Bohr atomic model and electron shells: Visual walkthrough

We study the simplest atom: hydrogen — one heavy positive lump (a proton) with one tiny negative speck (an electron) circling it.

Step 0 — The cast of characters (so no symbol surprises you)

Before any equation, meet everyone in the play. Look at the figure: the proton sits at the centre, the electron rides a circle of radius $r$ at speed $v$ .

Figure — Bohr atomic model and electron shells

Nothing else enters without a picture. Let's go.

Step 1 — Why the electron doesn't fly away: force balance

WHAT. We write down that the inward pull equals the inward force a circle demands.

WHY. Anything moving in a circle is constantly being yanked toward the centre — otherwise it would coast off in a straight line (Newton's first law). That required inward yank is called the centripetal force, and its size is $\frac{m v ^{2}}{r}$ . In this atom the only thing pulling inward is the electrical attraction between $+ e$ and $- e$ . So they must be equal, or the orbit couldn't hold.

PICTURE. In the figure the amber arrow (Coulomb pull, inward) is exactly the arrow the circle needs. If the amber arrow were too weak, the electron spirals out; too strong, it spirals in. Balance = steady circle.

Why the Coulomb law and not gravity? Gravity between a proton and electron is about $1 0^{39}$ times weaker — utterly negligible. Electricity is the tool because electricity is what actually holds the atom.

Step 2 — Bohr's brand-new rule: only some circles are allowed

WHAT. We add the one non-classical law: the electron's angular momentum may only be whole-number multiples of $ℏ$ .

WHY. Step 1 alone allows every radius — pick any $r$ , find a matching $v$ , done. But real atoms emit only specific colours, so only specific orbits can exist. Bohr's fix is to forbid all but a discrete ladder of orbits. "Angular momentum" $L = m v r$ measures how much spin-around-the-centre the electron carries (big mass, big speed, or big circle → more of it). Bohr declares this can only come in packets of $ℏ$ .

PICTURE. The figure shows three allowed circles ( $n = 1, 2, 3$ ) as solid cyan rings and the forbidden in-between circles as dashed grey. The electron may sit on a rung, never between rungs.

Step 3 — Combine the two rules to pin down the radius

WHAT. We substitute the $v$ from Step 2 into the balance from Step 1, then solve for $r$ .

WHY. We have two facts: a force fact (Step 1) and a quantization fact (Step 2). One equation, one unknown is what we want — so we eliminate the speed $v$ , leaving an equation in $r$ alone. This is just algebra, but the result is physics: only these radii survive.

PICTURE. The figure is a "flow": two boxes (Step 1, Step 2) pour into one, $v$ crosses out, and out drops the radius formula.

Put $v = \frac{n ℏ}{m r}$ into $\frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} = \frac{m v ^{2}}{r}$ :

$\frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} = \frac{m}{r} (\frac{n ℏ}{m r})^{2} = \frac{n ^{2} ℏ ^{2}}{m r ^{3}}$

Cancel one power of $r$ from each side and rearrange:

Step 4 — Add up the energy: kinetic plus potential

WHAT. We compute the electron's total energy $E = K E + P E$ .

WHY. Radius alone doesn't tell us about the light an atom emits. Colour comes from energy changes. So we need the energy of each orbit. Energy has two parts: kinetic (energy of motion) and potential (energy stored in the electrical attraction).

PICTURE. The figure is an energy bar chart for one orbit: a positive kinetic bar, a twice-as-tall negative potential bar, and their sum — a net negative bar. Negative means "trapped."

Kinetic energy is $K E = \frac{1}{2} m v^{2}$ . From Step 1, $m v^{2} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r}$ , so

$K E = \frac{1}{8 π ε _{0}} \frac{e ^{2}}{r} .$

Potential energy of two opposite charges is $P E = - \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r}$ (negative because opposite charges want to be together — you'd have to do work to pull them apart).

$E = K E + P E = \frac{1}{8 π ε _{0}} \frac{e ^{2}}{r} - \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r} = - \frac{1}{8 π ε _{0}} \frac{e ^{2}}{r} .$

Step 5 — Insert the quantized radius → the energy ladder

WHAT. Plug $r_{n} = a_{0} n^{2}$ into the energy of Step 4.

WHY. Energy depended on $r$ ; but only special $r$ values are allowed (Step 3). So energy is also a discrete ladder — exactly the ladder that produces spectral lines.

PICTURE. The figure is the famous energy-level ladder: rungs at $- 13.6, - 3.4, - 1.51, \dots$ eV, crowding toward $0$ at the top.

$E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2}} \frac{1}{n ^{2}} = - \frac{13.6 eV}{n ^{2}}$

Step 6 — Where the light comes from (the payoff)

WHAT. When the electron drops from a high rung $n_{i}$ to a low rung $n_{f}$ , the released energy leaves as a single flash of light — a photon.

WHY. This is why the whole model was believed: it predicts the exact colours hydrogen emits. A photon of energy $E$ has frequency $f = E / h$ (bigger energy = bluer light).

PICTURE. The figure shows a downward jump $n = 3 \to n = 2$ with an amber photon squirting out, and the height of the jump labelled as the photon's energy.

Step 7 — The edge cases (never get surprised)

WHAT. We check the extreme and degenerate situations so no scenario is a mystery.

WHY. A derivation you trust must survive its limits.

PICTURE. The figure lines up the three edge cases side by side.

The one-picture summary

Everything above, on one blueprint: force balance + quantization → radius ladder → energy ladder → emitted photon.

Recall Feynman retelling — say it like you'd tell a friend

An electron is stuck circling a proton. For the circle to hold, the electrical pull inward has to exactly match the "hold-me-in" force any circle needs — that's rule one. Bohr adds rule two: the electron is only allowed on certain circles, spaced like rungs on a ladder, because its "circling amount" must come in whole packets. Mix those two rules and the allowed circle sizes pop out: rung $n$ is $n^{2}$ times as wide as the smallest. Add up the energy of each rung — motion energy plus the (negative) togetherness energy — and you get a ladder of energies, all negative because the electron is trapped, crowding up toward zero. Zero means free. When the electron falls from a high rung to a low one, it coughs out the exact energy difference as one flash of light of one exact colour. That is why hydrogen glows in a fixed set of colours and not a smeared rainbow — and it's the same shell-and-energy logic that later decides whether a chunk of silicon conducts.

Recall Quick self-test

Which two facts pin the radius? ::: Force balance (Step 1) and quantization $m v r = n ℏ$ (Step 2).
How does radius scale with $n$ ? ::: As $n^{2}$ — $r_{n} = a_{0} n^{2}$ .
Sign of the total energy, and why? ::: Negative, because the electron is bound; you must add energy to free it.
Energy needed to ionize hydrogen from the ground state? ::: $13.6$ eV.
What produces a spectral line? ::: A downward jump $n_{i} \to n_{f}$ emitting a photon of $13.6 (1/ n_{f}^{2} - 1/ n_{i}^{2})$ eV.

Connections

Back to the parent: Bohr atomic model and electron shells
Quantum mechanical model of the atom — the successor that replaces fixed orbits with clouds.
Energy bands in solids — these single-atom rungs broaden into bands in a crystal.
Semiconductors and the band gap — the ionization idea becomes the band gap.
Conductors insulators and doping — the Hardware payoff.
Valence electrons and bonding · Atomic structure and the periodic table