Derivation of Bohr's radii and energies from electrostatics + quantization: Visual walkthrough

We build a hydrogen-like atom: a tiny heavy positive lump in the middle (the nucleus, charge $+ Z e$ ) and one light electron (charge $- e$ ) circling it. Nothing else. Let us start.

Step 1 — Draw the players and name every symbol

WHAT. Put the nucleus at the centre. Draw the electron on a circle of radius $r$ around it. The electron moves along the circle with speed $v$ .

WHY. Before any equation we must fix what each letter means on the picture, or the algebra later is just noise. Every symbol below is a thing you can point at in the figure.

PICTURE.

Nothing is quantized yet. Right now the electron could sit on any circle. Two rules will fix that.

Step 2 — Rule 1: the inward pull must equal the turning force

WHAT. Two forces live on this picture. The nucleus pulls the electron inward (opposite charges attract) — this is the Coulomb force. To keep any object moving in a circle, something must constantly pull it toward the centre — the centripetal requirement of circular motion. Here the Coulomb pull is that centripetal supply.

WHY. If the inward pull were too weak the electron would fly outward; too strong and it spirals in. A steady circle means the two are exactly equal. That equality is our first equation.

PICTURE.

Term by term: $ε_{0}$ is a fixed constant of nature (how "stiff" empty space is electrically). $Z e^{2}$ is "charge of nucleus times charge of electron". The $r^{2}$ on the bottom says the pull weakens with the square of distance. On the right, $m v^{2} / r$ is the exact inward force needed to bend a mass $m$ moving at speed $v$ around radius $r$ .

Multiply both sides by $r$ to clear the picture up — call it equation (A): $\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r}=mv^2 \tag{A}$

Why bother? (A) links $r$ and $v$ . But it holds for every radius — one equation, two unknowns. We need one more rule.

Step 3 — Rule 2: only special "step-sized" orbits are allowed

WHAT. Bohr's new postulate: the electron's angular momentum $m v r$ may only be a whole-number multiple of a fixed chunk $ℏ = h /2 π$ . See Quantization of Angular Momentum and Bohr Model Postulates.

$angular momentum m v r = whole number n \cdot fixed chunk ℏ = \frac{nh}{2 π}, n = 1, 2, 3, \dots$

WHY. Angular momentum $m v r$ measures "how much circling motion" the electron carries. Classically it can be anything. Bohr forbids the in-between values — only rungs $n = 1, 2, 3, \dots$ survive. That is what turns a smooth ramp of orbits into a staircase.

PICTURE. The de Broglie way to see it: the electron is a wave, and only whole numbers of wavelengths fit around the circle without cancelling themselves.

In the figure $n = 3$ fits neatly (green, closed wave); the dashed red curve shows a non-integer that meets itself out of phase and destroys itself — so nature does not allow it.

Solve this rule for $v$ — call it (B): $v=\frac{nh}{2\pi m r}\tag{B}$

Why solve for $v$ ? Because (A) still has both $r$ and $v$ in it. If we replace $v$ using (B), only $r$ remains — and $r$ is exactly what we want.

Step 4 — Substitute and unmask the radius

WHAT. Put (B) into (A). The speed $v$ vanishes and a clean formula for $r$ falls out.

WHY. Two equations, two unknowns $(r, v)$ — eliminate one to solve the other. This is the pivot of the whole derivation.

$\frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r} = m (\frac{nh}{2 π m r})^{2} = \frac{n ^{2} h ^{2}}{4 π ^{2} m r ^{2}}$

The right side squared the whole of (B): $n^{2}$ , $h^{2}$ on top; $4 π^{2} m r^{2}$ on the bottom (one $m$ cancelled with the $m$ out front). Now both sides have $1/ r$ or $1/ r^{2}$ — cancel one power of $r$ and rearrange:

PICTURE — the staircase of orbits.

Read the boxed result off the picture: because $n$ appears as $n^{2}$ , the circles are spaced $1, 4, 9, 16, \dots$ — they spread apart fast. Because $Z$ sits on the bottom, a stronger nucleus (bigger $Z$ ) squeezes every orbit inward.

Step 5 — The energy: kinetic and potential on one axis

WHAT. Total energy $E = K E + P E$ . Kinetic $K E = \frac{1}{2} m v^{2}$ (always positive — motion). Potential $P E = - \frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r}$ (negative — attraction; energy is released as the electron falls in, so the bound state sits below zero).

WHY. The atom's energy is the sum of these. We need to see how they stack.

PICTURE.

The green bar ( $K E$ ) rises above zero; the red bar ( $P E$ ) plunges below; the yellow bar ( $E$ , their sum) ends up below zero — the electron is bound.

Step 6 — The virial shortcut: $K E = - \frac{1}{2} P E$

WHAT. Equation (A) said $m v^{2} = \frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r}$ . Halve it: $K E = \frac{1}{2} m v^{2} = \frac{1}{2} \frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r}$ . But that last expression is exactly $- \frac{1}{2} P E$ . So $K E = - \frac{1}{2} P E$ .

WHY. This is the Virial Theorem for a $1/ r$ force. It lets us collapse three quantities into one: $E = K E + P E = - \frac{1}{2} P E + P E = \frac{1}{2} P E = - K E$

PICTURE — the exact heights.

The figure fixes the proportions: $K E$ is one unit up, $P E$ is two units down, and their sum $E$ lands one unit below zero — always half the potential, always negative.

So: $E = \frac{1}{2} P E = - \frac{Z e ^{2}}{8 π ε _{0} r}$

Step 7 — Insert the quantized radius: the energy ladder

WHAT. Put $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m Z e ^{2}}$ into $E = - \frac{Z e ^{2}}{8 π ε _{0} r}$ .

WHY. The energy formula still holds for any $r$ ; feeding in the allowed radii $r_{n}$ discretises the energy into levels.

$E_{n} = - \frac{Z e ^{2}}{8 π ε _{0}} \cdot \frac{π m Z e ^{2}}{ε _{0} n ^{2} h ^{2}} = - \frac{m Z ^{2} e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}}$

Watch the $Z$ : one factor from the Coulomb term, another because $r_{n}$ itself carries $1/ Z$ — together $Z^{2}$ .

PICTURE — the level ladder.

The rungs crowd together toward $E = 0$ as $n$ grows — because $1/ n^{2}$ shrinks fast. Level $n = 1$ is deepest ( $- 13.6 eV$ for H); the gaps to jump between rungs are what make the sharp lines of the hydrogen spectrum.

Step 8 — The edge case: $n \to \infty$ and ionization

WHAT. Let $n$ grow without bound. Then $r_{n} = a_{0} n^{2} / Z \to \infty$ (the orbit becomes enormous) and $E_{n} = - 13.6 Z^{2} / n^{2} \to 0$ (the level rises up to the zero line).

WHY. This is not a formula that "breaks" at the edge — it is the physical free-electron limit. $E = 0$ means the electron is barely held; push a hair more and it escapes. The energy needed to lift the ground electron all the way to $E = 0$ is the Ionization Energy: $E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6 eV (hydrogen) .$

PICTURE.

The rungs pile up against the $E = 0$ ceiling; the yellow arrow from $n = 1$ to the ceiling is the ionization energy — the depth of the deepest well.

The one-picture summary

Two rules go in on the left (force balance + quantization); one substitution merges them; out come the two boxed results on the right — radii climbing as $n^{2} / Z$ , energies deepening as $- Z^{2} / n^{2}$ .

Recall Feynman retelling of the whole walkthrough

Picture a stone whirled on a string. The string's inward pull keeps it circling — for the atom the "string" is the electric attraction of the tiny positive centre on the electron (Step 2). That gives one equation linking size and speed. Nature then adds a bizarre rule: the electron may only whirl at certain step-sized orbits, the ones where its wave closes on itself (Step 3). Put those two facts together, cancel the speed, and the size of every allowed circle pops out — spaced $1, 4, 9, 16, \dots$ (Step 4). Add up the electron's motion-energy and its pulled-in energy and, for this kind of pull, the total is always exactly half the negative attraction energy — so it sits below zero, meaning "trapped" (Steps 5–6). Feed the special sizes back in and you get a ladder of energies, deepest at the bottom, crowding toward the "free" line at the top (Step 7). Climb all the way to that line and the electron escapes — that climb is the ionization energy (Step 8). Every number in the Bohr atom comes from just those two pictures.

Active-recall

Which two pictures generate the entire derivation?

Force balance (Coulomb = centripetal) and quantized angular momentum (closed de Broglie wave).

In (B) we solved the quantization rule for which variable, and why?

For

v

, so we could delete it from (A) and be left with

r

alone.

Why does energy carry

Z^{2}

while radius carries only

Z

One

Z

from the Coulomb energy term, a second because

r_{n}

itself scales as

1/ Z

What does

n \to \infty

correspond to physically?

The electron becomes free:

r_{n} \to \infty

E_{n} \to 0^{-}

— the ionization limit.

For a

1/ r

force, how do KE, PE, E relate?

K E = - \frac{1}{2} P E

, so

E = \frac{1}{2} P E = - K E

(negative, bound).

Connections

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