS LIKE. The ladder converging onto the free line at $E = 0$ :

The one-picture summary

Everything above compressed: force balance → quantised ladder → radii $\propto n^{2} / Z$ → energies $\propto - Z^{2} / n^{2}$ , all landing on the ionisation edge.

Recall Feynman retelling of the whole walkthrough

We drew a positive nucleus and a tiny electron circling it, with a red arrow for the electric pull (Step 1). To keep circling, that pull must equal the "keep-me-turning" force — one equation linking speed and radius (Step 2). But that's one equation with two unknowns, so we can't solve it alone (Step 3). Bohr's magic rule saves us: the electron may only carry whole-number lumps $ℏ$ of circling-motion — a ladder, not a ramp. Plug that in and the radius pops out: allowed orbits sized $n^{2} / Z$ (Step 4). Then we asked for the orbit's total energy — motion energy plus the negative "trapped in a well" energy (Step 5). Using our force balance, the kinetic part is exactly half the well depth, so the total is negative: the electron has fallen partway into the funnel (Step 6). Feeding in the allowed radii turns that into an energy ladder, $- 13.6 Z^{2} / n^{2}$ , deep for strong nuclei and shallow for high steps (Step 7). Finally, climbing to the top of the ladder ( $n \to \infty$ ) frees the electron — the climb costs $13.6$ eV for hydrogen, exactly the measured ionisation energy (Step 8). Two boxes, one picture: radius up as $n^{2}$ , energy down as $- Z^{2} / n^{2}$ .

Recall Rebuild the logic yourself

Which equation links $v$ and $r$ ? ::: The force balance $k Z e^{2} / r = m v^{2}$ (Coulomb = centripetal). Why do we need a second equation? ::: Force balance has two unknowns ( $v$ and $r$ ); quantisation $m v r = n ℏ$ is the second. Where does the negative sign in $E_{n}$ come from? ::: From the potential energy $- k Z e^{2} / r$ (electron bound in an attractive well). Why is $E = - K E$ ? ::: Force balance makes $K E = k Z e^{2} /2 r$ while $E = - k Z e^{2} /2 r$ ; virial signature of an inverse-square force. What happens as $n \to \infty$ ? ::: $r \to \infty$ , $E \to 0$ ; the electron is free. The climb from $n = 1$ is the ionisation energy.

Connections

Parent topic — the formulas this page derives.
Rutherford model — the picture we started from in Step 1.
Planck's quantisation — the lump $ℏ$ behind Step 3.
de Broglie wavelength — a deeper reason the ladder in Step 3 exists.
Hydrogen spectrum & Rydberg formula — what the energy ladder predicts.
Ionisation energy — the $n \to \infty$ climb of Step 8.
Quantum mechanical model of atom — where these sharp orbits get replaced by clouds.